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class | __index_level_0__ int64 0 742k |
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$4.10 \cos ^{2}\left(45^{\circ}-\alpha\right)-\cos ^{2}\left(60^{\circ}+\alpha\right)-\cos 75^{\circ} \sin \left(75^{\circ}-2 \alpha\right)=\sin 2 \alpha$. | 4.10 Let's use formulas (4.17) and (4.27):
$$
\begin{aligned}
& \cos ^{2}\left(45^{\circ}-\alpha\right)-\cos ^{2}\left(60^{\circ}+\alpha\right)-\cos 75^{\circ} \sin \left(75^{\circ}-2 \alpha\right)= \\
& =\frac{1+\cos \left(90^{\circ}-2 \alpha\right)}{2}-\frac{1+\cos \left(120^{\circ}-2 \alpha\right)}{2}- \\
& -\frac{... | \sin2\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,608 |
$4.11 \frac{\operatorname{ctg}^{2} 2 \alpha-1}{2 \operatorname{ctg} 2 \alpha}-\cos 8 \alpha \operatorname{ctg} 4 \alpha=\sin 8 \alpha$.
$4.11 \frac{\cot^{2} 2 \alpha-1}{2 \cot 2 \alpha}-\cos 8 \alpha \cot 4 \alpha=\sin 8 \alpha$. | 4.11 Hint. Use the fact that
$$
\frac{\operatorname{ctg}^{2} 2 \alpha-1}{2 \operatorname{ctg} 2 \alpha}=\frac{1}{\operatorname{tg} 4 \alpha}=\operatorname{ctg} 4 \alpha
$$
and proceed with further transformations in the left-hand side of the identity. | Algebra | proof | Yes | Yes | olympiads | false | 49,609 | |
4.13 $\sin ^{6} \alpha+\cos ^{6} \alpha+3 \sin ^{2} \alpha \cos ^{2} \alpha=1$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
4.13 $\sin ^{6} \alpha+\cos ^{6} \alpha+3 \sin ^{2} \alpha \cos ^{2} \alpha=1$. | 4.13 Let's use the algebraic identity $(a+b)^{3}=$ $=a^{3}+3 a b(a+b)+b^{3}$ and the trigonometric identity $\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)^{3}=1$. Then we get
$\sin ^{6} \alpha+3 \sin ^{2} \alpha \cos ^{2} \alpha\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)+\cos ^{6} \alpha=1$,
from which
$\sin ^{6... | proof | Algebra | proof | Yes | Yes | olympiads | false | 49,611 |
$4.17 \frac{(1-\cos 2 \alpha) \cos \left(45^{\circ}+\alpha\right)}{2 \sin ^{2} 2 \alpha-\sin 4 \alpha}$. | 4.17 Using formulas (4.16) and (4.13), we have
$$
\begin{aligned}
& A=\frac{2 \sin ^{2} \alpha \cos \left(45^{\circ}+2 \alpha\right)}{2 \sin 2 \alpha(\sin 2 \alpha-\cos 2 \alpha)}= \\
& =\frac{\sin ^{2} \alpha \cos \left(45^{\circ}+2 \alpha\right)}{2 \sin \alpha \cos \alpha(\sin 2 \alpha-\cos 2 \alpha)}=\frac{\operato... | -\frac{\sqrt{2}}{4}\operatorname{tg}\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,614 |
$4.18 \cos \left(\frac{\pi}{6}-\frac{\alpha}{4}\right) \sin \left(\frac{\pi}{3}-\frac{\alpha}{4}\right) \sin \frac{\alpha}{4}$. | $4.18 \mathrm{~K}$ applying the formula (4.27) to the product of the first two factors, we get
$\cos \left(\frac{\pi}{6}-\frac{\alpha}{4}\right) \sin \left(\frac{\pi}{3}-\frac{\alpha}{4}\right)=\frac{1}{2}\left(\sin \left(\frac{\pi}{2}-\frac{\alpha}{2}\right)+\sin \frac{\pi}{6}\right)=\frac{1}{2}\left(\cos \frac{\alph... | \frac{1}{4}\sin\frac{3\alpha}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,615 |
$4.19 \frac{\sin ^{2}\left(\frac{\pi}{2}+\alpha\right)-\cos ^{2}\left(\alpha-\frac{\pi}{2}\right)}{\tan^{2}\left(\frac{\pi}{2}+\alpha\right)-\cot^{2}\left(\alpha-\frac{\pi}{2}\right)}$. | ### 4.19 We have
$$
\begin{aligned}
& \frac{\sin ^{2}\left(\frac{\pi}{2}+\alpha\right)-\cos ^{2}\left(\alpha-\frac{\pi}{2}\right)}{\tan ^{2}\left(\frac{\pi}{2}+\alpha\right)-\cot ^{2}\left(\alpha-\frac{\pi}{2}\right)}=\frac{\cos ^{2} \alpha-\sin ^{2} \alpha}{\cot ^{2} \alpha-\tan ^{2} \alpha}= \\
& =\frac{\cos 2 \alph... | \frac{1}{4}\sin^{2}2\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,616 |
4.22
$$
\frac{\cos ^{2}\left(\frac{5 \pi}{4}-2 \alpha\right)-\sin ^{2}\left(\frac{5 \pi}{4}-2 \alpha\right)}{\left(\cos \frac{\alpha}{2}+\sin \frac{\alpha}{2}\right)\left(\cos \left(2 \pi-\frac{\alpha}{2}\right)+\cos \left(\frac{\pi}{2}+\frac{\alpha}{2}\right)\right) \sin \alpha}
$$ | 4.22 Applying formula (4.14) to the numerator and then the reduction formula, we get $\cos \left(\frac{5 \pi}{2}-4 \alpha\right)=\sin 4 \alpha$. Transform the product of the first two factors in the denominator:
$$
\left(\cos \frac{\alpha}{2}+\sin \frac{\alpha}{2}\right)\left(\cos \frac{\alpha}{2}-\sin \frac{\alpha}{2... | 4\cos2\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,618 |
4.23
$$
\frac{4 \sin \left(\frac{5 \pi}{2}+\alpha\right)}{\operatorname{tg}^{2}\left(\frac{3 \pi}{2}-\alpha\right)-\operatorname{ctg}^{2}\left(\frac{3 \pi}{2}+\frac{\alpha}{2}\right)}
$$ | 4.23 We have
$$
\begin{aligned}
& \frac{4 \sin \left(\frac{5 \pi}{2}+\alpha\right)}{\operatorname{tg}^{2}\left(\frac{3 \pi}{2}-\alpha\right)-\operatorname{ctg}^{2}\left(\frac{3 \pi}{2}+\frac{\alpha}{2}\right)}=\frac{4 \cos \alpha}{\operatorname{ctg}^{2} \frac{\alpha}{2}-\operatorname{tg}^{2} \frac{\alpha}{2}}= \\
& =\... | \sin^2\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,619 |
$4.24 \frac{\cos \left(\frac{5 \pi}{2}-\alpha\right) \sin \left(\frac{\pi}{2}+\frac{\alpha}{2}\right)}{\left(2 \sin \frac{\pi-\alpha}{2}+\cos \left(\frac{3 \pi}{2}-\alpha\right)\right) \cos ^{2} \frac{\pi-\alpha}{4}}$.
Transform into a product (4.25-4.32): | 4.24 Find
$$
\frac{\cos \left(\frac{5 \pi}{2}-\alpha\right) \sin \left(\frac{\pi}{2}+\frac{\alpha}{2}\right)}{\left(2 \sin \frac{\pi-\alpha}{2}+\cos \left(\frac{3 \pi}{2}-\alpha\right)\right) \cos ^{2} \frac{\pi-\alpha}{4}}=
$$
$$
\begin{aligned}
& =\frac{\sin \alpha \cos \frac{\alpha}{2}}{\left(2 \cos \frac{\alpha}{... | 2\operatorname{tg}\frac{\alpha}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,620 |
$4.25 \sin 4 \alpha-2 \cos ^{2} 2 \alpha+1$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
$4.25 \sin 4 \alpha-2 \cos ^{2} 2 \alpha+1$. | 4.25 From formula (4.17), it follows that $2 \cos ^{2} 2 \alpha-1=\cos 4 \alpha$. Then the given expression will take the form $A=\sin 4 \alpha-\cos 4 \alpha$ and, therefore,
$A=\sin 4 \alpha-\sin \left(90^{\circ}-4 \alpha\right)=2 \cos 45^{\circ} \sin \left(4 \alpha-45^{\circ}\right)=$ $=\sqrt{2} \sin \left(4 \alpha-... | \sqrt{2}\sin(4\alpha-45) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,621 |
$4.26 \operatorname{tg} \frac{\alpha}{2}+\operatorname{ctg} \frac{\alpha}{2}+2$. | 4.26 Method I. We have
$$
\begin{aligned}
& \operatorname{tg} \frac{\alpha}{2}+\operatorname{ctg} \frac{\alpha}{2}+2=\frac{\sin ^{2} \frac{\alpha}{2}+\cos ^{2} \frac{\alpha}{2}}{\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}+2=\frac{2}{\sin \alpha}+2= \\
& =\frac{2(1+\sin \alpha)}{\sin \alpha}
\end{aligned}
$$
The sum ... | \frac{4\sin^{2}(\frac{\pi}{4}+\frac{\alpha}{2})}{\sin\alpha} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,622 |
$4.27 \frac{1}{\cos ^{4} \alpha}-\frac{1}{\sin ^{4} \alpha}$. | ### 4.27 We have
$$
\begin{aligned}
& \frac{1}{\cos ^{4} \alpha}-\frac{1}{\sin ^{4} \alpha}=\frac{\sin ^{4} \alpha-\cos ^{4} \alpha}{\sin ^{4} \alpha \cos ^{4} \alpha}= \\
& =\frac{\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)\left(\sin ^{2} \alpha-\cos ^{2} \alpha\right)}{\frac{1}{16} \sin ^{4} 2 \alpha}=-\frac{16 \... | -\frac{16\operatorname{ctg}2\alpha}{\sin^{3}2\alpha} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,623 |
$4.30 \sin 10 \alpha \sin 8 \alpha + \sin 8 \alpha \sin 6 \alpha - \sin 4 \alpha \sin 2 \alpha$. | ### 4.30 Method I. We have
$\sin 10 \alpha \sin 8 \alpha + \sin 8 \alpha \sin 6 \alpha - \sin 4 \alpha \sin 2 \alpha = \sin 8 \alpha \times$ $\times (\sin 10 \alpha + \sin 6 \alpha) - 2 \sin^2 2 \alpha \cos 2 \alpha = \sin 8 \alpha \cdot 2 \sin 8 \alpha \cdot$ $\cdot \cos 2 \alpha - 2 \sin^2 2 \alpha \cos 2 \alpha = 2... | 2\cos2\alpha\sin6\alpha\sin10\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,625 |
4.31 $\frac{\sin 13 \alpha + \sin 14 \alpha + \sin 15 \alpha + \sin 16 \alpha}{\cos 13 \alpha + \cos 14 \alpha + \cos 15 \alpha + \cos 16 \alpha}$.
