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values | question_type stringclasses 4
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3.039. $\cos 4 \alpha-\sin 4 \alpha \operatorname{ctg} 2 \alpha=-1$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
3.039. $\cos 4 \alpha-\sin 4 \alpha \cot 2 \alpha=-1$. | Solution.
$$
\begin{aligned}
& \cos 4 \alpha-\sin 4 \alpha \operatorname{ctg} 2 \alpha=\cos 4 \alpha-\sin 4 \alpha \cdot \frac{\cos 2 \alpha}{\sin 2 \alpha}= \\
& =\frac{\sin 2 \alpha \cos 4 \alpha-\cos 2 \alpha \sin 4 \alpha}{\sin 2 \alpha}=\frac{\sin (-2 \alpha)}{\sin 2 \alpha}=\frac{-\sin 2 \alpha}{\sin 2 \alpha}=-... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,531 |
3.040. $\frac{1-\cos 4 \alpha}{\cos ^{-2} 2 \alpha-1}+\frac{1+\cos 4 \alpha}{\sin ^{-2} 2 \alpha-1}=2$. | Solution.
$$
\begin{aligned}
& \frac{1-\cos 4 \alpha}{\cos ^{-2} 2 \alpha-1}+\frac{1+\cos 4 \alpha}{\sin ^{-2} 2 \alpha-1}=\frac{1-\cos 4 \alpha}{\frac{1}{\cos ^{2} 2 \alpha}-1}+\frac{1+\cos 4 \alpha}{\frac{1}{\sin ^{2} 2 \alpha}-1}= \\
& =\frac{(1-\cos 4 \alpha) \cos ^{2} 2 \alpha}{1-\cos ^{2} 2 \alpha}+\frac{(1+\cos... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,532 |
3.043. $\frac{1-2 \sin ^{2} \alpha}{1+\sin 2 \alpha}=\frac{1-\operatorname{tg} \alpha}{1+\operatorname{tg} \alpha}$.
3.043. $\frac{1-2 \sin ^{2} \alpha}{1+\sin 2 \alpha}=\frac{1-\tan \alpha}{1+\tan \alpha}$. | Solution.
Using the formulas
$$
1-2 \sin ^{2} x=\cos 2 x, \sin ^{2} x+\cos ^{2} x=1
$$
and
$\sin 2 x=2 \sin x \cos x$
we represent the left side of the equation in the form
$$
X=\frac{\cos 2 \alpha}{\cos ^{2} \alpha+\sin ^{2} \alpha+2 \sin \alpha \cos \alpha}=\frac{\cos 2 \alpha}{(\cos \alpha+\sin \alpha)^{2}}
$$... | proof | Algebra | proof | Yes | Yes | olympiads | false | 50,533 |
3.044. $\frac{\sin 2 \alpha+\sin 5 \alpha-\sin 3 \alpha}{\cos \alpha+1-2 \sin ^{2} 2 \alpha}=2 \sin \alpha$. | ## Solution.
$$
\begin{aligned}
& \frac{\sin 2 \alpha+(\sin 5 \alpha-\sin 3 \alpha)}{\cos \alpha+\left(1-2 \sin ^{2} 2 \alpha\right)}=\frac{2 \sin \alpha \cos \alpha+2 \cos 4 \alpha \sin \alpha}{\cos \alpha+\cos 4 \alpha}= \\
& =\frac{2 \sin \alpha(\cos \alpha+\cos 4 \alpha)}{\cos \alpha+\cos 4 \alpha}=2 \sin \alpha .... | 2\sin\alpha | Algebra | proof | Yes | Yes | olympiads | false | 50,534 |
3.045. $\frac{\operatorname{ctg}^{2} 2 \alpha-1}{2 \operatorname{ctg} 2 \alpha}-\cos 8 \alpha \operatorname{ctg} 4 \alpha=\sin 8 \alpha$.
3.045. $\frac{\cot^{2} 2 \alpha-1}{2 \cot 2 \alpha}-\cos 8 \alpha \cot 4 \alpha=\sin 8 \alpha$. | Solution.
$$
\frac{\operatorname{ctg}^{2} 2 \alpha-1}{2 \operatorname{ctg} 2 \alpha}-\cos 8 \alpha \operatorname{ctg} 4 \alpha=\frac{\frac{1}{\operatorname{tg}^{2} 2 \alpha}-1}{\frac{2}{\operatorname{tg} 2 \alpha}}-\cos 8 \alpha \operatorname{ctg} 4 \alpha=
$$
$$
\begin{aligned}
& =\frac{1-\operatorname{tg}^{2} 2 \al... | proof | Algebra | proof | Yes | Yes | olympiads | false | 50,535 |
3.047. $\operatorname{ctg}\left(45^{\circ}+2 \alpha\right)=\frac{\cos 4 \alpha}{1+\sin 4 \alpha}$.
3.047. $\operatorname{cot}\left(45^{\circ}+2 \alpha\right)=\frac{\cos 4 \alpha}{1+\sin 4 \alpha}$. | ## Solution.
Let $X=\operatorname{ctg}\left(45^{\circ}+2 \alpha\right)=\frac{1}{\operatorname{tg}\left(45^{\circ}+2 \alpha\right)}$.
Applying the formula $\operatorname{tg} \frac{x}{2}=\frac{1-\cos x}{\sin x}$, where $x \neq \pi+2 \pi n, n \in Z$, to the expression $\operatorname{tg}\left(45^{\circ}+2 \alpha\right)$,... | proof | Algebra | proof | Yes | Yes | olympiads | false | 50,536 |
3.048. $\frac{\left(\sin ^{2} \alpha+\tan^{2} \alpha+1\right)\left(\cos ^{2} \alpha-\cot^{2} \alpha+1\right)}{\left(\cos ^{2} \alpha+\cot^{2} \alpha+1\right)\left(\sin ^{2} \alpha+\tan^{2} \alpha-1\right)}=1$. | Solution.
$$
\begin{aligned}
& \frac{\left(\sin ^{2} \alpha+\tan ^{2} \alpha+1\right)\left(\cos ^{2} \alpha-\cot ^{2} \alpha+1\right)}{\left(\cos ^{2} \alpha+\cot ^{2} \alpha+1\right)\left(\sin ^{2} \alpha+\tan ^{2} \alpha-1\right)}= \\
& =\frac{\left(\sin ^{2} \alpha+\frac{\sin ^{2} \alpha}{\cos ^{2} \alpha}+1\right)... | proof | Algebra | proof | Yes | Yes | olympiads | false | 50,537 |
3.049. $\left(\frac{\sqrt{\tan \alpha}+\sqrt{\cot \alpha}}{\sin \alpha+\cos \alpha}\right)^{2}=\frac{2}{\sin 2 \alpha}$. | ## Solution.
$$
\begin{aligned}
& \left(\frac{\sqrt{\tan \alpha}+\sqrt{\cot \alpha}}{\sin \alpha+\cos \alpha}\right)^{2}=\frac{\tan \alpha+2 \sqrt{\tan \alpha \cot \alpha}+\cot \alpha}{\sin ^{2} \alpha+2 \sin \alpha \cos \alpha+\cos ^{2} \alpha}= \\
& =\frac{\frac{\sin \alpha}{\cos \alpha}+2+\frac{\cos \alpha}{\sin \a... | proof | Algebra | proof | Yes | Yes | olympiads | false | 50,538 |
3.050. $\sin ^{2}\left(45^{\circ}+\alpha\right)-\sin ^{2}\left(30^{\circ}-\alpha\right)-\sin 15^{\circ} \cos \left(15^{\circ}+2 \alpha\right)=\sin 2 \alpha$. | Solution.
$$
\begin{aligned}
& \sin ^{2}\left(45^{\circ}+\alpha\right)-\sin ^{2}\left(30^{\circ}-\alpha\right)-\sin 15^{\circ} \cos \left(15^{\circ}+2 \alpha\right)= \\
& =\frac{1-\cos \left(90^{\circ}+2 \alpha\right)}{2}-\frac{1-\cos \left(60^{\circ}-2 \alpha\right)}{2}-\frac{1}{2}\left(\sin (-2 \alpha)+\sin \left(30... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,539 |
3.051. $\sin ^{6} \alpha+\cos ^{6} \alpha+3 \sin ^{2} \alpha \cos ^{2} \alpha=1$. | ## Solution.
$$
\begin{aligned}
& \sin ^{6} \alpha+\cos ^{6} \alpha+3 \sin ^{2} \alpha \cos ^{2} \alpha= \\
& =\left(\sin ^{2} \alpha\right)^{3}+\left(\cos ^{2} \alpha\right)^{3}+3 \sin ^{2} \alpha \cos ^{2} \alpha= \\
& =\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)\left(\sin ^{4} \alpha-\sin ^{2} \alpha \cos ^{2} \... | proof | Algebra | proof | Yes | Yes | olympiads | false | 50,540 |
3.052. $\frac{\tan 3a}{\tan \alpha}=\frac{3-\tan^{2} \alpha}{1-3 \tan^{2} \alpha}$. | ## Solution.
$$
\frac{\tan 3 \alpha}{\tan \alpha}=\frac{\tan(2 \alpha+\alpha)}{\tan \alpha}=\frac{\frac{\tan 2 \alpha+\tan \alpha}{1-\tan 2 \alpha \tan \alpha}}{\tan \alpha}=\frac{\tan 2 \alpha+\tan \alpha}{(1-\tan 2 \alpha \tan \alpha) \tan \alpha}=
$$
$$
\begin{aligned}
& =\frac{\frac{2 \tan \alpha}{1-\tan^{2} \alp... | proof | Algebra | proof | Yes | Yes | olympiads | false | 50,541 |
3.053. $\sin \alpha \sin (x-\alpha)+\sin ^{2}\left(\frac{x}{2}-\alpha\right)=\sin ^{2} \frac{x}{2}$. | ## Solution.
