455(1,2,3) помогите пожалуйста )))

Question 1

Question 2

Решите задачу:

$1)\quad \star \; \; sin^2a=\frac{1}{2}(1-cos2a)\; \; \star$

$sin^2(\frac{\pi}{3}+x)+sin^2(\frac{\pi}{3}-x)+sin^2x=\\\\=\frac{1}{2}(1-cos(\frac{2\pi}{3}+2x))+\frac{1}{2}(1-cos(\frac{2\pi}{3}-2x)) +\frac{1}{2}(1-cos2x)=\\\\=\frac{3}{2}- \frac{1}{2} cos(\pi -\frac{\pi}{3}+2x)- \frac{1}{2} cos(\pi-\frac{\pi}{3}-2x)- \frac{1}{2} cos2x=\\\\=\frac{3}{2}- \frac{1}{2} cos(\pi -(\frac{\pi }{3}-2x))- \frac{1}{2} cos(\pi -(\frac{\pi}{3}+2x))- \frac{1}{2}cos2x =\\\\= \frac{3}{2}- \frac{1}{2}\cdot \Big (-cos(\frac{\pi}{3}-2x)-cos(\frac{\pi}{3}+2x)\Big )-\frac{1}{2}cos2x=$

$= \frac{3}{2}+ \frac{1}{2}\cdot 2cos(-2x)\cdot cos\frac{\pi}{3}-\frac{1}{2}cos2x=\\\\= \frac{3}{2} +cos2x\cdot \frac{1}{2}-\frac{1}{2}cos2x= \frac{3}{2} \\\\2)\quad \star \; \; cos^2 \alpha =\frac{1}{2}(1+cos2 \alpha )\; \; \star \\\\cos^2x+cos^2( \frac{2\pi}{3}-x)+cos^2(\frac{2\pi }{3}+x)=\\\\=\frac{1}{2}(1+cos2x)+ \frac{1}{2}(1+cos(\frac{4\pi }{3}-2x))+\frac{1}{2}(1+cos(\frac{4\pi}{3}+2x))=\\\\=\frac{3}{2}+ \frac{1}{2}cos2x+cos(\pi +(\frac{\pi}{3}-2x))+cos(\pi +(\frac{\pi}{3}+2x))=$

$= \frac{3}{2}+\frac{1}{2}cos2x -cos( \frac{\pi}{3}-2x)-cos(\frac{\pi}{3}+2x)=\\\\= \frac{3}{2}+ \frac{1}{2}cos2x-2\cdot cos\frac{\pi}{3}\cdot cos(-2x)=\\\\=\frac{3}{2}+ \frac{1}{2}cos2x-2\cdot \frac{1}{2}\cdot cos2x= \frac{3}{2} \\\\3)\; \; cos(x-y)\cdot (tgx\cdot tgy-1)+(1+tgx\cdot tgy)\cdot cos(x+y)=\\\\=cos(x-y)\cdot ( \frac{sinx}{cosx} \cdot \frac{siny}{cosy} -1)+(1+ \frac{sinx}{cosx} \cdot \frac{siny}{cosy} )\cdot cos(x+y)=$

$=cos(x-y)\cdot \frac{sinx\cdot siny-cosx\cdot cosy}{cosx\cdot cosy} +\frac{cosx\cdot cosy+sinx\cdot siny}{cosx\cdot cosy} \cdot cos(x+y)=$

$=cos(x-y)\cdot \Big (-\frac{cos(x+y)}{cosx\cdot cosy}\; \Big )+\frac{cos(x-y)}{cosx\cdot cosy} \cdot cos(x+y)=0$