4.32 $3 + 4 \cos 4 \alpha + \cos 8 \alpha$.
Prove the validity of the equalities (4.33-4.34):
4.33 $\left(\sin 160^{\circ} + \sin 40^{\circ}\right)\left(\sin 140^{\circ}... | 4.31 Given and e. Use formulas (4.19) and (4.21). Answer: $\operatorname{tg} \frac{29 \alpha}{2}$. | \operatorname{tg}\frac{29\alpha}{2} | Algebra | proof | Yes | Yes | olympiads | false | 49,626 |
$4.34 \frac{\cos 28^{\circ} \cos 56^{\circ}}{\sin 2^{\circ}}+\frac{\cos 2^{\circ} \cos 4^{\circ}}{\sin 28^{\circ}}=\frac{\sqrt{3} \sin 38^{\circ}}{4 \sin 2^{\circ} \sin 28^{\circ}}$
Calculate (4.35-4.38):
| 4.34 By combining the fractions on the left side of the equation, we get $\frac{\sin 28^{\circ} \cos 28^{\circ} \cos 56^{\circ}+\sin 2^{\circ} \cos 2^{\circ} \cos 4^{\circ}}{\sin 2^{\circ} \sin 28^{\circ}}=$ $=\frac{\sin 56^{\circ} \cos 56^{\circ}+\sin 4^{\circ} \cos 4^{\circ}}{2 \sin 2^{\circ} \sin 28^{\circ}}=\frac{\... | \frac{\sqrt{3}\sin38}{4\sin2\sin28} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,627 |
$4.35 \operatorname{tg} 435^{\circ}+\operatorname{tg} 375^{\circ}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
$4.35 \tan 435^{\circ}+\tan 375^{\circ}$. | ### 4.35 We have
$$
\begin{aligned}
& \tan 435^{\circ} + \tan 375^{\circ} = \tan 75^{\circ} + \tan 15^{\circ} = \frac{\sin 90^{\circ}}{\cos 75^{\circ} \cos 15^{\circ}} = \\
& = \frac{2}{\cos 90^{\circ} + \cos 60^{\circ}} = 4
\end{aligned}
$$
Here, reduction formulas, as well as (4.23) and (4.26), were used. | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,628 |
4.39 Given: $\operatorname{ctg} \alpha=\frac{3}{4}, \operatorname{ctg} \beta=\frac{1}{7}, 0<\alpha<\frac{\pi}{2}, 0<\beta<\frac{\pi}{2}$. Find $\alpha+\beta$. | 4.39 Note that $\operatorname{tg} \alpha=\frac{4}{3}, \operatorname{tg} \beta=7$. Then, by formula (4.11), we find
$$
\operatorname{tg}(\alpha+\beta)=\frac{\frac{4}{3}+7}{1-\frac{28}{3}}=-1
$$
Since by the condition $0<\alpha+\beta<\pi$, then $\alpha+\beta=\frac{3 \pi}{4}$. | \frac{3\pi}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,631 |
4.40 Find $\operatorname{ctg} 2 \alpha$, if it is known that $\sin \left(\alpha-90^{\circ}\right)=-\frac{2}{3}$ and $270^{\circ}<\alpha<360^{\circ}$. | 4.40 We have $\sin \left(\alpha-90^{\circ}\right)=-\sin \left(90^{\circ}-\alpha\right)=-\frac{2}{3}$, i.e., $\cos \alpha=\frac{2}{3}$.
Therefore, $\sin \alpha=-\sqrt{1-\frac{4}{9}}=-\frac{\sqrt{5}}{3}$ (since $\alpha$ is an angle in the IV quadrant). Next, we find
$$
\operatorname{ctg} 2 \alpha=\frac{\cos 2 \alpha}{\... | \frac{\sqrt{5}}{20} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,632 |
4.41 Prove that if $\alpha$ and $\beta$ satisfy the inequalities $0 \leqslant \alpha \leqslant \frac{\pi}{2}, 0 \leqslant \beta \leqslant \frac{\pi}{2}$ and $\operatorname{tg} \alpha=5, \operatorname{ctg} \beta=\frac{2}{3}$, then $\alpha+\beta=\frac{3 \pi}{4}$. | 4.41 We have $\operatorname{tg} \beta=\frac{3}{2}$. Since $\operatorname{tg}(\alpha+\beta)=\frac{\operatorname{tg} \alpha+\operatorname{tg} \beta}{1-\operatorname{tg} \alpha \operatorname{tg} \beta}$, then
$$
\operatorname{tg}(\alpha+\beta)=\frac{5+\frac{3}{2}}{1-\frac{15}{2}}=-1
$$
Considering that $0<\alpha+\beta<\... | \alpha+\beta=\frac{3\pi}{4} | Algebra | proof | Yes | Yes | olympiads | false | 49,633 |
4.42 The quantities $\alpha, \beta, \gamma$ form an arithmetic progression. Prove that
$$
\frac{\sin \alpha-\sin \gamma}{\cos \gamma-\cos \alpha}=\operatorname{ctg} \beta
$$ | 4.42 According to the property of an arithmetic progression, $\beta=\frac{\alpha+\gamma}{2}$. Using formulas (4.20) and (4.22), we transform the left side:
$$
\frac{\sin \alpha-\sin \gamma}{\cos \gamma-\cos \alpha}=\frac{2 \sin \frac{\alpha-\gamma}{2} \cos \frac{\alpha+\gamma}{2}}{2 \sin \frac{\alpha-\gamma}{2} \sin \... | proof | Algebra | proof | Yes | Yes | olympiads | false | 49,634 |
4.43 Given the fraction
$\frac{5}{1+\sqrt[3]{32 \cos ^{4} 15^{\circ}-10-8 \sqrt{3}}}$
Simplify the expression under the cube root, and then reduce the fraction.
## Group 6
Prove the identities (4.44-4.57):
| 4.43 Using formula (4.17), we find
$$
\begin{aligned}
& 32 \cos ^{4} 15^{\circ}-10-8 \sqrt{3}=32\left(\frac{1+\cos 30^{\circ}}{2}\right)^{2}-10-8 \sqrt{3}= \\
& =8\left(1+\frac{\sqrt{3}}{2}\right)^{2}-10-8 \sqrt{3}=4
\end{aligned}
$$
Next, we obtain
$$
\frac{5}{1+\sqrt[3]{4}}=\frac{1+4}{1+\sqrt[3]{4}}=\frac{(1+\sqrt... | 1-\sqrt[3]{4}+\sqrt[3]{16} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,635 |
$4.44 \operatorname{tg} 2 \alpha+\operatorname{ctg} 2 \alpha+\operatorname{tg} 6 \alpha+\operatorname{ctg} 6 \alpha=\frac{8 \cos ^{2} 4 \alpha}{\sin 12 \alpha}$.
