Question

Номер 3 и 4 пожалуйста!

Answer 1

$3)\; \; \frac{1+sin2a-cos2a}{11+sin2a+cos2a}=\frac{(1-cos2a)+sin2a}{(1+cos2a)+sin2a}=\frac{2sin^2a+2sina\cdot cosa}{2cos^2a+2sina\cdot cosa}=\\\\=\frac{2sina(sina+cosa)}{2cosa(cosa+sina)}=\frac{sina}{cosa}=tga$

$4)\; \; 4sin\frac{\pi }{24}\cdot cos\frac{\pi}{24}\cdot sin\frac{5\pi}{12}=2\cdot sin\frac{2\pi}{24}\cdot sin\frac{5\pi}{12}=\\\\=2\cdot sin\frac{\pi}{12}\cdot sin\frac{5\pi}{12}=2\cdot \frac{1}{2}\cdot (cos\frac{4\pi}{12}-cos\frac{6\pi}{12})=\\\\=cos\frac{\pi}{3}-cos\frac{\pi }{2}=\frac{1}{2}-0=\frac{1}{2}$

$\star \; \; sina\cdot cosa=\frac{1}{2}\cdot sin2a\; \; \star$