Question

Надо, чтобы получилось -sin2x

Answer 1

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

Answer 2

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