Using the formulas
$$
\sin A \sin B=\frac{1}{2}(\cos (A-B)-\cos (A+B))
$$
$\mathbf{h}$
$$
\sin ^{2} \frac{A}{2}=\frac{1-\cos A}{2}
$$
we represent the left side of the equation as
$$
\begin{aligned}
& X=\frac{1}{2}(\cos (2 \alpha-x)-\cos x)+\frac{1-\cos (x-2 \alpha)}{2}=\frac{\cos (x-2 \alpha)}{2}-\f... | proof | Algebra | proof | Yes | Yes | olympiads | false | 50,542 |
3.054. $\cos ^{2} \alpha-\sin ^{2} 2 \alpha=\cos ^{2} \alpha \cos 2 \alpha-2 \sin ^{2} \alpha \cos ^{2} \alpha$. | ## Solution.
$$
\begin{aligned}
& \cos ^{2} \alpha-\sin ^{2} 2 \alpha=\cos ^{2} \alpha-(\sin 2 \alpha)^{2}=\cos ^{2} \alpha-(2 \sin \alpha \cos \alpha)^{2}= \\
& =\cos ^{2} \alpha-4 \sin ^{2} \alpha \cos ^{2} \alpha=\cos ^{2} \alpha\left(1-4 \sin ^{2} \alpha\right)=
\end{aligned}
$$
$=\cos ^{2} \alpha\left(1-2 \sin ^... | proof | Algebra | proof | Yes | Yes | olympiads | false | 50,543 |
3.055. $\frac{\sin 7 \alpha}{\sin \alpha} - 2(\cos 2 \alpha + \cos 4 \alpha + \cos 6 \alpha) - 1 = 0$. | Solution.
$$
\begin{aligned}
& \frac{\sin 7 \alpha}{\sin \alpha}-2(\cos 2 \alpha+\cos 4 \alpha+\cos 6 \alpha)-1= \\
& =\frac{\sin (6 \alpha+\alpha)}{\sin \alpha}-2 \cos 2 \alpha-2 \cos 4 \alpha-2 \cos 6 \alpha-1= \\
& =\left(\frac{\sin (6 \alpha+\alpha)}{\sin \alpha}-2 \cos 6 \alpha\right)-2 \cos 2 \alpha-2 \cos 4 \al... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,544 |
3.056. $\sin ^{2} \alpha-\sin ^{2} \beta=\sin (\alpha+\beta) \sin (\alpha-\beta)$. | ## Solution.
$$
\begin{aligned}
& \sin ^{2} \alpha-\sin ^{2} \beta=(\sin \alpha-\sin \beta)(\sin \alpha+\sin \beta)= \\
& =2 \cos \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2} \cdot 2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}= \\
& =\left(2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha+\beta}{2}\rig... | proof | Algebra | proof | Yes | Yes | olympiads | false | 50,545 |
3.057. $\cos ^{4} x+\sin ^{2} y+\frac{1}{4} \sin ^{2} 2 x-1=\sin (y+x) \sin (y-x)$. | Solution.
$$
\begin{aligned}
& \cos ^{4} x+\sin ^{2} y+\frac{1}{4} \sin ^{2} 2 x-1=\cos ^{4} x+\sin ^{2} y+\frac{1}{4}(\sin 2 x)^{2}-1= \\
& =\cos ^{4} x+\sin ^{2} y+\frac{1}{4}(2 \sin x \cos x)^{2}-1= \\
& =\cos ^{4} x+\sin ^{2} y+\frac{1}{4} \cdot 4 \sin ^{2} x \cos ^{2} x-1= \\
& =\cos ^{4} x+\sin ^{2} y+\sin ^{2} ... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,546 |
3.061. $\frac{\tan \alpha+\tan \beta}{\tan(\alpha+\beta)}+\frac{\tan \alpha-\tan \beta}{\tan(\alpha-\beta)}+2 \tan^{2} \alpha=2 \cos^{-2} \alpha$. | ## Solution.
Since $\operatorname{tg} x+\operatorname{tg} y=\frac{\sin (x+y)}{\cos x \cos y}$ and $\operatorname{tg} x-\operatorname{tg} y=\frac{\sin (x-y)}{\cos x \cos y}$, where $x, y \neq \frac{\pi}{2}+\pi n, n \in Z$, the left side of the equation can be written as
$$
X=\frac{\frac{\sin (\alpha+\beta)}{\cos \alph... | 2\cos^{-2}\alpha | Algebra | proof | Yes | Yes | olympiads | false | 50,549 |
3.062. $1-\frac{1}{4} \sin ^{2} 2 \alpha+\cos 2 \alpha=\cos ^{2} \alpha+\cos ^{4} \alpha$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
3.062. $1-\frac{1}{4} \sin ^{2} 2 \alpha+\cos 2 \alpha=\cos ^{2} \alpha+\cos ^{4} \alpha$. | Solution.
$1-\frac{1}{4} \sin ^{2} 2 \alpha+\cos 2 \alpha=1-\frac{1}{4}(\sin 2 \alpha)^{2}+\cos 2 \alpha=$
$=1-\frac{1}{4}(2 \sin \alpha \cos \alpha)^{2}+\cos ^{2} \alpha-\sin ^{2} \alpha=1-\frac{1}{4} \cdot 4 \sin ^{2} \alpha \cos ^{2} \alpha+$
$+\cos ^{2} \alpha-\sin ^{2} \alpha=1+\cos ^{2} \alpha-\sin ^{2} \alpha... | Algebra | proof | Yes | Yes | olympiads | false | 50,550 | |
3.063. $1-\sin \left(\frac{\alpha}{2}-3 \pi\right)-\cos ^{2} \frac{\alpha}{4}+\sin ^{2} \frac{\alpha}{4}$. | Solution.
$$
\begin{aligned}
& 1-\sin \left(\frac{\alpha}{2}-3 \pi\right)-\cos ^{2} \frac{\alpha}{2}+\sin ^{2} \frac{\alpha}{2}=1+\sin \left(3 \pi-\frac{\alpha}{2}\right)-\cos ^{2} \frac{\alpha}{4}+\sin ^{2} \frac{\alpha}{4}= \\
& =2 \sin ^{2} \frac{\alpha}{4}+\sin \frac{\alpha}{2}=2 \sin ^{2} \frac{\alpha}{4}+\sin \l... | 2\sqrt{2}\sin\frac{\alpha}{4}\sin(\frac{\alpha+\pi}{4}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,551 |
3.067. $\cos \alpha\left(1+\cos ^{-1} \alpha+\operatorname{tg} \alpha\right)\left(1-\cos ^{-1} \alpha+\operatorname{tg} \alpha\right)$
3.067. $\cos \alpha\left(1+\cos ^{-1} \alpha+\tan \alpha\right)\left(1-\cos ^{-1} \alpha+\tan \alpha\right)$ | ## Solution.
$$
\begin{aligned}
& \cos \alpha\left(1+\cos ^{-1} \alpha+\operatorname{tg} \alpha\right)\left(1-\cos ^{-1} \alpha+\operatorname{tg} \alpha\right)= \\
& =\cos \alpha\left(1+\frac{1}{\cos \alpha}+\frac{\sin \alpha}{\cos \alpha}\right)\left(1-\frac{1}{\cos \alpha}+\frac{\sin \alpha}{\cos \alpha}\right)= \\
... | 2\sin\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,554 |
3.069. $\frac{1-\cos (8 \alpha-3 \pi)}{\tan 2 \alpha-\cot 2 \alpha}$. | Solution.
$$
\begin{aligned}
& \frac{1-\cos (8 \alpha-3 \pi)}{\tan 2 \alpha-\cot 2 \alpha}=\frac{1-\cos (3 \pi-8 \alpha)}{\frac{\sin 2 \alpha}{\cos 2 \alpha}-\frac{\cos 2 \alpha}{\sin 2 \alpha}}=\frac{(1-\cos (3 \pi-8 \alpha)) \sin 2 \alpha \cos 2 \alpha}{\sin ^{2} 2 \alpha-\cos ^{2} 2 \alpha}= \\
& =-\frac{(1-\cos (3... | -\frac{\sin8\alpha}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,556 |
3.070. $\cos \left(\frac{\pi}{6}-\frac{\alpha}{2}\right) \sin \left(\frac{\pi}{3}-\frac{\alpha}{2}\right) \sin \frac{\alpha}{2}$. | Solution.
$$
\begin{aligned}
& \cos \left(\frac{\pi}{6}-\frac{\alpha}{2}\right) \sin \left(\frac{\pi}{3}-\frac{\alpha}{2}\right) \sin \frac{\alpha}{2}=\frac{1}{2}\left(\sin \frac{\pi}{6}+\sin \left(\frac{\pi}{2}-\alpha\right)\right) \sin \frac{\alpha}{2}= \\
& =\frac{1}{2}\left(\frac{1}{2}+\cos \alpha\right) \sin \fra... | \frac{1}{4}\sin\frac{3}{2}\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,557 |
3.071. $\sin ^{2}\left(\frac{\alpha}{2}+2 \beta\right)-\sin ^{2}\left(\frac{\alpha}{2}-2 \beta\right)$. | ## Solution.
$$
\begin{aligned}
& \sin ^{2}\left(\frac{\alpha}{2}+2 \beta\right)-\sin ^{2}\left(\frac{\alpha}{2}-2 \beta\right)=\frac{1-\cos (\alpha+4 \beta)}{2}-\frac{1-\cos (\alpha-4 \beta)}{2}= \\
& =\frac{1}{2}-\frac{\cos (\alpha+4 \beta)}{2}-\frac{1}{2}+\frac{\cos (\alpha-4 \beta)}{2}=\frac{1}{2}(\cos (\alpha-4 \... | \sin\alpha\sin4\beta | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,558 |
3.074. $\sin ^{2}(\alpha+2 \beta)+\sin ^{2}(\alpha-2 \beta)-1$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
3.074. $\sin ^{2}(\alpha+2 \beta)+\sin ^{2}(\alpha-2 \beta)-1$. | Solution.
$$
\sin ^{2}(\alpha+2 \beta)+\sin ^{2}(\alpha-2 \beta)-1=\frac{1-\cos (2 \alpha+4 \beta)}{2}+
$$
$+\frac{1-\cos (2 \alpha-4 \beta)}{2}-1=\frac{1}{2}-\frac{\cos (2 \alpha+4 \beta)}{2}+\frac{1}{2}-\frac{\cos (2 \alpha-4 \beta)}{2}-1=$ $=-\frac{1}{2}(\cos (2 \alpha+4 \beta)+\cos (2 \alpha-4 \beta))=-\frac{1}{2... | -\cos2\alpha\cos4\beta | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,561 |
3.075. $(\cos \alpha-\cos 2 \beta)^{2}+(\sin \alpha+\sin 2 \beta)^{2}$. | ## Solution.