$4.44 \tan 2 \alpha+\cot 2 \alpha+\tan 6 \alpha+\cot 6 \alpha=\frac{8 \cos ^{2} 4 \alpha}{\sin 12 \alpha}$. | 4.44 Applying formulas (4.2), (4.3), (4.1), and (4.13) sequentially to the left side of the equation, we find
$$
A=\operatorname{tg} 2 \alpha+\operatorname{ctg} 2 \alpha+\operatorname{tg} 6 \alpha+\operatorname{ctg} 6 \alpha=\frac{\sin 2 \alpha}{\cos 2 \alpha}+\frac{\cos 2 \alpha}{\sin 2 \alpha}+
$$
$$
\begin{aligned... | \frac{8\cos^{2}4\alpha}{\sin12\alpha} | Algebra | proof | Yes | Yes | olympiads | false | 49,636 |
$4.45 \operatorname{tg} \alpha+\frac{1}{\cos \alpha}-1=\frac{\sqrt{2} \sin \frac{\alpha}{2}}{\sin \left(\frac{\pi}{4}-\frac{\alpha}{2}\right)}$. | ### 4.45 We have
$$
\operatorname{tg} \alpha + \frac{1}{\cos \alpha} - 1 = \frac{\sin \alpha + 1 - \cos \alpha}{\cos \alpha} = \frac{2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} + 2 \sin ^{2} \frac{\alpha}{2}}{\cos \alpha} =
$$
$$
\begin{gathered}
= \frac{2 \sin \frac{\alpha}{2} \left( \cos \frac{\alpha}{2} + \sin \... | \frac{\sqrt{2}\sin\frac{\alpha}{2}}{\sin(\frac{\pi}{4}-\frac{\alpha}{2})} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,637 |
$4.46 \frac{\operatorname{ctg}^{2}(2 \alpha-\pi)}{1+\operatorname{tg}^{2}\left(\frac{3 \pi}{2}-2 \alpha\right)}-3 \cos ^{2}\left(\frac{5 \pi}{2}-2 \alpha\right)=$
$=4 \sin \left(\frac{\pi}{6}-2 \alpha\right) \sin \left(\frac{\pi}{6}+2 \alpha\right)$. | 4.46 After transformations, the left side will take the following form:
$$
\begin{aligned}
& A=\cos ^{2} 2 \alpha-3 \sin ^{2} 2 \alpha= \\
& =4\left(\frac{1}{2} \cos 2 \alpha-\frac{\sqrt{3}}{2} \sin 2 \alpha\right)\left(\frac{1}{2} \cos 2 \alpha+\frac{\sqrt{3}}{2} \sin 2 \alpha\right)
\end{aligned}
$$
## Further, we ... | 4\sin(\frac{\pi}{6}-2\alpha)\sin(\frac{\pi}{6}+2\alpha) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,638 |
$4.47 \frac{\sin ^{2}\left(\frac{3 \pi}{2}-\alpha\right)\left(\tan^{2} \alpha-1\right) \cot\left(\alpha-\frac{5 \pi}{4}\right)}{\sin ^{2}\left(\frac{5 \pi}{4}+\alpha\right)}=2$. | 4.47 Note that
$$
\begin{aligned}
& \sin ^{2}\left(\frac{3 \pi}{2}-\alpha\right)=\cos ^{2} \alpha \\
& \operatorname{tg}^{2} \alpha-1=\frac{\sin ^{2} \alpha-\cos ^{2} \alpha}{\cos ^{2} \alpha}=\frac{(\sin \alpha-\cos \alpha)(\sin \alpha+\cos \alpha)}{\cos ^{2} \alpha} \\
& \sin \alpha-\cos \alpha=\sqrt{2} \sin \left(\... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,639 |
$4.50 \frac{1}{\cos ^{6} \alpha}-\tan^{6} \alpha=\frac{3 \tan^{2} \alpha}{\cos ^{2} \alpha}+1$. | 4.50 We have
\[
\begin{aligned}
& \frac{1}{\cos ^{6} \alpha}-\frac{\sin ^{6} \alpha}{\cos ^{6} \alpha}=\frac{\left(1-\sin ^{2} \alpha\right)\left(1+\sin ^{2} \alpha+\sin ^{4} \alpha\right)}{\cos ^{4} \alpha \cos ^{2} \alpha}= \\
& =\frac{1+\sin ^{2} \alpha+1-2 \cos ^{2} \alpha+\cos ^{4} \alpha}{\cos ^{4} \alpha}=\frac... | \frac{3\tan^{2}\alpha}{\cos^{2}\alpha}+1 | Algebra | proof | Yes | Yes | olympiads | false | 49,641 |
$4.51 \operatorname{ctg}^{2} \alpha+\operatorname{ctg}^{2} \beta-\frac{2 \cos (\beta-\alpha)}{\sin \alpha \sin \beta}+2=\frac{\sin ^{2}(\alpha-\beta)}{\sin ^{2} \alpha \sin ^{2} \beta}$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result direct... | 4.51 Transforming the left side, we get
\[
\begin{aligned}
& \operatorname{ctg}^{2} \alpha+\operatorname{ctg}^{2} \beta-\frac{2 \cos (\beta-\alpha)}{\sin \alpha \sin \beta}+2= \\
& =\frac{\cos ^{2} \alpha}{\sin ^{2} \alpha}+\frac{\cos ^{2} \beta}{\sin ^{2} \beta}-\frac{2 \cos \alpha \cos \beta}{\sin \alpha \sin \beta}... | proof | Algebra | proof | Yes | Yes | olympiads | false | 49,642 |
$4.53 \frac{1+\cos (2 \alpha-2 \pi)+\cos (4 \alpha+2 \pi)-\cos (6 \alpha-\pi)}{\cos (2 \pi-2 \alpha)+2 \cos ^{2}(2 \alpha+\pi)-1}=2 \cos 2 \alpha$.
$$
\cos \left(4 \alpha-\frac{9 \pi}{2}\right)
$$ | 4.53 Using reduction formulas, as well as (4.17) and (4.21), we get
$$
\begin{aligned}
& \frac{1+\cos 2 \alpha+\cos 4 \alpha+\cos 6 \alpha}{\cos 2 \alpha+2 \cos ^{2} 2 \alpha-1}=\frac{2 \cos ^{2} \alpha+2 \cos 5 \alpha \cos \alpha}{\cos 2 \alpha+\cos 4 \alpha}= \\
& =\frac{\cos \alpha(\cos \alpha+\cos 5 \alpha)}{\cos ... | 2\cos2\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,644 |
4.54
$$
\operatorname{ctg}\left(\frac{5 \pi}{4}+2 \alpha\right)\left(1-\cos \left(\frac{5 \pi}{2}+4 \alpha\right)\right)=\operatorname{tg} 4 \alpha
$$ | 4.54 Transform the left side:
$\frac{\sin 4 \alpha}{\operatorname{ctg}\left(\frac{\pi}{4}+2 \alpha\right) \cdot 2 \sin ^{2}\left(\frac{5 \pi}{4}+2 \alpha\right)}=$
$=\frac{\sin 4 \alpha}{2 \operatorname{ctg}\left(\frac{\pi}{4}+2 \alpha\right) \cdot \sin ^{2}\left(\frac{\pi}{4}+2 \alpha\right)}=$
$=\frac{\sin 4 \alph... | \operatorname{tg}4\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,645 |
$4.55 \cos ^{6} \alpha+\sin ^{6} \alpha=\frac{5+3 \cos 4 \alpha}{8}$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
$4.55 \cos ^{6} \alpha+\sin ^{6} \alpha=\frac{5+3 \cos 4 \alpha}{8}$. | 4.55 Using the formula for the sum of cubes, we get $\cos ^{6} \alpha+\sin ^{6} \alpha=\left(\cos ^{2} \alpha+\sin ^{2} \alpha\right)\left(\cos ^{4} \alpha-\cos ^{2} \alpha \sin ^{2} \alpha+\sin ^{4} \alpha\right)=$
$$
\begin{aligned}
& =\left(\frac{1+\cos 2 \alpha}{2}\right)^{2}-\frac{1}{4} \sin ^{2} 2 \alpha+\left(\... | \frac{1}{8}(5+3\cos4\alpha) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,646 |
$4.56 \operatorname{ctg}\left(30^{\circ}-\alpha\right) \operatorname{ctg}\left(150^{\circ}-\alpha\right) \operatorname{ctg}\left(270^{\circ}+\alpha\right)=\operatorname{tg} 3 \alpha$. | 4.56 Transform the left side of the identity:
$$
\begin{aligned}
& \operatorname{ctg}\left(30^{\circ}-\alpha\right) \operatorname{ctg}\left(150^{\circ}-\alpha\right) \operatorname{ctg}\left(270^{\circ}+\alpha\right)= \\
& =-\operatorname{tg} \alpha \operatorname{tg}\left(60^{\circ}+\alpha\right) \operatorname{tg}\left... | \operatorname{tg}3\alpha | Algebra | proof | Yes | Yes | olympiads | false | 49,647 |
$4.58 \sqrt{\frac{\cos 2 \alpha}{\cot^{2} \alpha-\tan^{2} \alpha}} ; 90^{\circ}<\alpha<135^{\circ}$. | ### 4.58 We have
$\frac{\cos 2 \alpha}{\cot^{2} \alpha - \tan^{2} \alpha} = \frac{\cos 2 \alpha \sin^{2} \alpha \cos^{2} \alpha}{(\cos^{2} \alpha - \sin^{2} \alpha)(\cos^{2} \alpha + \sin^{2} \alpha)} = \frac{1}{4} \sin^{2} 2 \alpha$.
Since by the condition $90^{\circ} < \alpha < 135^{\circ}$, then $180^{\circ} < 2 \... | -\frac{1}{2}\sin2\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,648 |
$4.59 \frac{\cos 2 \alpha-\cos 6 \alpha+\cos 10 \alpha-\cos 14 \alpha}{\sin 2 \alpha+\sin 6 \alpha+\sin 10 \alpha+\sin 14 \alpha}$. | ### 4.59 Finding
\[
\begin{aligned}
& \frac{\cos 2 \alpha - \cos 6 \alpha + \cos 10 \alpha - \cos 14 \alpha}{\sin 2 \alpha + \sin 6 \alpha + \sin 10 \alpha + \sin 14 \alpha} = \\
& = \frac{2 \sin 8 \alpha \sin 6 \alpha - 2 \sin 8 \alpha \sin 2 \alpha}{2 \sin 8 \alpha \cos 6 \alpha + 2 \sin 8 \alpha \cos 2 \alpha} = \f... | \tan2\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,649 |
$4.60 \frac{4 \sin \left(4 \alpha-\frac{\pi}{2}\right)}{\operatorname{ctg}^{2}\left(2 \alpha-\frac{3 \pi}{2}\right)-\operatorname{tg}^{2}\left(2 \alpha+\frac{5 \pi}{2}\right)}-1$. | 4.60 We have
$$
\begin{aligned}
& \frac{4 \sin \left(4 \alpha-\frac{\pi}{2}\right)}{\operatorname{ctg}^{2}\left(2 \alpha-\frac{3 \pi}{2}\right)-\operatorname{tg}^{2}\left(2 \alpha+\frac{5 \pi}{2}\right)}-1=\frac{-4 \cos 4 \alpha}{\operatorname{tg}^{2} 2 \alpha-\operatorname{ctg}^{2} 2 \alpha}-1= \\
& =\frac{-4 \cos 4 ... | -\cos^{2}4\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,650 |
4.61 $\frac{\cos \left(4 \alpha-\frac{\pi}{2}\right) \sin \left(\frac{5 \pi}{2}+2 \alpha\right)}{(1+\cos 2 \alpha)(1+\cos 4 \alpha)}$. | 4.61 Using reduction formulas first, and then (4.18) and (4.17), we find
$\frac{\cos \left(4 \alpha-\frac{\pi}{2}\right) \sin \left(\frac{5 \pi}{2}+2 \alpha\right)}{(1+\cos 2 \alpha)(1+\cos 4 \alpha)}=\frac{\sin 4 \alpha \cos 2 \alpha}{(1+\cos 2 \alpha)(1+\cos 4 \alpha)}=$
$=\frac{\sin 2 \alpha}{1+\cos 2 \alpha} \cdo... | \tan\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,651 |
$4.62 \cos ^{4} 2 \alpha-6 \cos ^{2} 2 \alpha \sin ^{2} 2 \alpha+\sin ^{4} 2 \alpha$. | ### 4.62 Finding
\[
\begin{aligned}
& \cos ^{4} 2 \alpha-6 \cos ^{2} 2 \alpha \sin ^{2} 2 \alpha+\sin ^{4} 2 \alpha= \\
& =\left(\cos ^{2} 2 \alpha-\sin ^{2} 2 \alpha\right)^{2}-\sin ^{2} 4 \alpha=\cos ^{2} 4 \alpha-\sin ^{2} 4 \alpha=\cos 8 \alpha
\end{aligned}
\] | \cos8\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,652 |
$4.63 \frac{\tan 615^{\circ}-\tan 555^{\circ}}{\tan 795^{\circ}+\tan 735^{\circ}}$. | 4.63 Using reduction formulas, as well as (4.23) and (4.24), we get
$$
\frac{\operatorname{tg} 75^{\circ}-\operatorname{tg} 15^{\circ}}{\operatorname{tg} 75^{\circ}+\operatorname{tg} 15^{\circ}}=\frac{\sin 60^{\circ} \cos 75^{\circ} \cos 15^{\circ}}{\cos 75^{\circ} \cos 15^{\circ} \sin 90^{\circ}}=\frac{\sqrt{3}}{2}
$... | \frac{\sqrt{3}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,653 |
$4.64 \frac{3 \cos ^{2}\left(\alpha+270^{\circ}\right)-\sin ^{2}\left(\alpha-270^{\circ}\right)}{3 \sin ^{2}\left(\alpha-90^{\circ}\right)-\cos ^{2}\left(\alpha+90^{\circ}\right)}$. | 4.64 Let's use the reduction formulas and transform the numerator and the denominator separately:
$$
\begin{aligned}
& 3 \sin ^{2} \alpha-\cos ^{2} \alpha=\frac{3-3 \cos 2 \alpha-1-\cos 2 \alpha}{2}=1-2 \cos 2 \alpha= \\
& =2\left(\frac{1}{2}-\cos 2 \alpha\right)=2\left(\cos 60^{\circ}-\cos 2 \alpha\right)= \\
& =4 \s... | \operatorname{tg}(\alpha+30)\operatorname{tg}(\alpha-30) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,654 |
$4.65 \sin ^{2}\left(135^{\circ}-2 \alpha\right)-\sin ^{2}\left(210^{\circ}-2 \alpha\right)-\sin 195^{\circ} \cos \left(165^{\circ}-4 \alpha\right)$. | 4.65 To the first two terms, we apply formula (4.16), and to the third, $-(4.27):$
$\sin ^{2}\left(135^{\circ}-2 \alpha\right)-\sin ^{2}\left(210^{\circ}-2 \alpha\right)-\sin 195^{\circ} \cos \left(165^{\circ}-4 \alpha\right)=$
$=\frac{1}{2}\left(1-\cos \left(270^{\circ}-4 \alpha\right)-1+\cos \left(420^{\circ}-4 \al... | \sin4\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,655 |
$4.66 \sin \left(\frac{5 \pi}{2}+4 \alpha\right)-\sin ^{6}\left(\frac{5 \pi}{2}+2 \alpha\right)+\cos ^{6}\left(\frac{7 \pi}{2}-2 \alpha\right)$.
