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742k
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