$(\cos \alpha-\cos 2 \beta)^{2}+(\sin \alpha+\sin 2 \beta)^{2}=$
$=\cos ^{2} \alpha-2 \cos \alpha \cos 2 \beta+\cos ^{2} 2 \beta+\sin ^{2} \alpha+2 \sin \alpha \sin 2 \beta+\sin ^{2} 2 \beta=$
$=\left(\cos ^{2} \alpha+\sin ^{2} \alpha\right)+\left(\cos ^{2} 2 \beta+\sin ^{2} 2 \beta\right)-2(\cos \alpha... | 4\sin^{2}\frac{\alpha+2\beta}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,562 |
3.076. $\frac{(1-\cos 2 \alpha) \cos \left(45^{\circ}+2 \alpha\right)}{2 \sin ^{2} 2 \alpha-\sin 4 \alpha}$. | ## Solution.
$$
\begin{aligned}
& \frac{(1-\cos 2 \alpha) \cos \left(45^{\circ}+2 \alpha\right)}{2 \sin ^{2} 2 \alpha-\sin 4 \alpha}=\frac{\left(1-1+2 \sin ^{2} \alpha\right) \cdot \frac{\sqrt{2}}{2}(\cos 2 \alpha-\sin 2 \alpha)}{2 \sin ^{2} 2 \alpha-2 \sin 2 \alpha \cos 2 \alpha}= \\
& =\frac{\sqrt{2} \sin ^{2} \alph... | -\frac{\sqrt{2}}{4}\tan\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,563 |
3.077. $\cos ^{2}\left(\frac{3}{8} \pi-\frac{\alpha}{4}\right)-\cos ^{2}\left(\frac{11}{8} \pi+\frac{\alpha}{4}\right)$. | Solution.
$$
\begin{aligned}
& \cos ^{2}\left(\frac{3}{8} \pi-\frac{\alpha}{4}\right)-\cos ^{2}\left(\frac{11}{8} \pi+\frac{\alpha}{4}\right)=\frac{1+\cos \left(\frac{3 \pi}{4}-\frac{\alpha}{2}\right)}{2}-\frac{1+\cos \left(\frac{11}{4} \pi+\frac{\alpha}{2}\right)}{2}= \\
& =\frac{1+\cos \left(\frac{4 \pi-\pi}{4}-\fra... | \frac{\sqrt{2}}{2}\sin\frac{\alpha}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,564 |
3.078. $\operatorname{ctg}\left(45^{\circ}-\frac{\alpha}{2}\right)+\operatorname{ctg}\left(135^{\circ}-\frac{\alpha}{2}\right)$. | Solution.
$$
\begin{aligned}
& \operatorname{ctg}\left(45^{\circ}-\frac{\alpha}{2}\right)+\operatorname{ctg}\left(135^{\circ}-\frac{\alpha}{2}\right)=\frac{\cos \left(45^{\circ}-\frac{\alpha}{2}\right)}{\sin \left(45^{\circ}-\frac{\alpha}{2}\right)}+\frac{\cos \left(135^{\circ}-\frac{\alpha}{2}\right)}{\sin \left(135^... | 2\operatorname{tg}\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,565 |
3.079. $\frac{1+\operatorname{cot} 2 \alpha \operatorname{cot} \alpha}{\operatorname{tan} \alpha+\operatorname{cot} \alpha}$. | ## Solution.
$\frac{1+\operatorname{ctg} 2 \alpha \operatorname{ctg} \alpha}{\operatorname{tg} \alpha+\operatorname{ctg} \alpha}=\frac{1+\frac{\operatorname{ctg}^{2} \alpha-1}{2 \operatorname{ctg} \alpha} \cdot \operatorname{ctg} \alpha}{\frac{1}{\operatorname{ctg} \alpha}+\operatorname{ctg} \alpha}=\frac{\frac{2+\ope... | \frac{\operatorname{ctg}\alpha}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,566 |
3.081. $\sin ^{2}\left(\alpha-\frac{3 \pi}{2}\right)\left(1-\operatorname{tg}^{2} \alpha\right) \operatorname{tg}\left(\frac{\pi}{4}+\alpha\right) \cos ^{-2}\left(\frac{\pi}{4}-\alpha\right)$. | Solution.
$$
\begin{aligned}
& \sin ^{2}\left(\alpha-\frac{3 \pi}{2}\right)\left(1-\operatorname{tg}^{2} \alpha\right) \operatorname{tg}\left(\frac{\pi}{4}+\alpha\right) \cos ^{-2}\left(\frac{\pi}{4}-\alpha\right)= \\
& =\left(\sin \left(\frac{3}{2} \pi-\alpha\right)\right)^{2}\left(1-\operatorname{tg}^{2} \alpha\righ... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,567 |
3.082. $1-\frac{1}{1-\sin ^{-1}\left(2 \alpha+\frac{3}{2} \pi\right)}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
3.082. $1-\frac{1}{1-\sin ^{-1}\left(2 \alpha+\frac{3}{2} \pi\right)}$. | Solution.
$$
\begin{aligned}
& 1-\frac{1}{1-\sin ^{-1}\left(2 \alpha+\frac{3}{2} \pi\right)}=1-\frac{1}{1-\frac{1}{\sin \left(2 \alpha+\frac{3 \pi}{2}\right)}}=1-\frac{1}{1-\frac{1}{\sin \left(\frac{3 \pi}{2}+2 \alpha\right)}}= \\
& =1-\frac{1}{1-\frac{1}{-\cos 2 \alpha}}=1-\frac{1}{1+\frac{1}{\cos 2 \alpha}}=1-\frac{... | \frac{1}{2\cos^{2}\alpha} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,568 |
3.085. $1-\frac{1}{1-\sin ^{-1}\left(\frac{\pi}{2}+\alpha\right)}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
3.085. $1-\frac{1}{1-\sin ^{-1}\left(\frac{\pi}{2}+\alpha\right)}$. | ## Solution.
$$
\begin{aligned}
& 1-\frac{1}{1-\sin ^{-1}\left(\frac{\pi}{2}+\alpha\right)}=1-\frac{1}{1-\frac{1}{\sin \left(\frac{\pi}{2}+\alpha\right)}}=1-\frac{1}{1-\frac{1}{\cos \alpha}}=1-\frac{1}{\frac{\cos \alpha-1}{\cos \alpha}}= \\
& =1-\frac{\cos \alpha}{\cos \alpha-1}=\frac{\cos \alpha-1 \cdot \cos \alpha}{... | 0.5\sin^{-2}\frac{\alpha}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,570 |
3.086. $\frac{1-\operatorname{tg}(\pi-2 \alpha) \operatorname{tg} \alpha}{\operatorname{tg}\left(\frac{3}{2} \pi-\alpha\right)+\operatorname{tg} \alpha}$. | ## Solution.
$$
\begin{aligned}
& \frac{1-\operatorname{tg}(\pi-2 \alpha) \operatorname{tg} \alpha}{\operatorname{tg}\left(\frac{3}{2} \pi-\alpha\right)+\operatorname{tg} \alpha}=\frac{1+\operatorname{tg} 2 \alpha \operatorname{tg} \alpha}{\operatorname{ctg} \alpha+\operatorname{tg} \alpha}=\frac{1+\frac{2 \operatorna... | \frac{\operatorname{tg}2\alpha}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,571 |
3.088. $\frac{\operatorname{ctg}\left(270^{\circ}-\alpha\right)}{1-\operatorname{tg}^{2}\left(\alpha-180^{\circ}\right)} \cdot \frac{\operatorname{ctg}^{2}\left(360^{\circ}-\alpha\right)-1}{\operatorname{ctg}\left(180^{\circ}+\alpha\right)}$. | ## Solution.
$\frac{\operatorname{ctg}\left(270^{\circ}-\alpha\right)}{1-\operatorname{tg}^{2}\left(\alpha-180^{\circ}\right)} \cdot \frac{\operatorname{ctg}^{2}\left(360^{\circ}-\alpha\right)-1}{\operatorname{ctg}\left(180^{\circ}+\alpha\right)}=\frac{\operatorname{tg} \alpha}{1-\operatorname{tg}^{2} \alpha} \cdot \f... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,573 |
3.090. $\frac{\left(1+\operatorname{tg}^{2}\left(\alpha-90^{\circ}\right)\right)\left(\sin ^{-2}\left(\alpha-270^{\circ}\right)-1\right)}{\left(1+\operatorname{ctg}^{2}\left(\alpha+270^{\circ}\right)\right) \cos ^{-2}\left(\alpha+90^{\circ}\right)}$.
3.090. $\frac{\left(1+\tan^{2}\left(\alpha-90^{\circ}\right)\right)\... | Solution.
$$
\begin{aligned}
& \frac{\left(1+\operatorname{tg}^{2}\left(\alpha-90^{\circ}\right)\right)\left(\sin ^{-2}\left(\alpha-270^{\circ}\right)-1\right)}{\left(1+\operatorname{ctg}^{2}\left(\alpha+270^{\circ}\right)\right) \cos ^{-2}\left(\alpha+90^{\circ}\right)}= \\
& =\frac{\left(1+\left(\operatorname{tg}\le... | \sin^{2}\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,574 |
3.094. $\frac{\cos ^{2}\left(2 \alpha-90^{\circ}\right)+\operatorname{ctg}^{2}\left(90^{\circ}+2 \alpha\right)+1}{\sin ^{2}\left(2 a-270^{\circ}\right)+\operatorname{tg}^{2}\left(270^{\circ}+2 \alpha\right)+1}$. | Solution.
$$
\frac{\cos ^{2}\left(2 \alpha-90^{\circ}\right)+\operatorname{ctg}^{2}\left(90^{\circ}+2 \alpha\right)+1}{\sin ^{2}\left(2 \alpha-270^{\circ}\right)+\operatorname{tg}^{2}\left(270^{\circ}+2 \alpha\right)+1}=
$$
$$
\begin{aligned}
& =\frac{\left(\cos \left(90^{\circ}-2 \alpha\right)\right)^{2}+\left(\oper... | \operatorname{tg}^{2}2\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,576 |
3.098. $\sin \left(2 \alpha-\frac{3}{2} \pi\right)+\cos \left(2 \alpha-\frac{8}{3} \pi\right)+\cos \left(\frac{2}{3} \pi+2 \alpha\right)$. | ## Solution.