Transform into a product (4.67-4.73):
Translate the text above into English, please retain the line breaks and format of the source text, and output the translation result d... | 4.66 Instruction. Use reduction formulas, as well as the formula for the difference of cubes.
Answer: $\quad \frac{1}{8} \sin 4 \alpha \sin 8 \alpha$. | \frac{1}{8}\sin4\alpha\sin8\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,656 |
$4.67 \frac{1}{\sqrt{3}} \sin 4 \alpha+1-2 \cos ^{2} 2 \alpha$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
$4.67 \frac{1}{\sqrt{3}} \sin 4 \alpha+1-2 \cos ^{2} 2 \alpha$. | 4.67 Since $\cos 4 \alpha=2 \cos ^{2} 2 \alpha-1$, the given expression is transformed as follows:
$$
\begin{aligned}
& \frac{1}{\sqrt{3}} \sin 4 \alpha-\cos 4 \alpha=\frac{2}{\sqrt{3}}\left(\frac{1}{2} \sin 4 \alpha-\frac{\sqrt{3}}{2} \cos 4 \alpha\right)= \\
& =\frac{2}{\sqrt{3}} \sin \left(4 \alpha-60^{\circ}\right... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,657 | |
$4.68 \sqrt{1+\sin \frac{\alpha}{2}}-\sqrt{1-\sin \frac{\alpha}{2}}$, where $0<\alpha<180^{\circ}$. | 4.68 Preliminarily note that $0<\frac{\alpha}{4} \leqslant 45^{\circ},-45^{\circ} \leqslant-\frac{\alpha}{4}<0$, and $0 \leqslant 45^{\circ}-\frac{\alpha}{4}<45^{\circ}$. The expressions under the square roots are transformed as follows:
$$
1+\sin \frac{\alpha}{2}=1+\cos \left(90^{\circ}-\frac{\alpha}{2}\right)=2 \cos... | 2\sin\frac{\alpha}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,658 |
$4.69 \frac{\sin ^{2}(\alpha+\beta)-\sin ^{2} \alpha-\sin ^{2} \beta}{\sin ^{2}(\alpha+\beta)-\cos ^{2} \alpha-\cos ^{2} \beta}$. | 4.69 Instruction. Use the formulas for reducing the degree.
Answer: $-\operatorname{tg} \alpha \operatorname{tg} \beta$. | -\operatorname{tg}\alpha\operatorname{tg}\beta | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,659 |
4.70 $1-\sin ^{2} \alpha-\sin ^{2} \beta+2 \sin \alpha \sin \beta \cos (\alpha-\beta)$. | 4.70 We have
\[
\begin{aligned}
& \sin ^{2} \alpha+\sin ^{2} \beta-(\cos (\alpha-\beta)-\cos (\alpha+\beta)) \cos (\alpha-\beta)= \\
& =\frac{1}{2}-\frac{1}{2} \cos 2 \alpha+\frac{1}{2}-\frac{1}{2} \cos 2 \beta-\cos ^{2}(\alpha-\beta)+\frac{1}{2} \cos 2 \alpha+\frac{1}{2} \cos 2 \beta= \\
& =1-\cos ^{2}(\alpha-\beta)=... | \cos^2(\alpha-\beta) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,660 |
$4.73 \frac{1-2 \sin ^{2} \alpha}{2 \operatorname{tg}\left(\frac{5 \pi}{4}+\alpha\right) \cos ^{2}\left(\frac{\pi}{4}+\alpha\right)}-\operatorname{tg} \alpha+\sin \left(\frac{\pi}{2}+\alpha\right)-\cos \left(\alpha-\frac{\pi}{2}\right)$.
Prove the validity of the equalities (4.74-4.78): | ### 4.73 We have
$\frac{\cos 2 \alpha}{2 \sin \left(\frac{\pi}{4}+\alpha\right) \cos \left(\frac{\pi}{4}+\alpha\right)}-\operatorname{tg} \alpha+\cos \alpha-\sin \alpha=$ $\frac{\cos 2 \alpha}{\cos 2 \alpha}-\operatorname{tg} \alpha+\cos \alpha-\sin \alpha=$ $=\frac{\cos \alpha-\sin \alpha+\cos \alpha(\cos \alpha-\sin... | \frac{2\sqrt{2}\cos(\frac{\pi}{4}+\alpha)\cos^{2}\frac{} | Algebra | proof | Yes | Yes | olympiads | false | 49,661 |
$4.74 \frac{\cos 64^{\circ} \cos 4^{\circ}-\cos 86^{\circ} \cos 26^{\circ}}{\cos 71^{\circ} \cos 41^{\circ}-\cos 49^{\circ} \cos 19^{\circ}}=-1$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
$4.74 \frac{\cos 64^{\circ} \co... | 4.74 Transform the numerator and denominator separately:
$$
\frac{\cos 60^{\circ}+\cos 68^{\circ}-\cos 60^{\circ}-\cos 112^{\circ}}{2}=\cos 68^{\circ}
$$
$$
\frac{\cos 30^{\circ}+\cos 112^{\circ}-\cos 30^{\circ}-\cos 68^{\circ}}{2}=-\cos 68^{\circ}
$$.
Thus, the quotient is -1. | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,662 | |
$4.75 \operatorname{ctg} 10^{\circ} \operatorname{ctg} 50^{\circ} \operatorname{ctg} 70^{\circ}=\operatorname{ctg} 30^{\circ}$. | 4.75 In the left part of the supposed equality, we will proceed to the calculation of the numerator and the denominator. We have $\cos 10^{\circ} \cos 50^{\circ} \cos 70^{\circ}=\frac{1}{2}\left(\cos 60^{\circ}+\cos 40^{\circ}\right) \cos 70^{\circ}=$
$$
=\frac{1}{2}\left(\frac{1}{2} \cos 70^{\circ}+\frac{1}{2} \cos 1... | \cot30 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,663 |
$4.76 \sin 20^{\circ} \sin 40^{\circ} \sin 60^{\circ} \sin 80^{\circ}=\frac{3}{16}$. | 4.76 First, we use formula (4.25), and then (4.27):
$\sin 20^{\circ} \sin 40^{\circ} \sin 60^{\circ} \sin 80^{\circ}=\frac{\sqrt{3}}{2} \sin 20^{\circ} \sin 40^{\circ} \sin 80^{\circ}=$
$=\frac{\sqrt{3}}{4} \sin 80^{\circ}\left(\cos 20^{\circ}-\cos 60^{\circ}\right)=$
$=\frac{\sqrt{3}}{4}\left(\sin 80^{\circ} \cos 2... | \frac{3}{16} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,664 |
$4.77 \operatorname{tg} 9^{\circ}+\operatorname{tg} 15^{\circ}-\operatorname{tg} 27^{\circ}-\operatorname{ctg} 27^{\circ}+\operatorname{ctg} 9^{\circ}+\operatorname{ctg} 15^{\circ}=8$. | ### 4.77 We have
$$
\begin{aligned}
& \tan 9^{\circ} + \tan 81^{\circ} - (\tan 27^{\circ} + \tan 63^{\circ}) + \tan 15^{\circ} + \tan 75^{\circ} = \\
& = \frac{2}{\cos 90^{\circ} + \cos 72^{\circ}} - \frac{2}{\cos 90^{\circ} + \cos 36^{\circ}} + \frac{2}{\cos 90^{\circ} + \cos 60^{\circ}} = \\
& = \frac{2}{\cos 72^{\c... | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,665 |
$4.79 \cos \frac{3 \pi}{5} \cos \frac{6 \pi}{5}$. | 4.79 Method I. Let $A=\cos \frac{3 \pi}{5} \cos \frac{6 \pi}{5}$. Multiplying both sides of this equation by $2 \sin \frac{3 \pi}{5}$, we get
$2 A \sin \frac{3 \pi}{5}=2 \sin \frac{3 \pi}{5} \cos \frac{3 \pi}{5} \cos \frac{6 \pi}{5} ; 2 A \sin \frac{3 \pi}{5}=\sin \frac{6 \pi}{5} \cos \frac{6 \pi}{5} ;$
$2 A \sin \fr... | \frac{1}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,667 |
$4.80 \operatorname{tg}\left(\frac{5 \pi}{4}+x\right)+\operatorname{tg}\left(\frac{5 \pi}{4}-x\right)$, if $\operatorname{tg}\left(\frac{3 \pi}{2}+x\right)=\frac{3}{4}$. | 4.80 We find
\[
\begin{aligned}
& \operatorname{tg}\left(\frac{5 \pi}{4}+x\right)+\operatorname{tg}\left(\frac{5 \pi}{4}-x\right)=\frac{\sin \frac{5 \pi}{2}}{\cos \left(\frac{5 \pi}{4}+x\right) \cos \left(\frac{5 \pi}{4}-x\right)}= \\
& =\frac{2}{\cos \frac{5 \pi}{2}+\cos 2 x}=\frac{2}{\cos 2 x}
\end{aligned}
\]
Sinc... | -\frac{50}{7} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,668 |
$4.81 \sin \frac{\alpha+\beta}{2}$ and $\cos \frac{\alpha+\beta}{2}$, if $\sin \alpha+\sin \beta=-\frac{21}{65}, \cos \alpha+$
$$
+\cos \beta=-\frac{27}{65} ; \frac{5 \pi}{2}<\alpha<3 \pi \text { and }-\frac{\pi}{2}<\beta<0
$$ | ### 4.81 From the condition, it follows that
$$
2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}=-\frac{21}{65}, 2 \cos \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}=-\frac{27}{65}
$$
from which $\operatorname{tg} \frac{\alpha+\beta}{2}=\frac{7}{9}$.
