Question

Докажите тригонометрическое тождество

Answer 1

$\frac{sin^2(\frac{3\pi}{2}+ \alpha )}{ctg^2( \alpha -2\pi )}+\frac{sin^2(- \alpha )}{ctg^2( \alpha -\frac{3\pi}{2})}=\frac{(-cos \alpha )^2}{ctg^2 \alpha }+\frac{sin^2 \alpha }{(-tg \alpha )^2}=\\\\=\frac{cos^2 \alpha \cdot sin^2 \alpha }{cos^2 \alpha }+\frac{sin^2 \alpha \cdot cos^2 \alpha }{sin^2 \alpha }=sin^2 \alpha +cos^2 \alpha =1\\\\\\P.S.\; sin(\frac{3\pi}{2}+ \alpha )=-cos \alpha \\\\ctg( \alpha -2\pi )=-ctg(2\pi- \alpha )=-ctg(- \alpha )=ctg \alpha \\\\sin(- \alpha )=-sin \alpha$

$ctg( \alpha -\frac{3\pi}{2})=-ctg(\frac{3\pi}{2}- \alpha )=-tg \alpha$