Let
$$
\begin{aligned}
& X=\sin \left(2 \alpha-\frac{3}{2} \pi\right)+\cos \left(2 \alpha-\frac{8}{3} \pi\right)+\cos \left(\frac{2}{3} \pi+2 \alpha\right)= \\
& =-\sin \left(\frac{3}{2} \pi-2 \alpha\right)+\cos \left(\frac{8}{3} \pi-2 \alpha\right)+\cos \left(\frac{2}{3} \pi+2 \alpha\right) \\
& -\sin \... | 0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,577 |
3.099. $\frac{4 \sin ^{2}(\alpha-5 \pi)-\sin ^{2}(2 \alpha+\pi)}{\cos ^{2}\left(2 \alpha-\frac{3}{2} \pi\right)-4+4 \sin ^{2} \alpha}$. | ## Solution.
$$
\begin{aligned}
& \frac{4 \sin ^{2}(\alpha-5 \pi)-\sin ^{2}(2 \alpha+\pi)}{\cos ^{2}\left(2 \alpha-\frac{3}{2} \pi\right)-4+4 \sin ^{2} \alpha}=\frac{4(-\sin (5 \pi-\alpha))^{2}-(\sin (\pi+2 \alpha))^{2}}{\left(\cos \left(\frac{3}{2} \pi-2 \alpha\right)\right)^{2}-4+4 \sin ^{2} \alpha}= \\
& =\frac{4 \... | -\tan^{4}\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,578 |
3.100. $\sin ^{2}\left(\frac{9}{8} \pi+\alpha\right)-\sin ^{2}\left(\frac{17}{8} \pi-\alpha\right)$. | Solution.
$$
\begin{aligned}
& \sin ^{2}\left(\frac{9}{8} \pi+\alpha\right)-\sin ^{2}\left(\frac{17}{8} \pi-\alpha\right)=\left(\sin \left(\frac{8 \pi+\pi}{8}+\alpha\right)\right)^{2}-\left(\sin \left(\frac{16 \pi+\pi}{8}-\alpha\right)\right)^{2}= \\
& =\left(\sin \left(\pi+\left(\frac{\pi}{8}+\alpha\right)\right)\rig... | \frac{1}{\sqrt{2}}\sin2\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,579 |
3.101. $\operatorname{ctg}(4 \alpha-\pi)\left(\cos ^{4}\left(\frac{5}{4} \pi-2 \alpha\right)-\sin ^{4}\left(\frac{9}{4} \pi-2 \alpha\right)\right)$. | Solution.
$$
\begin{aligned}
& \operatorname{ctg}(4 \alpha-\pi)\left(\cos ^{4}\left(\frac{5}{4} \pi-2 \alpha\right)-\sin ^{4}\left(\frac{9}{4} \pi-2 \alpha\right)\right)= \\
& =-\operatorname{ctg}(\pi-4 \alpha)\left(\left(\cos \left(\frac{4 \pi+\pi}{4}-2 \alpha\right)\right)^{4}-\left(\sin \left(\frac{8 \pi+\pi}{4}-2 ... | 4\cos2\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,580 |
3.103.
$$
\frac{\operatorname{tg}\left(\frac{5}{4} \pi-\alpha\right)(1+\sin 2 \alpha)}{\cos \left(\frac{5}{2} \pi-2 \alpha\right)}
$$
3.103.
$$
\frac{\tan\left(\frac{5}{4} \pi-\alpha\right)(1+\sin 2 \alpha)}{\cos \left(\frac{5}{2} \pi-2 \alpha\right)}
$$ | Solution.
$$
\frac{\operatorname{tg}\left(\frac{5}{4} \pi-\alpha\right)(1+\sin 2 \alpha)}{\cos \left(\frac{5}{2} \pi-2 \alpha\right)}=\frac{\operatorname{tg}\left(\frac{4 \pi+\pi}{4}-\alpha\right)(1+\sin 2 \alpha)}{\cos \left(\frac{4 \pi+\pi}{2}-2 \alpha\right)}=
$$
$$
\begin{aligned}
& =\frac{\operatorname{tg}\left(... | \operatorname{ctg}2\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,581 |
3.105. $\frac{\sin 6 \alpha}{\sin 2 \alpha}+\frac{\cos (6 \alpha-\pi)}{\cos 2 \alpha}$. | ## Solution.
$\frac{\sin 6 \alpha}{\sin 2 \alpha}+\frac{\cos (6 \alpha-\pi)}{\cos 2 \alpha}=\frac{\sin 6 \alpha}{\sin 2 \alpha}+\frac{\cos (\pi-6 \alpha)}{\cos 2 \alpha}=\frac{\sin 6 \alpha}{\sin 2 \alpha}+\frac{\cos 6 \alpha}{\cos 2 \alpha}=$
$=\frac{\sin 6 \alpha \cos 2 \alpha-\cos 6 \alpha \sin 2 \alpha}{\sin 2 \al... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,582 |
3.106.
$$
\frac{1+\cos (4 \alpha-2 \pi)+\cos \left(4 \alpha-\frac{\pi}{2}\right)}{1+\cos (4 \alpha+\pi)+\cos \left(4 \alpha+\frac{3}{2} \pi\right)}
$$ | Solution.
$$
\begin{aligned}
& \frac{1+\cos (4 \alpha-2 \pi)+\cos \left(4 \alpha-\frac{\pi}{2}\right)}{1+\cos (4 \alpha+\pi)+\cos \left(4 \alpha+\frac{3}{2} \pi\right)}=\frac{1+\cos (2 \pi-4 \alpha)+\cos \left(\frac{\pi}{2}-4 \alpha\right)}{1+\cos (\pi+4 \alpha)+\cos \left(\frac{3}{2} \pi+4 \alpha\right)}= \\
& =\frac... | \operatorname{ctg}2\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,583 |
3.107. $\frac{\sin (2 \alpha+2 \pi)+2 \sin (4 \alpha-\pi)+\sin (6 \alpha+4 \pi)}{\cos (6 \pi-2 \alpha)+2 \cos (4 \alpha-\pi)+\cos (6 \alpha-4 \pi)}$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
3.107. $\frac{\sin (2 \alpha+2 \p... | Solution.
$$
\begin{aligned}
& \frac{\sin (2 \alpha+2 \pi)+2 \sin (4 \alpha-\pi)+\sin (6 \alpha+4 \pi)}{\cos (6 \pi-2 \alpha)+2 \cos (4 \alpha-\pi)+\cos (6 \alpha-4 \pi)}= \\
& =\frac{\sin (2 \pi+2 \alpha)-2 \sin (\pi-4 \alpha)+\sin (4 \pi+6 \alpha)}{\cos (6 \pi-2 \alpha)+2 \cos (\pi-4 \alpha)+\cos (4 \pi-6 \alpha)}=\... | \tan4\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,584 |
3.108. $\frac{4 \sin \left(\frac{5}{2} \pi+\alpha\right)}{\operatorname{tg}^{2}\left(\frac{3}{2} \pi-\frac{\alpha}{2}\right)-\operatorname{ctg}^{2}\left(\frac{3}{2} \pi+\frac{\alpha}{2}\right)}$.
3.108. $\frac{4 \sin \left(\frac{5}{2} \pi+\alpha\right)}{\tan^{2}\left(\frac{3}{2} \pi-\frac{\alpha}{2}\right)-\cot^{2}\le... | Solution.
$$
\begin{aligned}
& \frac{4 \sin \left(\frac{5}{2} \pi+\alpha\right)}{\operatorname{tg}^{2}\left(\frac{3}{2} \pi-\frac{\alpha}{2}\right)-\operatorname{ctg}^{2}\left(\frac{3}{2} \pi+\frac{\alpha}{2}\right)}=\frac{4 \sin \left(\frac{4 \pi+\pi}{2}+\alpha\right)}{\left(\operatorname{tg}\left(\frac{3}{2} \pi-\fr... | \sin^{2}\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,585 |
3.110. $\frac{\cos 3 \alpha+\cos 4 \alpha+\cos 5 \alpha}{\sin 3 \alpha+\sin 4 \alpha+\sin 5 \alpha}$. | ## Solution.
$$
\begin{aligned}
& \frac{\cos 3 \alpha + \cos 4 \alpha + \cos 5 \alpha}{\sin 3 \alpha + \sin 4 \alpha + \sin 5 \alpha} = \frac{2 \cos 4 \alpha \cos \alpha + \cos 4 \alpha}{2 \sin 4 \alpha \cos \alpha + \sin 4 \alpha} = \frac{\cos 4 \alpha (2 \cos \alpha + 1)}{\sin 4 \alpha (2 \cos \alpha + 1)} = \\
& = ... | \operatorname{ctg}4\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,586 |
3.111.
$$
\frac{\cos ^{2}\left(\frac{5}{2} \pi-2 \alpha\right)+4 \cos ^{2}\left(\frac{7}{2} \pi-\alpha\right)-4}{1+\cos (4 \alpha-\pi)-8 \sin ^{2}(5 \pi-\alpha)}
$$ | Solution.
$$
\begin{aligned}
& \frac{\cos ^{2}\left(\frac{5}{2} \pi-2 \alpha\right)+4 \cos ^{2}\left(\frac{7}{2} \pi-\alpha\right)-4}{1+\cos (4 \alpha-\pi)-8 \sin ^{2}(5 \pi-\alpha)}= \\
& =\frac{\left(\cos \left(\frac{4 \pi+\pi}{2}-2 \alpha\right)\right)^{2}+4\left(\cos \left(\frac{6 \pi+\pi}{2}-\alpha\right)\right)^... | \frac{1}{2}\operatorname{ctg}^{4}\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,587 |
3.113. $\frac{1+\cos \alpha+\cos 2 \alpha+\cos 3 \alpha}{\cos \alpha+2 \cos ^{2} \alpha-1}$. | ## Solution.
$$
\begin{aligned}
& \frac{1+\cos \alpha+\cos 2 \alpha+\cos 3 \alpha}{\cos \alpha+2 \cos ^{2} \alpha-1}=\frac{1+\cos 2 \alpha+(\cos \alpha+\cos 3 \alpha)}{\cos \alpha+2 \cos ^{2} \alpha-1}= \\
& =\frac{1+\cos 2 \alpha+2 \cos 2 \alpha \cos \alpha}{\cos \alpha+2 \cos ^{2} \alpha-1}=\frac{1+2 \cos ^{2} \alph... | 2\cos\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,588 |
3.114. $\sin 4 \alpha-2 \cos ^{2} 2 \alpha+1$.
Translate the above text into English, keeping the original text's line breaks and format, and output the translation result directly.
