[tex] \frac{1-8cos^2 \frac{ \alpha }{2}*sin^2 \frac{ \alpha }{2} }{cos3 \alpha +cos \alpha }*(2cos^2 \frac{ \alpha }{2}-1)[/tex]. Created by Ладушки123. algebra-ru

tex]...

hote

Verified answer

$\frac{1-8Cos^2 \frac{x}{2}*Sin^2 \frac{x}{2}}{Cos3x+Cosx}*(2Cos^2 \frac{x}{2}-1)=$

$= \frac{1-2(2Cos \frac{x}{2}Sin \frac{x}{2})^2}{4Cos^3x-3Cosx+Cosx}*(2( \frac{1+Cosx}{2})-1))=$

$= \frac{1-2Sin^2x}{2Cosx*(2Cos^2x-1)}*Cosx=$

$= \frac{1-2Sin^2x}{2(2Cos^2x-1)}= \frac{Cos2x}{2(2* \frac{1+Cos2x}{2}-1)}=$

$= \frac{Cos2x}{2(1+Cos2x-1)}= \frac{Cos2x}{2(Cos2x)}= \frac{1}{2}$

2 votes Thanks 2

m11m

Verified answer

$\frac{1-2(2sin \frac{ \alpha }{2}cos \frac{ \alpha}{2})^2}{2cos \frac{3 \alpha+ \alpha}{2}cos \frac{3 \alpha- \alpha}{2}}*(2cos^2 \frac{ \alpha}{2}-cos^2 \frac{ \alpha}{2}-sin^2 \frac{ \alpha}{2})= \\ \\ 
= \frac{1-2(sin(2* \frac{ \alpha}{2}))^2}{2cos2 \alpha*cos \alpha}*(cos^2 \frac{ \alpha}{2}-sin^2 \frac{ \alpha}{2})= \\ \\ 
= \frac{1-2sin^2 \alpha}{2cos2 \alpha*cos \alpha}*cos(2* \frac{ \alpha}{2})= \\ \\ 
= \frac{cos^2 \alpha+sin^2 \alpha-2sin^2 \alpha}{2cos2 \alpha}= \\ \\ 
$
$= \frac{cos^2 \alpha-sin^2 \alpha}{2cos2 \alpha}= \frac{cos2 \alpha}{2cos2 \alpha}= \frac{1}{2}$

2 votes Thanks 2

Answers & Comments

Verified answer

Verified answer

рекомендуемые вопросы

Helpful Links

Helpful Social