Question

Докажите тождество:
2cos2x-1 sin3x-sinx 1
------------- + ---------------- = -------
2sinxcosx cos3x+cosx sin2x​

Answer 1

$\frac{2cos^2x-1}{2\, sinx\, cosx}+\frac{sin3x-sinx}{cos3x+cosx}=\frac{cos2x}{sin2x}+\frac{2sinx\cdot cos2x}{2\, cos2x\cdot cosx}=\frac{cos2x}{sin2x}+\frac{sinx}{cosx}=\\\\=\frac{cos2x\cdot cosx+sinx\cdot sin2x}{sin2x\cdot cosx}=\frac{cos(2x-x)}{sin2x\cdot cosx}=\frac{cosx}{sin2x\cdot cosx}=\frac{1}{sin2x}\\\\\\\frac{1}{sin2x}=\frac{1}{sin2x}\\\\\\\star \; \; \; \cos(\alpha -\beta )=cos\alpha \cdot cos\beta +sin\alpha \cdot sin\beta$