3.114. $\sin 4 \alpha-2 \cos ^{2} 2 \alpha+1$. | Solution.
$$
\begin{aligned}
& \sin 4 \alpha-\cos 4 \alpha=\sin 4 \alpha-\sin \left(90^{\circ}-4 \alpha\right)=2 \cos 45^{\circ} \sin \left(4 \alpha-45^{\circ}\right)= \\
& =2 \cdot \frac{\sqrt{2}}{2} \cdot \sin \left(4 \alpha-45^{\circ}\right)=\sqrt{2} \sin \left(4 \alpha-45^{\circ}\right)
\end{aligned}
$$
Answer: $... | \sqrt{2}\sin(4\alpha-45) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,589 |
3.115. $\operatorname{tg} \frac{\alpha}{2}+\operatorname{ctg} \frac{\alpha}{2}+2$.
3.115. $\tan \frac{\alpha}{2}+\cot \frac{\alpha}{2}+2$. | Solution.
$$
\begin{aligned}
& \tan \frac{\alpha}{2}+\cot \frac{\alpha}{2}+2=\frac{\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}}+\frac{\cos \frac{\alpha}{2}}{\sin \frac{\alpha}{2}}+2=\frac{\sin ^{2} \frac{\alpha}{2}+\cos ^{2} \frac{\alpha}{2}}{\cos \frac{\alpha}{2} \sin \frac{\alpha}{2}}+2= \\
& =\frac{1}{\sin \frac{\... | 4\sin^{2}(\frac{\pi}{4}+\frac{\alpha}{2})\sin^{-1}\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,590 |
3.117. $\frac{\tan^{4} \alpha-\tan^{6} \alpha}{\cot^{4} \alpha-\cot^{2} \alpha}$. | ## Solution.
$$
\begin{aligned}
& \frac{\tan^{4} \alpha-\tan^{6} \alpha}{\cot^{4} \alpha-\cot^{2} \alpha}=\frac{\tan^{4} \alpha-\tan^{6} \alpha}{\frac{1}{\tan^{4} \alpha}-\frac{1}{\tan^{2} \alpha}}=\frac{\tan^{4} \alpha\left(1-\tan^{2} \alpha\right)}{\frac{1-\tan^{2} \alpha}{\tan^{4} \alpha}}= \\
& =\frac{\tan^{4} \al... | \tan^{8}\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,592 |
3.118. $1-3 \operatorname{tg}^{2}\left(\alpha+270^{\circ}\right)$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
3.118. $1-3 \operatorname{tg}^{2}\left(\alpha+270^{\circ}\right)$. | Solution.
$$
\begin{aligned}
& 1-3 \operatorname{tg}^{2}\left(\alpha+270^{\circ}\right)=1-3\left(\operatorname{tg}\left(270^{\circ}+\alpha\right)\right)^{2}=1-3 \operatorname{ctg}^{2} \alpha=4\left(\frac{1}{4}-\frac{3}{4} \operatorname{ctg}^{2} \alpha\right)= \\
& =4\left(\frac{1}{2}-\frac{\sqrt{3}}{2} \operatorname{c... | \frac{4\sin(\alpha-60)\sin(\alpha+60)}{\sin^{2}\alpha} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,593 |
3.119. $1-3 \operatorname{tg}^{2}\left(\alpha-180^{\circ}\right)$.
Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly.
3.119. $1-3 \operatorname{tg}^{2}\left(\alpha-180^{\circ}\right)$. | ## Solution.
$1-3 \operatorname{tg}^{2}\left(\alpha-180^{\circ}\right)=1-3\left(-\operatorname{tg}\left(180^{\circ}-\alpha\right)\right)^{2}=$
$$
\begin{aligned}
& =1-3 \operatorname{tg}^{2} \alpha=4\left(\frac{1}{4}-\frac{3}{4} \operatorname{tg}^{2} \alpha\right)=4\left(\frac{1}{2}-\frac{\sqrt{3}}{2} \operatorname{t... | 4\sin(30-\alpha)\sin(30+\alpha)\cos^{-2}\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,594 |
3.120. $\operatorname{tg}^{2}\left(\alpha-\frac{3}{2} \pi\right)-\operatorname{ctg}^{2}\left(\alpha+\frac{3}{2} \pi\right)$. | ## Solution.
$$
\begin{aligned}
& \operatorname{tg}^{2}\left(\alpha-\frac{3}{2} \pi\right)-\operatorname{ctg}^{2}\left(\alpha+\frac{3}{2} \pi\right)=\left(-\operatorname{tg}\left(\frac{3}{2} \pi-\alpha\right)\right)^{2}-\left(\operatorname{ctg}\left(\frac{3}{2} \pi+\alpha\right)\right)^{2}= \\
& =\operatorname{ctg}^{2... | 4\cos2\alpha\sin^{-2}2\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,595 |
3.121. $3 \sin ^{2}\left(\alpha-270^{\circ}\right)-\cos ^{2}\left(\alpha+270^{\circ}\right)$.
Translate the text above into English, please retain the original text's line breaks and format, and output the translation result directly.
3.121. $3 \sin ^{2}\left(\alpha-270^{\circ}\right)-\cos ^{2}\left(\alpha+270^{\... | ## Solution.
$$
\begin{aligned}
& 3 \sin ^{2}\left(\alpha-270^{\circ}\right)-\cos ^{2}\left(\alpha+270^{\circ}\right)=3\left(-\sin \left(270^{\circ}-\alpha\right)\right)^{2}-\left(\cos \left(270^{\circ}+\alpha\right)\right)^{2}= \\
& =3\left(\sin \left(270^{\circ}-\alpha\right)\right)^{2}-\left(\cos \left(270^{\circ}+... | 4\cos(30+\alpha)\cos(30-\alpha) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,596 |
3.124. $3-4 \cos ^{2}\left(\frac{3}{2} \pi-\alpha\right)$. | Solution.
$$
\begin{aligned}
& 3-4 \cos ^{2}\left(\frac{3}{2} \pi-\alpha\right)=3-4 \cdot \frac{1+\cos (3 \pi-2 \alpha)}{2}= \\
& =3-2-2 \cos (3 \pi-2 \alpha)=1-2 \cos (3 \pi-2 \alpha)= \\
& =1+2 \cos 2 \alpha=2\left(\frac{1}{2}+\cos 2 \alpha\right)=2\left(\cos \frac{\pi}{3}+\cos 2 \alpha\right)= \\
& =2 \cdot 2 \cos ... | 4\cos(\frac{\pi}{6}+\alpha)\cos(\frac{\pi}{6}-\alpha) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,597 |
3.126. $1+\cos \left(\frac{\pi}{2}+3 \alpha\right)-\sin \left(\frac{3}{2} \pi-3 \alpha\right)+\operatorname{ctg}\left(\frac{5}{2} \pi+3 \alpha\right)$. | ## Solution.
$$
1+\cos \left(\frac{\pi}{2}+3 \alpha\right)-\sin \left(\frac{3}{2} \pi-3 \alpha\right)+\operatorname{ctg}\left(\frac{5}{2} \pi+3 \alpha\right)=
$$
$$
\begin{aligned}
& =1-\sin 3 \alpha+\cos 3 \alpha-\operatorname{tg} 3 \alpha=1-\sin 3 \alpha+\cos 3 \alpha-\frac{\sin 3 \alpha}{\cos 3 \alpha}= \\
& =\fra... | \frac{2\sqrt{2}\cos^{2}\frac{3\alpha}{2}\sin(\frac{\pi}{4}-3\alpha)}{\cos3\alpha} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,598 |
3.127. $1+\cos \left(2 \alpha+270^{\circ}\right)+\sin \left(2 \alpha+450^{\circ}\right)$. | ## Solution.
$$
\begin{aligned}
& 1+\cos \left(2 \alpha+270^{\circ}\right)+\sin \left(2 \alpha+450^{\circ}\right)=1+\cos \left(270^{\circ}+2 \alpha\right)+\sin \left(450^{\circ}+2 \alpha\right)= \\
& =1+\sin 2 \alpha+\cos 2 \alpha=\cos ^{2} \alpha+\sin ^{2} \alpha+2 \sin \alpha \cos \alpha+\cos ^{2} \alpha-\sin ^{2} \... | 2\sqrt{2}\cos\alpha\cos(45-\alpha) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,599 |
3.128. $1-\cos \left(2 \alpha-270^{\circ}\right)+\sin \left(2 \alpha+270^{\circ}\right)$. | Solution.
$1-\cos \left(2 \alpha-270^{\circ}\right)+\sin \left(2 \alpha+270^{\circ}\right)=1-\cos \left(270^{\circ}-2 \alpha\right)+\sin \left(270^{\circ}+2 \alpha\right)=$ $=1+\sin 2 \alpha-\cos 2 \alpha=\sin ^{2} \alpha+\cos ^{2} \alpha+2 \sin \alpha \cos \alpha-\left(\cos ^{2} \alpha-\sin ^{2} \alpha\right)=$ $=(\s... | 2\sqrt{2}\sin\alpha\cos(45-\alpha) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,600 |
3.129. $\sin \left(\frac{5}{2} \pi-2 \alpha\right)+2 \sin ^{2}\left(2 \alpha-\frac{3}{2} \pi\right)-1$. | ## Solution.
$$
\begin{aligned}
& \sin \left(\frac{5}{2} \pi-2 \alpha\right)+2 \sin ^{2}\left(2 \alpha-\frac{3}{2} \pi\right)-1=\sin \left(\frac{5}{2} \pi-2 \alpha\right)+2 \sin ^{2}\left(\frac{3}{2} \pi-2 \alpha\right)-1= \\
& =\cos 2 \alpha+2 \cos ^{2} 2 \alpha-1=\cos 2 \alpha+\cos 4 \alpha=2 \cos 3 \alpha \cos \alp... | 2\cos\alpha\cos3\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,601 |
3.133. $2 \cos ^{2}\left(\frac{\alpha}{2}-\frac{3 \pi}{2}\right)+\sqrt{3} \cos \left(\frac{5}{2} \pi-\alpha\right)-1$. | ## Solution.
$$
\begin{aligned}
& 2 \cos ^{2}\left(\frac{\alpha}{2}-\frac{3 \pi}{2}\right)+\sqrt{3} \cos \left(\frac{5}{2} \pi-\alpha\right)-1= \\
& =2 \cos ^{2}\left(\frac{3 \pi}{2}-\frac{\alpha}{2}\right)+\sqrt{3} \cos \left(\frac{5}{2} \pi-\alpha\right)-1= \\
& =2 \sin ^{2} \frac{\alpha}{2}-\sqrt{3} \sin \alpha-1=1... | 2\sin(\alpha-\frac{\pi}{6}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,603 |
3.134. $\frac{\sin 4 \alpha+\sin 5 \alpha+\sin 6 \alpha}{\cos 4 \alpha+\cos 5 \alpha+\cos 6 \alpha}$. | Solution.