Since $2 \pi<\alpha+\beta<3 \pi$, then $\pi<\frac{\a... | \sin\frac{\alpha+\beta}{2}=-\frac{7}{\sqrt{130}},\cos\frac{\alpha+\beta}{2}=-\frac{9}{\sqrt{130}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,669 |
4.82 Let $A, B, C$ be the interior angles of a triangle. Prove that
$$
\operatorname{tg} \frac{A}{2} \operatorname{tg} \frac{B}{2}+\operatorname{tg} \frac{B}{2} \operatorname{tg} \frac{C}{2}+\operatorname{tg} \frac{C}{2} \operatorname{tg} \frac{A}{2}=1
$$ | 4.82 Since \(A + B + C = \pi\), then \(\frac{A}{2} + \frac{B}{2} + \frac{C}{2} = \frac{\pi}{2}\).
## Next, we find
\[
\begin{aligned}
& \operatorname{tg} \frac{A}{2}\left(\operatorname{tg} \frac{B}{2} + \operatorname{tg} \frac{C}{2}\right) + \operatorname{tg} \frac{B}{2} \operatorname{tg} \frac{C}{2} = \operatorname{... | proof | Algebra | proof | Yes | Yes | olympiads | false | 49,670 |
4.83 It is known that $\frac{\sin (\alpha+\beta)}{\sin (\alpha-\beta)}=\frac{p}{q}$. Find $\operatorname{ctg} \beta$. | ### 4.83 We have
$\frac{\sin \alpha \cos \beta + \cos \alpha \sin \beta}{\sin \alpha \cos \beta - \cos \alpha \sin \beta} = \frac{p}{q}$, or $\frac{\operatorname{ctg} \beta + \operatorname{ctg} \alpha}{\operatorname{ctg} \beta - \operatorname{ctg} \alpha} = \frac{p}{q}$.
Therefore, $\operatorname{ctg} \beta = \frac{p... | \operatorname{ctg}\beta=\frac{p+q}{p-q}\operatorname{ctg}\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,671 |
4.84 Check that $\operatorname{tg} 20^{\circ}+4 \sin 20^{\circ}=\sqrt{3}$. | 4.84 Applying formulas (4.2), (4.13), and (4.19), we get
$$
\begin{aligned}
& A=\tan 20^{\circ}+4 \sin 20^{\circ}=\frac{\sin 20^{\circ}+4 \sin 20^{\circ} \cos 20^{\circ}}{\cos 20^{\circ}}= \\
& =\frac{\sin 20^{\circ}+2 \sin 40^{\circ}}{\cos 20^{\circ}}=\frac{(\sin 20^{\circ}+\sin 40^{\circ})+\sin 40^{\circ}}{\cos 20^{... | \sqrt{3} | Algebra | proof | Yes | Yes | olympiads | false | 49,672 |
4.85 Find the value of $\operatorname{tg} \frac{x}{2}$, if $\sin x-\cos x=1.4$. | 4.85 It is convenient to use formulas (4.28) and (4.29), considering that they are valid only for $x \neq \pi(2 n+1), n \in Z$. However, in this case, $x$ cannot take these values. Indeed, if $x=\pi(2 n+1)$, then
$$
\sin (\pi(2 n+1))-\cos (\pi(2 n+1))=0-(-1) \neq 1.4
$$
Expressing $\sin x$ and $\cos x$ in terms of $\... | \operatorname{tg}\frac{x}{2}=2\operatorname{tg}\frac{x}{2}=3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,673 |
4.86 It is known that $\sin \alpha - \cos \alpha = n$. Find $\sin ^{3} \alpha - \cos ^{3} \alpha$. | 4.86 Instruction. Sequentially square and cube both sides of the equality $\sin \alpha - \cos \alpha = n$.
Answer: $\frac{3 n - n^{3}}{2}$. | \frac{3n-n^{3}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,674 |
4.87 Find $\cos 2 \alpha$, if it is known that $2 \operatorname{ctg}^{2} \alpha+7 \operatorname{ctg} \alpha+3=0$ and the number $\alpha$ satisfies the inequalities: a) $\frac{3 \pi}{2}<\alpha<\frac{7 \pi}{4}$; b) $\frac{7 \pi}{4}<\alpha<2 \pi$ | 4.87 Solving the quadratic equation with respect to $\operatorname{ctg} \alpha$, we find $(\operatorname{ctg} \alpha)_{1}=-\frac{1}{2}\left((\operatorname{tg} \alpha)_{1}=-2\right)$ and $(\operatorname{ctg} \alpha)_{2}=-3\left((\operatorname{tg} \alpha)_{2}=-\frac{1}{3}\right)$.
Now, using the condition of the problem... | (\cos2\alpha)_{1}=-\frac{3}{5};(\cos2\alpha)_{2}=\frac{4}{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,675 |
4.88 Prove that the expression
$$
\frac{1-2 \sin ^{2}\left(\alpha-\frac{3 \pi}{2}\right)+\sqrt{3} \cos \left(2 \alpha+\frac{3 \pi}{2}\right)}{\sin \left(\frac{\pi}{6}-2 \alpha\right)}
$$
does not depend on $\alpha$, where $\alpha \neq \frac{\pi n}{2}+\frac{\pi}{12}$. | 4.88 After applying the reduction formulas, we get
\[
\begin{aligned}
& \frac{1-2 \cos ^{2} \alpha+\sqrt{3} \sin 2 \alpha}{\sin \left(\frac{\pi}{6}-2 \alpha\right)}=\frac{-2 \cos 2 \alpha+\sqrt{3} \sin 2 \alpha}{\sin \left(\frac{\pi}{6}-2 \alpha\right)}= \\
& =\frac{2\left(\frac{\sqrt{3}}{2} \sin 2 \alpha-\frac{1}{2} ... | -2 | Algebra | proof | Yes | Yes | olympiads | false | 49,676 |
4.89 It is known that $\sin \alpha + \sin \beta = 2 \sin (\alpha + \beta), \alpha + \beta \neq 2 \pi n (n \in \mathbb{Z})$. Find $\operatorname{tg} \frac{\alpha}{2} \operatorname{tg} \frac{\beta}{2}$. | 4.89 We have
$$
2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}=4 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha+\beta}{2}
$$
from which
$\frac{\cos \frac{\alpha-\beta}{2}}{\cos \frac{\alpha+\beta}{2}}=2$.
Therefore,
$$
\operatorname{tg} \frac{\alpha}{2} \operatorname{tg} \frac{\beta}{2}=\frac{2 \sin \frac... | \frac{1}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,677 |
4.90 Show that if $p$ is constant, then the function
$$
f(\alpha)=\frac{p \cos ^{3} \alpha-\cos 3 \alpha}{\cos \alpha}+\frac{p \sin ^{3} \alpha+\sin 3 \alpha}{\sin \alpha}
$$ | 4.90 By bringing the terms to a common denominator, we get
$\frac{p\left(\cos ^{3} \alpha \sin \alpha+\sin ^{3} \alpha \cos \alpha\right)+(\sin 3 \alpha \cos \alpha-\cos 3 \alpha \sin \alpha)}{\sin \alpha \cos \alpha}=$
$=\frac{p \sin \alpha \cos \alpha+\sin 2 \alpha}{\sin \alpha \cos \alpha}=p+2$. | p+2 | Algebra | proof | Yes | Yes | olympiads | false | 49,678 |
5.1 $\cos \left(\frac{\pi}{3}-3 x\right)=-\frac{1}{2}$.
