Question

Надо, чтобы получилось -sin2x

Answer 1

$\frac{sin(x-\pi)cos(x+2\pi)sin(4\pi-x) }{sin(\frac{\pi}{2} +x)ctg(2\pi-x)ctg(\frac{3\pi}{2} +x)} =\frac{-sin(x)cos(x)(-sin(x))}{cos(x)(-ctg(x))(-tg(x))} =\frac{sin^2(x)}{ctg(x)*tg(x)}=sin^2(x)$

Answer 2

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