$\frac{\sin 4 \alpha+\sin 5 \alpha+\sin 6 \alpha}{\cos 4 \alpha+\cos 5 \alpha+\cos 6 \alpha}=\frac{(\sin 4 \alpha+\sin 6 \alpha)+\sin 5 \alpha}{(\cos 4 \alpha+\cos 6 \alpha)+\cos 5 \alpha}=$
$=\frac{2 \sin 5 \alpha \cos \alpha+\sin 5 \alpha}{2 \cos 5 \alpha \cos \alpha+\cos 5 \alpha}=\frac{\sin 5 \alpha(2 \... | \tan5\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,604 |
3.135. $-\cos 5 \alpha \cos 4 \alpha-\cos 4 \alpha \cos 3 \alpha+2 \cos ^{2} 2 \alpha \cos \alpha$. | Solution.
$-\cos 5 \alpha \cos 4 \alpha-\cos 4 \alpha \cos 3 \alpha+2 \cos ^{2} 2 \alpha \cos \alpha=$
$=-\cos 4 \alpha(\cos 5 \alpha+\cos 3 \alpha)+2 \cos ^{2} 2 \alpha \cos \alpha=$
$=-\cos 4 \alpha \cdot 2 \cos 4 \alpha \cos \alpha+2 \cos ^{2} 2 \alpha \cos \alpha=$
$=-2 \cos ^{2} 4 \alpha \cos \alpha+2 \cos ^{2... | 2\cos\alpha\sin2\alpha\sin6\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,605 |
3.136. $\sin 10 \alpha \sin 8 \alpha+\sin 8 \alpha \sin 6 \alpha-\sin 4 \alpha \sin 2 \alpha$. | Solution.
$\sin 10 \alpha \sin 8 \alpha+\sin 8 \alpha \sin 6 \alpha-\sin 4 \alpha \sin 2 \alpha .=$
$=\sin 8 \alpha(\sin 10 \alpha+\sin 6 \alpha)-\sin 4 \alpha \sin 2 \alpha=$
$=2 \sin 8 \alpha \sin 8 \alpha \cos 2 \alpha-\sin 4 \alpha \sin 2 \alpha=$
$=2 \sin ^{2} 8 \alpha \cos 2 \alpha-\sin 4 \alpha \sin 2 \alpha=... | 2\cos2\alpha\sin6\alpha\sin10\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,606 |
3.137. $\frac{\cos 7 \alpha-\cos 8 \alpha-\cos 9 \alpha+\cos 10 \alpha}{\sin 7 \alpha-\sin 8 \alpha-\sin 9 \alpha+\sin 10 \alpha}$. | ## Solution.
$$
\begin{aligned}
& \frac{\cos 7 \alpha - \cos 8 \alpha - \cos 9 \alpha + \cos 10 \alpha}{\sin 7 \alpha - \sin 8 \alpha - \sin 9 \alpha + \sin 10 \alpha} = \frac{(\cos 10 \alpha + \cos 7 \alpha) - (\cos 9 \alpha + \cos 8 \alpha)}{(\sin 10 \alpha + \sin 7 \alpha) - (\sin 9 \alpha + \sin 8 \alpha)} = \\
& ... | \operatorname{ctg}\frac{17\alpha}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,607 |
3.138. $\sin 5 \alpha-\sin 6 \alpha-\sin 7 \alpha+\sin 8 \alpha$. | Solution.
$$
\begin{aligned}
& \sin 5 \alpha-\sin 6 \alpha-\sin 7 \alpha+\sin 8 \alpha=(\sin 8 \alpha+\sin 5 \alpha)-(\sin 7 \alpha+\sin 6 \alpha)= \\
& =2 \sin \frac{13 \alpha}{2} \cos \frac{3 \alpha}{2}-2 \sin \frac{13 \alpha}{2} \cos \frac{\alpha}{2}=2 \sin \frac{13 \alpha}{2}\left(\cos \frac{3 \alpha}{2}-\cos \fra... | -4\sin\frac{\alpha}{2}\sin\alpha\sin\frac{13\alpha}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,608 |
3.140. $\frac{\sin 13 \alpha+\sin 14 \alpha+\sin 15 \alpha+\sin 16 \alpha}{\cos 13 \alpha+\cos 14 \alpha+\cos 15 \alpha+\cos 16 \alpha}$. | Solution.
$$
\begin{aligned}
& \frac{\sin 13 \alpha+\sin 14 \alpha+\sin 15 \alpha+\sin 16 \alpha}{\cos 13 \alpha+\cos 14 \alpha+\cos 15 \alpha+\cos 16 \alpha}= \\
& =\frac{(\sin 16 \alpha+\sin 13 \alpha)+(\sin 15 \alpha+\sin 14 \alpha)}{(\cos 16 \alpha+\cos 13 \alpha)+(\cos 15 \alpha+\cos 14 \alpha)}= \\
& =\frac{2 \s... | \tan\frac{29\alpha}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,609 |
3.141. $\sin 2 \alpha+\sin 4 \alpha+\sin 6 \alpha$. | Solution.
$$
\begin{aligned}
& \sin 2 \alpha+\sin 4 \alpha+\sin 6 \alpha .=(\sin 2 \alpha+\sin 4 \alpha)+\sin 2(3 \alpha)= \\
& =2 \sin 3 \alpha \cos \alpha+2 \sin 3 \alpha \cos 3 \alpha=2 \sin 3 \alpha(\cos \alpha+\cos 3 \alpha)= \\
& =2 \sin 3 \alpha \cdot 2 \cos 2 \alpha \cos \alpha=4 \sin 3 \alpha \cos 2 \alpha \c... | 4\sin3\alpha\cos2\alpha\cos\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,610 |
3.144. $3+4 \cos 4 \alpha+\cos 8 \alpha$.
3.144. $3+4 \cos 4 \alpha+\cos 8 \alpha$. | ## Solution.
$$
\begin{aligned}
& 3+4 \cos 4 \alpha+\cos 8 \alpha=3+4\left(2 \cos ^{2} 2 \alpha-1\right)+2 \cos ^{2} 4 \alpha-1= \\
& =3+8 \cos ^{2} 2 \alpha-4+2\left(2 \cos ^{2} 2 \alpha-1\right)^{2}-1= \\
& =8 \cos ^{2} 2 \alpha+2\left(4 \cos ^{4} 2 \alpha-4 \cos ^{2} 2 \alpha+1\right)-2= \\
& =8 \cos ^{2} 2 \alpha+... | 8\cos^{4}2\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,611 |
3.174. $\sin 2 \alpha+\sin 4 \alpha-\sin 6 \alpha$. | Solution.
$\sin 2 \alpha+\sin 4 \alpha-\sin 6 \alpha=\sin 2 \alpha+\sin 4 \alpha-\sin 2(3 \alpha)=$
$=2 \sin 3 \alpha \cos \alpha-2 \sin 3 \alpha \cos 3 \alpha=2 \sin 3 \alpha(\cos \alpha-\cos 3 \alpha)=$
$=2 \sin 3 \alpha \cdot(-2 \sin 2 \alpha \sin (-\alpha))=4 \sin 3 \alpha \sin 2 \alpha \sin \alpha$.
Answer: $4... | 4\sin3\alpha\sin2\alpha\sin\alpha | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,613 |
3.148. $\left(\sin 160^{\circ}+\sin 40^{\circ}\right)\left(\sin 140^{\circ}+\sin 20^{\circ}\right)+\left(\sin 50^{\circ}-\sin 70^{\circ}\right) \times$
$$
\times\left(\sin 130^{\circ}-\sin 110^{\circ}\right)=1
$$ | Solution.
$\left(\sin 160^{\circ}+\sin 40^{\circ}\right)\left(\sin 140^{\circ}+\sin 20^{\circ}\right)+\left(\sin 50^{\circ}-\sin 70^{\circ}\right)\left(\sin 130^{\circ}-\sin 110^{\circ}\right)=$
$$
\begin{aligned}
& =\left(\sin \left(180^{\circ}-20^{\circ}\right)+\sin 40^{\circ}\right)\left(\sin \left(180^{\circ}-40^... | 1 | Algebra | proof | Yes | Yes | olympiads | false | 50,614 |
3.149. $\left(\cos 34^{\circ}\right)^{-1}+\left(\tan 56^{\circ}\right)^{-1}=\cot 28^{\circ}$. | ## Solution.
$$
\begin{aligned}
& \left(\cos 34^{\circ}\right)^{-1}+\left(\operatorname{tg} 56^{\circ}\right)^{-1}=\frac{1}{\cos 34^{\circ}}+\operatorname{ctg} 56^{\circ}=\frac{1}{\cos \left(90^{\circ}-56^{\circ}\right)}+\operatorname{ctg} 56^{\circ}= \\
& =\frac{1}{\sin 56^{\circ}}+\frac{\cos 56^{\circ}}{\sin 56^{\ci... | proof | Algebra | proof | Yes | Yes | olympiads | false | 50,615 |
3.150. $\frac{\cos 28^{\circ} \cos 56^{\circ}}{\sin 2^{\circ}}+\frac{\cos 2^{\circ} \cos 4^{\circ}}{\sin 28^{\circ}}=\frac{\sqrt{3} \sin 38^{\circ}}{4 \sin 2^{\circ} \sin 28^{\circ}}$. | ## Solution.
$$
\begin{aligned}
& \frac{\cos 28^{\circ} \cos 56^{\circ}}{\sin 2^{\circ}}+\frac{\cos 2^{\circ} \cos 4^{\circ}}{\sin 28^{\circ}}=\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{4 \sin 28^{\circ} \cos 28^{\cir... | \frac{\sqrt{3}\sin38}{4\sin2\sin28} | Algebra | proof | Yes | Yes | olympiads | false | 50,616 |
3.152. $\left(\cos 70^{\circ}+\cos 50^{\circ}\right)\left(\cos 310^{\circ}+\cos 290^{\circ}\right)+\left(\cos 40^{\circ}+\cos 160^{\circ}\right) \times$
$$
\times\left(\cos 320^{\circ}-\cos 380^{\circ}\right)=1
$$ | ## Solution.