5.2 $(\cos x-1)\left(3-\operatorname{ctg}^{2} \frac{x}{2}\right)=0$. | 5.1 Writing the equation in the form $\cos \left(3 x-\frac{\pi}{3}\right)=-\frac{1}{2}$ and using formula (5.2), we have
$$
3 x-\frac{\pi}{3}= \pm \arccos \left(-\frac{1}{2}\right)+2 \pi n
$$
Since $\arccos \left(-\frac{1}{2}\right)=\frac{2 \pi}{3}$, we obtain two series of roots:
1) $3 x-\frac{\pi}{3}=\frac{2 \pi}{... | x_{1}=\frac{\pi}{3}+\frac{2\pin}{3};x_{2}=-\frac{\pi}{9}+\frac{2\pin}{3},n\in\boldsymbol{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,679 |
$5.3 \cos 3 x-\sin x=\sqrt{3}(\cos x-\sin 3 x)$. | 5.3 Dividing both sides of the equation by 2 and considering that $\frac{1}{2}=\cos \frac{\pi}{3}=\sin \frac{\pi}{6}, \mathrm{a} \frac{\sqrt{3}}{2}=\sin \frac{\pi}{3}=\cos \frac{\pi}{6}$, we have
$\frac{1}{2} \cos 3 x+\frac{\sqrt{3}}{2} \sin 3 x=\frac{\sqrt{3}}{2} \cos x+\frac{1}{2} \sin x$,
$\cos \frac{\pi}{3} \cos ... | x_{1}=\frac{\pi}{8}(4k+1),x_{2}=\frac{\pi}{12}+\pik,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,680 |
5.4 $2(\cos 4 x-\sin x \cos 3 x)=\sin 4 x+\sin 2 x$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
5.4 $2(\cos 4 x-\sin x \cos 3 x)=\sin 4 x+\sin 2 x$. | 5.4 Let's rewrite the equation as
$\cos 4 x=\sin x \cos 3 x+0.5(\sin 4 x+\sin 2 x)$.
Transform the right side of the equation:
$\sin x \cos 3 x+0.5 \cdot 2 \sin 3 x \cos x=\sin 4 x$
Thus, $\cos 4 x=\sin 4 x$, from which $\operatorname{tg} 4 x=1 ; 4 x=\frac{\pi}{4}+\pi k, x=$ $=\frac{\pi}{16}(4 k+1)$ (division by $\... | \frac{\pi}{16}(4k+1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,681 |
$5.5 \sin 3 x+\sin 5 x=\sin 4 x$. | 5.5 Applying the formula for the sum of sines (4.19), after transformations we get
$2 \sin 4 x \cos x=\sin 4 x ; \sin 4 x(2 \cos x-1)=0$.
Thus:
1) $\sin 4 x=0 ; 4 x=\pi k ; x_{1}=\frac{\pi k}{4}$;
2) $\cos x=\frac{1}{2} ; x_{2}= \pm \frac{\pi}{3}+2 \pi k=\frac{\pi}{3}(6 k \pm 1)$.
Answer: $\quad x_{1}=\frac{\pi k}{4... | x_{1}=\frac{\pik}{4},x_{2}=\frac{\pi}{3}(6k\1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,682 |
5.6. $\sin \left(15^{\circ}+x\right)+\sin \left(45^{\circ}-x\right)=1$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
5.6. $\sin \left(15^{\circ}+x\right)+\sin \left(45^{\circ}-x\right)=1$. | 5.6 Using formula (4.19), we get $2 \sin 30^{\circ} \times$ $x \cos \left(15^{\circ}-x\right)=1 ;$ since $\sin 30^{\circ}=0.5$, then $\cos \left(15^{\circ}-x\right)=1$, or $\cos \left(x-15^{\circ}\right)=1$, from which by formula (5.9) we find $x-15^{\circ}=360^{\circ} k$
Answer: $x=15^{\circ}+360^{\circ} k, k \in Z$. | 15+360k,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,683 |
$5.7 \cos t \sin \left(\frac{\pi}{2}+6 t\right)+\cos \left(\frac{\pi}{2}-t\right) \sin 6 t=\cos 6 t+\cos 4 t$. | 5.7 Using the reduction formulas (4.21) and (4.10), we have $\cos t \cos 6 t+\sin t \sin 6 t=2 \cos 5 t \cos t$ $\cos 5 t=2 \cos 5 t \cos t ; \cos 5 t(1-2 \cos t)=0$.
From this, we obtain:
1) $\cos 5 t=0 ; 5 t=\frac{\pi}{2}+\pi k, t_{1}=\frac{\pi}{10}(2 k+1)$;
2) $\cos t=\frac{1}{2} ; t_{2}= \pm \frac{\pi}{3}+2 \pi k... | t_{1}=\frac{\pi}{10}(2k+1),t_{2}=\\frac{\pi}{3}+2\pik,k\inZ | Algebra | proof | Yes | Yes | olympiads | false | 49,684 |
$5.8 \sin 9 x=2 \sin 3 x$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
$5.8 \sin 9 x=2 \sin 3 x$. | 5.8 Representing $2 \sin 3 x$ as $\sin 3 x + \sin 3 x$, we have $\sin 9 x - \sin 3 x = \sin 3 x, 2 \cos 6 x \sin 3 x = \sin 3 x$
From this, we obtain:
1) $\sin 3 x = 0, 3 x = \pi k ; x_{1} = \frac{\pi k}{3}$;
$$
\text { 2) } \cos 6 x = \frac{1}{2}, 6 x = \pm \frac{\pi}{3} + 2 \pi k ; x_{2} = \frac{\pi}{18}(6 k \pm 1... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,685 | |
5.9 $\sin z+\sin 2 z+\sin 3 z=\cos z+\cos 2 z+\cos 3 z$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
5.9 $\sin z+\sin 2 z+\sin 3 z=\cos z+\cos 2 z+\cos 3 z$. | 5.9 Noting that $\sin z + \sin 3z = 2 \sin 2z \cos z$, and $\cos z + \cos 3z = 2 \cos 2z \cos z$, we have
$2 \sin 2z \cos z + \sin 2z = 2 \cos 2z \cos z + \cos 2z$
$\sin 2z (2 \cos z + 1) = \cos 2z (2 \cos z + 1)$
$(2 \cos z + 1)(\sin 2z - \cos 2z) = 0$.
Thus:
1) $2 \cos z + 1 = 0, \cos z = -\frac{1}{2}; z_1 = \fr... | z_1=\frac{2\pi}{3}(3k\1),z_2=\frac{\pi}{4}(4k+1),k\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,686 |
5.10 $1-\cos 6 x=\tan 3 x$.
5.11 $1+\sin 2 x=(\cos 3 x+\sin 3 x)^{2}$. | 5.10 We have
$2 \sin ^{2} 3 x=\operatorname{tg} 3 x, \cos 3 x \neq 0 ; 2 \sin ^{2} 3 x=\frac{\sin 3 x}{\cos 3 x} ；$
$2 \sin ^{2} 3 x \cos 3 x=\sin 3 x$
1) $\sin 3 x=0,3 x=\pi k ; x_{1}=\frac{\pi k}{3}$;
2) $2 \sin 5 x \cos 3 x=1 ; \sin 6 x=1,6 x=\frac{\pi}{2}+2 \pi k ; x_{2}=\frac{\pi}{12}(4 k+1)$.
Solution: $\quad... | x_{1}=\frac{\pik}{3},x_{2}=\frac{\pi}{12}(4k+1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,687 |
$5.12 \cos ^{2} 3 x+\cos ^{2} 4 x+\cos ^{2} 5 x=1.5$ | 5.12 Let's use the formula for reducing the power (4.17):
$2\left(\cos ^{2} 3 x+\cos ^{2} 4 x+\cos ^{2} 5 x\right)=3$
$1+\cos 6 x+1+\cos 8 x+1+\cos 10 x=3$
$\cos 6 x+\cos 8 x+\cos 10 x=0 ; \cos 8 x+2 \cos 8 x \cos 2 x=0 ;$
1) $\cos 8 x=0 ; 8 x=\frac{\pi}{2}+\pi k ; x_{1}=\frac{\pi}{16}(2 k+1)$;
2) $\cos 2 x=-\frac{... | x_{1}=\frac{\pi}{16}(2k+1),x_{2}=\frac{\pi}{3}(3k\1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,688 |
$5.14 \sin 3 x+\sin 5 x=2\left(\cos ^{2} 2 x-\sin ^{2} 3 x\right)$. | 5.14 To the left side of the equation, we apply formula (4.19), and to the right side, we apply formulas (4.17) and (4.16):
$2 \sin 4 x \cos x=2\left(\frac{1+\cos 4 x}{2}-\frac{1-\cos 6 x}{2}\right)$.
$2 \sin 4 x \cos x=\cos 4 x+\cos 6 x ; \sin 4 x \cos x=\cos 5 x \cos x$
1) $\cos x=0 ; x_{1}=\frac{\pi}{2}+\pi k$
2)... | x_{1}=\frac{\pi}{2}(2k+1),x_{2}=\frac{\pi}{18}(4k+1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,689 |
$5.15 \operatorname{ctg} t-\sin t=2 \sin ^{2} \frac{t}{2}$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
$5.15 \operatorname{ctg} t-\sin t=2 \sin ^{2} \frac{t}{2}$. | 5.15 Replacing $2 \sin ^{2} \frac{t}{2}$ with $1-\cos t$, we have
$\operatorname{ctg} t-\sin t=1-\cos t ; \operatorname{ctg} t-1=\sin t-\cos t, \sin t \neq 0$;
$\frac{\cos t-\sin t}{\sin t}=\sin t-\cos t,(\sin t-\cos t)(\sin t+1)=0$.
From this, we obtain:
1) $\sin t-\cos t=0, \operatorname{tg} t=1 ; t_{1}=\frac{\pi... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,690 | |
$5.16 \cos ^{3} x+\cos ^{2} x-4 \cos ^{2} \frac{x}{2}=0$. | 5.16 By replacing $2 \cos ^{2} \frac{x}{2}$ with $1+\cos x$ and factoring the left side of the equation, we get
$$
\begin{aligned}
& \cos ^{3} x+\cos ^{2} x-2(1+\cos x)=0 \\
& \cos ^{2} x(\cos x+1)-2(1+\cos x)=0 ;(1+\cos x)\left(\cos ^{2} x-2\right)=0 \\
& \cos ^{2} x-2 \neq 0, \text { since }|\cos x| \leqslant 1 ; 1+... | \pi(2k+1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,691 |
$5.18 \sin ^{4} x+\cos ^{4} x=\cos ^{2} 2 x+0.25$. | 5.18 Let's use the identity $1=\left(\sin ^{2} x+\cos ^{2} x\right)^{2}$, from which $\sin ^{4} x+\cos ^{4} x=1-2 \sin ^{2} x \cos ^{2} x=1-\frac{1}{2} \sin ^{2} 2 x$.
