Question

Помогите решить подробно 1)cos 3п/8*cosп/8-sin3п/8*sinп/8 всё разделить на tg(п/4+B)
2)sin²(x+y)+sin²(x-y)/2cos²xcos²y=tg²x+tg²y

Answer 1

$1)\; \; \frac{cos\frac{3\pi }{8}\cdot cos\frac{\pi}{8}-sin \frac{3\pi}{8}\cdot sin\frac{\pi}{8}}{tg(\frac{\pi}{4}+\beta)}=\frac{cos(\frac{3\pi}{8}+\frac{\pi }{8})}{tg(\frac{\pi}{4}+\beta)}=\frac{cos\frac{\pi}{2}}{tg(\frac{\pi}{4}+\beta )}=\frac{0}{tg(\frac{\pi}{4}+\beta )}=0\\\\2)\; \; \frac{sin^2(x+y)+sin^2(x-y)}{2cos^2x\cdot cos^2y}=\\\\=\frac{(sinx\, cosy+cosx\, siny)^2+(sinx\, cosy-cosx\, siny)^2}{2cos^2x\, cos^2y}=\\\\=\frac{sin^2x\, cos^2y+cos^2x\, sin^2y+sin^2x\, cos^2y+cos^2x\, sin^2y}{2cos^2x\, cos^2y}=\\\\=\frac{2sin^2x\, cos^2y}{2cos^2x\, cos^2y}+\frac{2cos^2x\, sin^2y}{2cos^2x\, cos^2y}=\frac{sin^2x}{cos^2x}+\frac{sin^2y}{cos^2y}=tg^2x+tg^2y$