$\left(\cos 70^{\circ}+\cos 50^{\circ}\right)\left(\cos 310^{\circ}+\cos 290^{\circ}\right)+\left(\cos 40^{\circ}+\cos 160^{\circ}\right) \times$ $\times\left(\cos 320^{\circ}-\cos 380^{\circ}\right)=\left(\cos 70^{\circ}+\cos 50^{\circ}\right)\left(\cos \left(360^{\circ}-50^{\circ}\right)+\cos \left(360^... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,618 | |
3.153. $\sin ^{2} \frac{\pi}{8}+\cos ^{2} \frac{3 \pi}{8}+\sin ^{2} \frac{5 \pi}{8}+\cos ^{2} \frac{7 \pi}{8}$. | Solution.
$$
\begin{aligned}
& \sin ^{2} \frac{\pi}{8}+\cos ^{2} \frac{3 \pi}{8}+\sin ^{2} \frac{5 \pi}{8}+\cos ^{2} \frac{7 \pi}{8}= \\
& =\frac{1-\cos \frac{\pi}{4}}{2}+\frac{1+\cos \frac{3 \pi}{4}}{2}+\frac{1-\cos \frac{5 \pi}{4}}{2}+\frac{1+\cos \frac{7 \pi}{4}}{2}=
\end{aligned}
$$
$$
\begin{aligned}
& =\frac{4-... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,619 |
3.154. $\operatorname{tg} 435^{\circ}+\operatorname{tg} 375^{\circ}$.
Translate the above text into English, please retain the original text's line breaks and format, and output the translation result directly.
3.154. $\operatorname{tan} 435^{\circ}+\operatorname{tan} 375^{\circ}$. | ## Solution.
$$
\begin{aligned}
& \tan 435^{\circ}+\tan 375^{\circ}=\tan\left(450^{\circ}-15^{\circ}\right)+\tan\left(360^{\circ}+15^{\circ}\right)= \\
& =\cot 15^{\circ}+\tan 15^{\circ}=\frac{\cos 15^{\circ}}{\sin 15^{\circ}}+\frac{\sin 15^{\circ}}{\cos 15^{\circ}}=\frac{\cos ^{2} 15^{\circ}+\sin ^{2} 15^{\circ}}{\si... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,620 |
3.155. $\operatorname{tg} 255^{\circ}-\operatorname{tg} 195^{\circ}$.
The translation of the above text into English, retaining the original text's line breaks and format, is as follows:
3.155. $\operatorname{tg} 255^{\circ}-\operatorname{tg} 195^{\circ}$. | ## Solution.
$$
\begin{aligned}
& \tan 255^{\circ}-\tan 195^{\circ}=\tan\left(270^{\circ}-15^{\circ}\right)-\tan\left(180^{\circ}+15^{\circ}\right)= \\
& =\cot 15^{\circ}-\tan 15^{\circ}=\frac{\cos 15^{\circ}}{\sin 15^{\circ}}-\frac{\sin 15^{\circ}}{\cos 15^{\circ}}=\frac{\cos ^{2} 15^{\circ}-\sin ^{2} 15^{\circ}}{\si... | 2\sqrt{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,621 |
3.158. $\sin \left(2 \alpha+\frac{5}{4} \pi\right)$, if $\operatorname{tg} \alpha=\frac{2}{3}$. | ## Solution.
$$
\sin \left(2 \alpha+\frac{5}{4} \pi\right)=\sin \left(\frac{4 \pi+\pi}{4}+2 \alpha\right)=\sin \left(\pi+\left(2 \alpha+\frac{\pi}{4}\right)\right)=-\sin \left(2 \alpha+\frac{\pi}{4}\right)=
$$
$=-\sin 2 \alpha \cos \frac{\pi}{4}-\cos 2 \alpha \sin \frac{\pi}{4}=-\frac{\sqrt{2}}{2} \sin 2 \alpha-\frac... | -\frac{17\sqrt{2}}{26} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,622 |
3.159. $\cos \left(2 \alpha+\frac{7}{4} \pi\right)$, if $\operatorname{ctg} \alpha=\frac{2}{3}$. | ## Solution.
$$
\begin{aligned}
& \cos \left(2 \alpha+\frac{7}{4} \pi\right)=\cos \left(\frac{8 \pi-\pi}{4}+2 \alpha\right)=\cos \left(2 \pi+\left(2 \alpha-\frac{\pi}{4}\right)\right)= \\
& =\cos \left(2 \alpha-\frac{\pi}{4}\right)=\cos 2 \alpha \cos \frac{\pi}{4}+\sin 2 \alpha \sin \frac{\pi}{4}=\frac{\sqrt{2}}{2} \c... | \frac{7\sqrt{2}}{26} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,623 |
3.160. $\frac{5}{6+7 \sin 2 \alpha}$, if $\operatorname{tg} \alpha=0.2$. | ## Solution.
$$
\begin{aligned}
& \frac{5}{6+7 \sin 2 \alpha}=\frac{5}{6+\frac{14 \tan \alpha}{1+\tan^{2} \alpha}}=\frac{5+5 \tan^{2} \alpha}{6+6 \tan^{2} \alpha+14 \tan \alpha}= \\
& =\frac{5+5 \cdot 0.04}{6+6 \cdot 0.04+14 \cdot 0.2}=\frac{65}{113}
\end{aligned}
$$
Answer: $\frac{65}{113}$. | \frac{65}{113} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,624 |
3.162. $\sin \alpha, \quad$ if $\sin \frac{\alpha}{2}+\cos \frac{\alpha}{2}=1.4$. | ## Solution.
$\sin \frac{\alpha}{2}+\cos \frac{\alpha}{2}=1.4 \Rightarrow$
$\Rightarrow \sin ^{2} \frac{\alpha}{2}+2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}+\cos ^{2} \frac{\alpha}{2}=1.96, \quad\left(\sin ^{2} \frac{\alpha}{2}+\cos ^{2} \frac{\alpha}{2}\right)+\sin \alpha=1.96$,
$1+\sin \alpha=1.96$.
Then $\si... | \sin\alpha=0.96 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,625 |
3.163. $\sin 2 \alpha$, if $\sin \alpha - \cos \alpha = p$. | Solution.
$\sin \alpha-\cos \alpha=p \quad \Rightarrow$
$\Rightarrow \sin ^{2} \alpha-2 \sin \alpha \cos \alpha+\cos ^{2} \alpha=p^{2}, \quad 1-\sin 2 \alpha=p^{2}$,
from which $\sin 2 \alpha=1-p^{2}$.
Answer: $1-p^{2}$. | 1-p^{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,626 |
3.167. Find the number $\alpha \in\left(\frac{\pi}{2}, \pi\right)$, if it is known that $\operatorname{tg} 2 \alpha=-\frac{12}{5}$. | Solution.
$$
\frac{2 \tan \alpha}{1-\tan^{2} \alpha}=-\frac{12}{5}, \quad \frac{2 \tan \alpha}{1-\tan^{2} \alpha}+\frac{12}{5}=0 \Rightarrow
$$
$\Rightarrow 6 \tan^{2} \alpha-5 \tan \alpha-6=0$, from which $(\tan \alpha)_{1}=\frac{3}{2}$, which does not fit the solution of the problem, since by condition the angle be... | \pi-\arctan\frac{2}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,628 |
3.169. Find the number $\beta\left(\frac{\pi}{2}<\beta<\pi\right)$, if it is known that $\operatorname{tg}(\alpha+\beta)=\frac{9}{19}$ and $\operatorname{tg} \alpha=-4$. | Solution.
$$
\operatorname{tg}(\alpha+\beta)=\frac{\operatorname{tg} \alpha+\operatorname{tg} \beta}{1-\operatorname{tg} \alpha \operatorname{tg} \beta}=\frac{9}{19} .
$$
Since $\operatorname{tg} \alpha=-4$, then $\frac{-4+\operatorname{tg} \beta}{1+4 \operatorname{tg} \beta}=\frac{9}{19}, \operatorname{tg} \beta=-4$... | \beta=\pi-\operatorname{arctg}5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,629 |
3.170. Find $\sin ^{4} \alpha+\cos ^{4} \alpha$, if it is known that $\sin \alpha-\cos \alpha=\frac{1}{2}$. | ## Solution.
$$
\sin ^{4} \alpha+\cos ^{4} \alpha=\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)^{2}-2 \sin ^{2} \alpha \cos ^{2} \alpha=1-2 \sin ^{2} \alpha \cos ^{2} \alpha
$$
By squaring both sides of the equation $\sin \alpha-\cos \alpha=\frac{1}{2}$, we get
$$
\sin ^{2} \alpha-2 \sin \alpha \cos \alpha+\cos ^{2... | \frac{23}{32} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,630 |
3.171. Given: $\operatorname{ctg} \alpha=\frac{3}{4}, \operatorname{ctg} \beta=\frac{1}{7}, 0<\alpha<\frac{\pi}{2}, 0<\beta<\frac{\pi}{2}$. Find $\alpha+\beta$. | ## Solution.
$$
\begin{aligned}
& \operatorname{ctg} \alpha+\operatorname{ctg} \beta=\frac{3}{4}+\frac{1}{7}, \quad \frac{\cos \alpha}{\sin \alpha}+\frac{\cos \beta}{\sin \beta}=\frac{25}{28}, \\
& \frac{\sin \beta \cos \alpha+\cos \beta \sin \alpha}{\sin \alpha \sin \beta}=\frac{25}{28} .