## Then we get
$1-\frac{1}{2} \sin ^{2} 2 x=1-\sin ^{2} 2 x+\frac{1}{4} ; 1-2 \sin ^{2} 2 x=0 ;$
$\cos 4 x=0, 4 x=\frac{\pi}{2}+\pi k ; x=\frac{\pi}{... | \frac{\pi}{8}(2k+1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,692 |
5.19 $7+4 \sin x \cos x+1,5(\tan x+\cot x)=0$.
The translation is as follows:
5.19 $7+4 \sin x \cos x+1.5(\tan x+\cot x)=0$. | 5.19 The equation is defined for $\sin x \neq 0, \cos x \neq 0$. Transform its left side:
$$
\begin{aligned}
& 7+2 \sin 2 x+\frac{3}{2}\left(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right)= \\
& =7+2 \sin 2 x+\frac{3\left(\sin ^{2} x+\cos ^{2} x\right)}{2 \sin x \cos x}=7+2 \sin 2 x+\frac{3}{\sin 2 x}
\end{aligned}... | (-1)^{k+1}\cdot\frac{\pi}{12}+\frac{\pik}{2},k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,693 |
$5.20 \operatorname{tg} x+\operatorname{ctg} x=\frac{2}{\cos 4 x}$.
$5.20 \tan x + \cot x = \frac{2}{\cos 4x}$. | 5.20 The equation is defined for $\sin x \neq 0, \cos x \neq 0, \cos 4 x \neq 0$.
$$
\text { Since } \operatorname{tg} x+\operatorname{ctg} x=\frac{2}{\sin 2 x} \text {, then } \frac{2}{\sin 2 x}=\frac{2}{\cos 4 x} \text {, from which }
$$
$$
\sin 2 x=\cos 4 x, \cos \left(\frac{\pi}{2}-2 x\right)-\cos 4 x=0
$$
$$
\s... | \frac{\pi}{12}(4k+1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,694 |
$5.22 \sin ^{2} x-2 \sin x \cos x=3 \cos ^{2} x$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
$5.22 \sin ^{2} x-2 \sin x \cos x=3 \cos ^{2} x$. | 5.22 Since the values of $x$ for which $\cos x=0$ are not solutions of the equation, we can divide both sides by $\cos ^{2} x$, obtaining:
$$
\begin{aligned}
& \operatorname{tg}^{2} x-2 \operatorname{tg} x=3 ; \operatorname{tg}^{2} x-2 \operatorname{tg} x-3=0 \\
& \operatorname{tg} x=1 \pm \sqrt{1+3}=1 \pm 2 ; \operat... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,695 | |
$5.26 \operatorname{ctg} x-\operatorname{tg} x=2\left(\frac{1}{\operatorname{tg} x+1}+\frac{1}{\operatorname{tg} x-1}\right)=4$ :
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
$5.26 \cot x - \tan x = 2\left(\frac{1}{\tan x + 1} +... | 5.26 We have
$$
\begin{aligned}
& \frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}+2 \frac{\tan x-1+\tan x+1}{\tan^{2} x-1}=4 \\
& \sin x \neq 0, \cos x \neq 0,|\tan x| \neq 1 \\
& \frac{\cos^{2} x-\sin^{2} x}{\sin x \cos x}+\frac{4 \tan x}{\tan^{2} x-1}=4 ; \frac{2 \cos 2 x}{\sin 2 x}-2 \tan 2 x=4 \\
& \cot 2 x-\tan 2 x=2... | \frac{\pi}{16}(4k+1),k\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,696 |
$5.27 \sin ^{2}\left(\frac{\pi}{8}+t\right)=\sin t+\sin ^{2}\left(\frac{\pi}{8}-t\right)$. | 5.27 Instruction. Use the power reduction formula (4.16).
Omвem: $t_{1}=\pi k k_{2}=\frac{\pi}{4}(8 k \pm 1), k \in Z$. | t_{1}=\pik,t_{2}=\frac{\pi}{4}(8k\1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,697 |
$5.28 \operatorname{tg}\left(x-15^{\circ}\right) \operatorname{ctg}\left(x+15^{\circ}\right)=\frac{1}{3}$. | 5.28 Let's move on to sines and cosines and use formula (4.27):
$$
\begin{aligned}
& \frac{2 \sin \left(x-15^{\circ}\right) \cos \left(x+15^{\circ}\right)}{2 \cos \left(x-15^{\circ}\right) \sin \left(x+15^{\circ}\right)}=\frac{1}{3} \\
& \cos \left(x-15^{\circ}\right) \neq 0 ; \sin \left(x+15^{\circ}\right) \neq 0 \\
... | 45(4k+1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,698 |
$5.29 \sin ^{3} z \cos z-\sin z \cos ^{3} z=\frac{\sqrt{2}}{8}$. | ### 5.29 We have
$2 \sin z \cos z\left(\sin ^{2} z-\cos ^{2} z\right)=\frac{\sqrt{2}}{4} ;-\sin 2 z \cos 2 z=\frac{\sqrt{2}}{4} ;$
$$
\sin 4 z=-\frac{\sqrt{2}}{2} ; 4 z=(-1)^{k+1} \frac{\pi}{4}+\pi k ; z=(-1)^{k+1} \frac{\pi}{16}+\frac{\pi k}{4}
$$
Answer: $z=(-1)^{k+1} \frac{\pi}{16}+\frac{\pi k}{4}, k \in Z$. | (-1)^{k+1}\frac{\pi}{16}+\frac{\pik}{4},k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,699 |
$5.30 \operatorname{tg} x \operatorname{tg} 20^{\circ}+\operatorname{tg} 20^{\circ} \operatorname{tg} 40^{\circ}+\operatorname{tg} 40^{\circ} \operatorname{tg} x=1$.
## Group B
Solve the equations ( $5.31-5.58$ ): | 5.30 After factoring out $\operatorname{tg} x$ and further transformations, we get
$$
\begin{aligned}
& \operatorname{tg} x\left(\operatorname{tg} 20^{\circ}+\operatorname{tg} 40^{\circ}\right)=1-\operatorname{tg} 20^{\circ} \operatorname{tg} 40^{\circ} ; \operatorname{tg} x=\frac{1-\operatorname{tg} 20^{\circ} \opera... | 30+180\cdotk,k\in\boldsymbol{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,700 |
$5.31 \operatorname{tg} 6 x \cos 2 x-\sin 2 x-2 \sin 4 x=0$. | 5.31 Here $\cos 6 x \neq 0$. Therefore,
$\sin 6 x \cos 2 x - \sin 2 x \cos 6 x = 2 \sin 4 x \cos 6 x$
$\sin 4 x = 2 \sin 4 x \cos 6 x$
from which we get:
1) $\sin 4 x = 0 ; x = \frac{\pi k}{4}$, but for $k = 2 l + 1$ we have $\cos 6 x = 0$, so $x_{1} = \frac{\pi k}{2}$
2) $\cos 6 x = \frac{1}{2} ; x_{2} = \frac{\pi}... | x_{1}=\frac{\pik}{2},x_{2}=\frac{\pi}{18}(6k\1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,701 |
$5.34 \operatorname{tg}\left(120^{\circ}+3 x\right)-\operatorname{tg}\left(140^{\circ}-x\right)=2 \sin \left(80^{\circ}+2 x\right)$. | ### 5.34 We have
$$
\operatorname{tg}\left(120^{\circ}+3 x\right)+\operatorname{tg}\left(40^{\circ}+x\right)=2 \sin \left(80^{\circ}+2 x\right)
$$
Let $40^{\circ}+x=y$; then
$\operatorname{tg} 3 y+\operatorname{tg} y=2 \sin 2 y ; \cos 3 y \neq 0$, from which it follows that $\cos y \neq 0$; $\frac{\sin 4 y}{\cos y \... | 60\cdotk-40,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,703 |
$5.35 \operatorname{tg} 2 x-\operatorname{ctg} 3 x+\operatorname{ctg} 5 x=0$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
$5.35 \tan 2x - \cot 3x + \cot 5x = 0$. | ### 5.35 We have
$$
\begin{aligned}
& \frac{\sin 2 x}{\cos 2 x} - \frac{\cos 3 x}{\sin 3 x} + \frac{\cos 5 x}{\sin 5 x} = 0 \\
& \sin 3 x \neq 0, \sin 5 x \neq 0, \cos 2 x \neq 0 \\
& \frac{\sin 2 x \sin 3 x - \cos 2 x \cos 3 x}{\cos 2 x \sin 3 x} + \frac{\cos 5 x}{\sin 5 x} = 0
\end{aligned}
$$
$-\frac{\cos 5 x}{\co... | x_{1}=\frac{\pi}{10}(2k+1),x_{2}=\frac{\pi}{6}(2k+1),k\in\mathbb{Z} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,704 |
$5.36 \operatorname{tg} x \frac{3-\operatorname{tg}^{2} x}{1-3 \operatorname{tg}^{2} x}=\sin 6 x$. | 5.36 Here $\operatorname{tg} x \neq \pm \frac{1}{\sqrt{3}}, \cos x \neq 0$. Transform the left side of the equation:
$$
\begin{aligned}
& \operatorname{tg} x \frac{3 \cos ^{2} x-\sin ^{2} x}{\cos ^{2} x-3 \sin ^{2} x}=\operatorname{tg} x \frac{2 \cos ^{2} x+\cos 2 x}{\cos 2 x-2 \sin ^{2} x}= \\
& =\operatorname{tg} x ... | x_{1}=\frac{\pik}{3},x_{2}=\frac{\pi}{12}(2k+1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,705 |
$5.37 \operatorname{tg}\left(35^{\circ}+x\right) \operatorname{ctg}\left(10^{\circ}-x\right)=\frac{2}{3}$. | 5.37 Instruction. Put $y=35^{\circ}+x$ and solve the equation $\operatorname{tg} y \operatorname{tg}\left(45^{\circ}+y\right)=\frac{2}{3}$.