\end{aligned}
$$
Given that... | \alpha+\beta=\frac{3\pi}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,631 |
3.172. Find $\operatorname{ctg} 2 \alpha$, if it is known that $\sin \left(\alpha-90^{\circ}\right)=-\frac{2}{3}$ and $270^{\circ}<\alpha<360^{\circ}$. | Solution.
$$
\begin{aligned}
& \sin \left(\alpha-90^{\circ}\right)=-\frac{2}{3}, \quad-\sin \left(90^{\circ}-\alpha\right)=-\frac{2}{3}, \quad \sin \left(90^{\circ}-\alpha\right)=\frac{2}{3} \\
& \cos \alpha=\frac{2}{3}, \quad \cos ^{2} \alpha=\frac{4}{9} \\
& 1-\sin ^{2} \alpha=\frac{4}{9}, \quad \sin ^{2} \alpha=\fr... | \frac{\sqrt{5}}{20} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,632 |
3.173. Prove that if $\alpha$ and $\beta$ satisfy the inequalities $0<\alpha<\frac{\pi}{2}, 0<\beta<\frac{\pi}{2}$ and $\cos \alpha=\frac{7}{\sqrt{50}}, \operatorname{tg} \beta=\frac{1}{3}$, then $\alpha+2 \beta=\frac{\pi}{4}$. | Solution.
$$
\begin{aligned}
& \operatorname{tg}(\alpha+2 \beta)=\frac{\operatorname{tg} \alpha+\operatorname{tg} 2 \beta}{1-\operatorname{tg} \alpha \operatorname{tg} 2 \beta}=\frac{\frac{\sin \alpha}{\cos \alpha}+\frac{2 \operatorname{tg} \beta}{1-\operatorname{tg}^{2} \beta}}{1-\frac{\sin \alpha}{\cos \alpha} \cdot... | proof | Algebra | proof | Yes | Yes | olympiads | false | 50,633 |
3.174. Find $\operatorname{tg} 2 \alpha$, if it is known that $\cos \left(\alpha-90^{\circ}\right)=0.2$ and $90^{\circ}<\alpha<180^{\circ}$. | ## Solution.
$$
\begin{aligned}
& \cos \left(\alpha-90^{\circ}\right)=\cos \left(90^{\circ}-\alpha\right)=\sin \alpha=0.2, \sin ^{2} \alpha=0.04 \\
& 1-\cos ^{2} \alpha=0.04, \cos ^{2} \alpha=0.96=\frac{24}{25}
\end{aligned}
$$
$$
\begin{aligned}
& \cos \alpha=-\sqrt{\frac{24}{25}}=-\frac{2 \sqrt{6}}{5} \text { when ... | -\frac{4\sqrt{6}}{23} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,634 |
3.175. Prove that if $\alpha$ and $\beta$ satisfy the inequalities
$$
0 \leq \alpha \leq \frac{\pi}{2}, 0 \leq \beta \leq \frac{\pi}{2} \text { and } \operatorname{tg} \alpha=5, \operatorname{ctg} \beta=\frac{2}{3} \text {, then } \alpha+\beta=\frac{3 \pi}{4} \text {. }
$$ | ## Solution.
$$
\operatorname{ctg} \beta=\frac{2}{3}, \frac{\cos \beta}{\sin \beta}=\frac{2}{3}, \frac{\cos ^{2} \beta}{\sin ^{2} \beta}=\frac{4}{9}, \frac{1-\sin ^{2} \beta}{\sin ^{2} \beta}=\frac{4}{9}, \sin ^{2} \beta=\frac{9}{13},
$$
from which, for $0 \leq \beta \leq \frac{\pi}{2}$, we have
$$
\sin \beta=\frac{... | \alpha+\beta=\frac{3\pi}{4} | Algebra | proof | Yes | Yes | olympiads | false | 50,635 |
3.176. Given: $\operatorname{ctg} \alpha=4, \operatorname{ctg} \beta=\frac{5}{3}, 0<\alpha<\frac{\pi}{2}, 0<\beta<\frac{\pi}{2}$. Find $\alpha+\beta$. | Solution.
$$
\operatorname{ctg} \alpha=4, \frac{\cos \alpha}{\sin \alpha}=4, \frac{\cos ^{2} \alpha}{\sin ^{2} \alpha}=16, \frac{1-\sin ^{2} \alpha}{\sin ^{2} \alpha}=16
$$
from which $\sin ^{2} \alpha=\frac{1}{17}$, hence for $0<\alpha<\frac{\pi}{2}$ we have
$$
\sin \alpha=\frac{1}{\sqrt{17}} ; \operatorname{ctg} \... | \alpha+\beta=\frac{\pi}{4} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,636 |
3.177. Calculate $(1+\operatorname{ctg} \alpha)(1+\operatorname{ctg} \beta)$, if $\alpha+\beta=\frac{3 \pi}{4}$. | Solution.
$$
\begin{aligned}
& (1+\operatorname{ctg} \alpha)(1+\operatorname{ctg} \beta)=\left(1+\frac{\cos \alpha}{\sin \alpha}\right)\left(1+\frac{\cos \beta}{\sin \beta}\right)=\frac{\sin \alpha+\cos \alpha}{\sin \alpha} \times \\
& \times \frac{\sin \beta+\cos \beta}{\sin \beta}=\frac{\cos \alpha \cos \beta+\sin \... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,637 |
3.178. Calculate $(1+\operatorname{tg} \alpha)(1+\operatorname{tg} \beta)$, if $\alpha+\beta=\frac{\pi}{4}$. | Solution.
$$
(1+\operatorname{tg} \alpha)(1+\operatorname{tg} \beta)=\left(1+\frac{\sin \alpha}{\cos \alpha}\right)\left(1+\frac{\sin \beta}{\cos \beta}\right)=\frac{\cos \alpha+\sin \alpha}{\cos \alpha} \times
$$
$$
\begin{aligned}
& \times \frac{\cos \beta+\sin \beta}{\cos \beta}=\frac{\cos \alpha \cos \beta+\sin \... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,638 |
3.180. Show that the expression $\frac{\sin \alpha+\operatorname{tg} \alpha}{\cos \alpha+\operatorname{ctg} \alpha}$ is non-negative in its domain of definition. | Solution.
$$
\frac{\sin \alpha+\tan \alpha}{\cos \alpha+\cot \alpha}=\frac{\sin \alpha+\tan \alpha}{\cos \alpha+\frac{1}{\tan \alpha}}=
$$
$$
\begin{aligned}
& =\frac{\frac{2 \tan \frac{\alpha}{2}}{1+\tan^{2} \frac{\alpha}{2}}+\frac{2 \tan \frac{\alpha}{2}}{1-\tan^{2} \frac{\alpha}{2}}}{\frac{1-\tan^{2} \frac{\alpha}... | proof | Algebra | proof | Yes | Yes | olympiads | false | 50,639 |
3.182. Prove that $\cos 2 - \cos 8 < 0$. | ## Solution.
$\cos 2-\cos 8=-2 \sin 5 \sin (-3)=2 \sin 5 \sin 3$.
Since $\frac{3 \pi}{2} < 5 < \frac{7 \pi}{4}$, we have $\sin 5 < 0$. Also, $\sin 3 > 0$. Therefore, $2 \sin 5 \sin 3 < 0$ and $\cos 2-\cos 8 < 0$, which is what we needed to prove. | proof | Inequalities | proof | Yes | Yes | olympiads | false | 50,641 |
3.183. The quantities $\alpha, \beta, \gamma$ in the given order form an arithmetic progression. Prove that $\frac{\sin \alpha-\sin \gamma}{\cos \gamma-\cos \alpha}=\operatorname{ctg} \beta$. | ## Solution.
According to the property of the terms of an arithmetic progression
$$
a_{k}=\frac{a_{k-1}+a_{k+1}}{2}, k=2,3, \ldots, n-1
$$
therefore
$$
\beta=\frac{\alpha+\gamma}{2} \text{.}
$$
Then
$$
\frac{\sin \alpha-\sin \gamma}{\cos \gamma-\cos \alpha}=\frac{2 \cos \frac{\alpha+\gamma}{2} \sin \frac{\alpha-\... | proof | Algebra | proof | Yes | Yes | olympiads | false | 50,642 |
3.184. 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. | Solution.
$$
\frac{5}{1+\sqrt[3]{32 \cos ^{4} 15^{\circ}-10-8 \sqrt{3}}}=\frac{5}{1+\sqrt[3]{32\left(\cos ^{2} 15^{\circ}\right)^{2}-10-8 \sqrt{3}}}=
$$
$$
\begin{aligned}
& =\frac{5}{1+\sqrt[3]{32\left(\frac{1+\cos 30^{\circ}}{2}\right)^{2}-10-8 \sqrt{3}}}=\frac{5}{1+\sqrt[3]{8\left(1+\frac{\sqrt{3}}{2}\right)^{2}-1... | 1-\sqrt[3]{4}+\sqrt[3]{16} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,643 |
4.002. The sum of the first and fifth terms of an arithmetic progression is 5/3, and the product of the third and fourth terms is 65/72. Find the sum of the first 17 terms of this progression. | Solution.
We have $\left\{\begin{array}{l}a_{1}+a_{5}=\frac{5}{3}, \\ a_{3} \cdot a_{4}=\frac{65}{72} .\end{array}\right.$
Using formula (4.1), we find
$$
\begin{aligned}
& \left\{\begin{array} { l }
{ a _ { 1 } + a _ { 1 } + 4 d = \frac { 5 } { 3 } , } \\
{ ( a _ { 1 } + 2 d ) ( a _ { 1 } + 3 d ) = \frac { 6 5 } {... | \frac{119}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,644 |
4.003. In a shooting competition, for each miss in a series of 25 shots, the shooter received penalty points: for the first miss - one penalty point, and for each subsequent miss - half a point more than for the previous one. How many times did the shooter hit the target, having received 7 penalty points? | Solution.
Let $a_{1}=1$ be the first term of the arithmetic progression, $d=\frac{1}{2}$ be its common difference, and $S_{n}=7$ be the sum of the first $n$ terms of this progression, where $n$ is the number of terms. Using formula (4.5), we have
$$
\frac{2+(n-1) \cdot \frac{1}{2}}{2} \cdot n=7, n^{2}+3 n-28=0
$$
fr... | 21 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,645 |
4.004. Find the first three terms $a_{1}, a_{2}, a_{3}$ of an arithmetic progression, given that $a_{1}+a_{3}+a_{5}=-12$ and $a_{1} a_{3} a_{5}=80$. | Solution.
From the condition we have $\left\{\begin{array}{l}a_{1}+a_{3}+a_{5}=-12, \\ a_{1} \cdot a_{3} \cdot a_{5}=80 .\end{array}\right.$
Using formula (4.1), we get
$$
\begin{aligned}
& \left\{\begin{array}{l}
a_{1}+a_{1}+2 d+a_{1}+4 d=-12, \\
a_{1}\left(a_{1}+2 d\right)\left(a_{1}+4 d\right)=80
\end{array} \Lef... | 2,-1,-4;-10,-7,-4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 50,646 |
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