Answer: $\quad x_{1}=\operatorname{arctg} \frac{1}{3}-35^{\circ}+180^{\circ} \cdot k, x_{2}=-\operatorname{arctg} 2-35^{\circ}+180^{\circ} \cdot k, k \in Z$. | x_{1}=\operatorname{arctg}\frac{1}{3}-35+180\cdotk,x_{2}=-\operatorname{arctg}2-35+180\cdotk,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,706 |
$5.38 \frac{1}{\cos ^{4} z}=\frac{160}{9}-\frac{2(\operatorname{ctg} 2 z \operatorname{ctg} z+1)}{\sin ^{2} z}$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
$5.38 \frac{1}{\cos ^{4} z}=\frac{160}{9}-\frac{2(\cot 2 z \cot z+1)}{... | 5.38 Here $\cos z \neq 0, \sin z \neq 0$. Notice that
$\operatorname{ctg} 2 z \operatorname{ctg} z+1=\frac{\cos 2 z \cos z+\sin 2 z \sin z}{\sin 2 z \sin z}=$
$=\frac{\cos z}{2 \sin z \cos z \sin z}=\frac{1}{2 \sin ^{2} z}$.
Then we obtain the equation
$\frac{1}{\sin ^{4} z}+\frac{1}{\cos ^{4} z}=\frac{160}{9} ; \s... | \frac{\pi}{6}(3k\1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,707 |
$5.39 \sin 2 x+2 \operatorname{ctg} x=3$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
$5.39 \sin 2 x+2 \operatorname{ctg} x=3$. | 5.39 Using the formula $\sin 2 x=\frac{2 \operatorname{tg} x}{1+\operatorname{tg}^{2} x}$, we get
$$
\frac{2 \operatorname{tg} x}{1+\operatorname{tg}^{2} x}+\frac{2}{\operatorname{tg} x}=3 ; 4 \operatorname{tg}^{2} x+2=3 \operatorname{tg} x+3 \operatorname{tg}^{3} x
$$
Notice that $\operatorname{tg} x=1$ is a solutio... | \frac{\pi}{4}(4k+1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,708 |
$5.43 \cos \left(22^{\circ}-t\right) \cos \left(82^{\circ}-t\right)+\cos \left(112^{\circ}-t\right) \cos \left(172^{\circ}-t\right)=$ $=0.5(\sin t+\cos t)$. | 5.43 Transform the left side of the equation:
$$
\begin{aligned}
& \frac{1}{2} \cos \left(104^{\circ}-2 t\right)+\frac{1}{2} \cos 60^{\circ}+\frac{1}{2} \cos \left(284^{\circ}-2 t\right)+\frac{1}{2} \cos 60^{\circ}= \\
& =-\frac{1}{2} \sin \left(14^{\circ}-2 t\right)+\frac{1}{4}+\frac{1}{2} \sin \left(14^{\circ}-2 t\r... | t_{1}=360\cdotk,t_{2}=90(4k+1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,710 |
$5.44 \operatorname{tg} 5 x-2 \operatorname{tg} 3 x=\operatorname{tg}^{2} 3 x \operatorname{tg} 5 x$. | 5.44 Here $\cos 3 x \neq 0, \cos 5 x \neq 0$. Move $\operatorname{tg} 3 x$ to the right side of the equation:
$$
\begin{aligned}
& \operatorname{tg} 5 x-\operatorname{tg} 3 x=\operatorname{tg} 3 x+\operatorname{tg}^{2} 3 x \operatorname{tg} 5 x \\
& \operatorname{tg} 5 x-\operatorname{tg} 3 x=\operatorname{tg} 3 x(1+\... | \pik,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,711 |
$5.45 \operatorname{tg}^{4} x+\operatorname{ctg}^{4} x=\frac{82}{9}(\operatorname{tg} x \operatorname{tg} 2 x+1) \cos 2 x$. | 5.45 Note that $\operatorname{tg}^{4} x+\operatorname{ctg}^{4} x=\left(\operatorname{tg}^{2} x+\operatorname{ctg}^{2} x\right)^{2}-2$. Transform the right side of the equation:
$(\operatorname{tg} x \operatorname{tg} 2 x+1) \cos 2 x=\left(\operatorname{tg} x \frac{2 \operatorname{tg} x}{1-\operatorname{tg}^{2} x}+1\ri... | \frac{\pi}{6}(3k\1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,712 |
$5.46 \operatorname{ctg} x-\operatorname{tg} x-2 \operatorname{tg} 2 x-4 \operatorname{tg} 4 x+8=0$. | 5.46 Noting that $\operatorname{ctg} x-\operatorname{tg} x=\frac{1-\operatorname{tg}^{2} x}{\operatorname{tg} x}=2 \operatorname{ctg} 2 x$, we transform the left side of the equation:
$2 \operatorname{ctg} 2 x-2 \operatorname{tg} 2 x-4 \operatorname{tg} 4 x+8=4 \operatorname{ctg} 4 x-4 \operatorname{tg} 4 x+8=8 \opera... | \frac{\pi}{32}(4k+3),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,713 |
$5.47 \operatorname{tg}^{4} x+\operatorname{ctg}^{4} x+\operatorname{tg}^{2} x-\operatorname{ctg}^{2} x=\frac{106}{9}$.
Note: In this context, $\operatorname{tg}$ and $\operatorname{ctg}$ are the notations for tangent (tan) and cotangent (cot) respectively. The equation can be rewritten using more common notations as... | 5.47 Instruction. Put $\operatorname{tg}^{2} x-\operatorname{ctg}^{2} x=z$; then $\operatorname{tg}^{4} x+$ $+\operatorname{ctg}^{4} x=z^{2}+2$.
Answer: $x_{1}=\frac{\pi}{3}(3 k \pm 1), x_{2}= \pm \frac{1}{2} \arccos \frac{\sqrt{157}-6}{11}+\pi k, k \in Z$. | x_{1}=\frac{\pi}{3}(3k\1),x_{2}=\\frac{1}{2}\arccos\frac{\sqrt{157}-6}{11}+\pik,k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,714 |
$5.48 \frac{3(\cos 2 x+\operatorname{ctg} 2 x)}{\operatorname{ctg} 2 x-\cos 2 x}-2(\sin 2 x+1)=0$.
$5.49(\cos x-\sin x)^{2}+\cos ^{4} x-\sin ^{4} x=0.5 \sin 4 x$. | 5.48 We have
$$
\frac{3 \cos 2 x\left(1+\frac{1}{\sin 2 x}\right)}{\cos 2 x\left(\frac{1}{\sin 2 x}-1\right)}-2(\sin 2 x+1)=0, \cos 2 x \neq 0, \sin 2 x \neq 0
$$
$$
\frac{3(\sin 2 x+1)}{1-\sin 2 x}-2(\sin 2 x+1)=0
$$
$\sin 2 x \neq -1$, since in this case $\cos 2 x=0$;
$$
\frac{3}{1-\sin 2 x}=2, \sin 2 x=-\frac{1}... | (-1)^{k+1}\frac{\pi}{12}+\frac{\pik}{2},k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,715 |
$5.50 \sin 6 x+2=2 \cos 4 x$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
$5.50 \sin 6 x+2=2 \cos 4 x$. | 5.50 We have
$\sin 6 x=2 \cos 4 x-2 ; \sin 6 x=-4 \sin ^{2} 2 x$.
Subtract $\sin 2 x$ from both sides of the equation:
$\sin 6 x-\sin 2 x=-4 \sin ^{2} 2 x-\sin 2 x$
$2 \sin 2 x \cos 4 x=-\sin 2 x(4 \sin 2 x+1)$.
Thus:
1) $\sin 2 x=0 ; 2 x=\pi k ; x_{1}=\frac{\pi k}{2}$
2) $2 \cos 4 x+4 \sin 2 x+1=0 ; 2-4 \sin ^{2}... | x_{1}=\frac{\pik}{2},x_{2}=(-1)^{k+1}\frac{\pi}{12}+\frac{\pik}{2},k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,716 |
$5.51 \frac{\cos ^{2}\left(\frac{\pi}{2}-2 t\right)}{1+\cos 2 t}=\frac{1}{\cos ^{2} 2 t}-1$. | ### 5.51 We have
$$
\begin{aligned}
& \frac{\sin ^{2} 2 t}{1+\cos 2 t}=\frac{1}{\cos ^{2} 2 t}-1, \cos 2 t \neq 0, \cos 2 t \neq-1 \\
& \frac{\sin ^{2} 2 t}{1+\cos 2 t}=\operatorname{tg}^{2} 2 t
\end{aligned}
$$
1) $\sin 2 t=0 ; 2 t=2 \pi k$, since $\cos 2 t=-1$ when $2 t=\pi(2 k+1)$;
2) $\cos ^{2} 2 t=1+\cos 2 t ; \... | t_{1}=\pik,t_{2}=\\frac{1}{2}\arccos\frac{1-\sqrt{5}}{2}+\pik,k\inZ | Algebra | proof | Yes | Yes | olympiads | false | 49,717 |
$5.52 \operatorname{ctg} x-\operatorname{tg} x=\sin x+\cos x$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
$5.52 \operatorname{ctg} x-\operatorname{tg} x=\sin x+\cos x$. | 5.52 Here $\sin x \neq 0, \cos x \neq 0$. We have:
$$
\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}=\sin x+\cos x
$$
$\cos ^{2} x-\sin ^{2} x=\sin x \cos x(\sin x+\cos x)$
$(\sin x+\cos x)(\cos x-\sin x-\sin x \cos x)=0 ;$
1) $\sin x+\cos x=0 ; \operatorname{tg} x=-1 ; x_{1}=\frac{\pi}{4}(4 k-1)$;
2) $\cos x-\sin x-\... | x_{1}=\frac{\pi}{4}(4k-1),x_{2}=(-1)^{k}\arcsin\frac{\sqrt{2}-2}{2}+\frac{\pi}{4}(4k+1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,718 |
$5.54 \frac{\sin ^{2} t-\tan^{2} t}{\cos ^{2} t-\cot^{2} t}+2 \tan^{3} t+1=0$. | 5.54 Here $\sin t \neq 0, \cos t \neq 0$. Transform the expression
$$
\begin{aligned}
& \frac{\sin ^{2} t-\operatorname{tg}^{2} t}{\cos ^{2} t-\operatorname{ctg}^{2} t} ; \text { we have } \\
& \frac{\operatorname{tg}^{2} t\left(\cos ^{2} t-1\right)}{\operatorname{ctg}^{2} t\left(\sin ^{2} t-1\right)}=\frac{\operatorn... | \frac{\pi}{4}(4k-1),k\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 49,720 |
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