Question

Помгите решить,с подробным решением! Последнее (е) можно не решать,главное первые 2,заранее спасибо!

Answer 1

$a)\ \frac{6cos^234-3}{cos169*cos79}=\frac{3(2cos^234-1)}{\frac{cos(169+79)+cos(169-79)}{2}}= \frac{3*cos(2*34)*2}{cos248+cos90}=\frac{6cos68}{cos(180+68)+0}=\\ =\frac{6cos68}{-cos68}=-6\\ b)\ (cos^2\frac{3\pi}{8}-cos^2\frac{\pi}{8})*cos\frac{39\pi}{4}=(1+cos(2*\frac{3\pi}{8})-(1+cos(2*\frac{\pi}{8}))*\\ *cos(\frac{40\pi}{4}-\frac{\pi}{4})=(1+cos\frac{3\pi}{4}-1-cos\frac{\pi}{4})*cos(10\pi-\frac{\pi}{4})=\\ =(-\frac{\sqrt2}{2}-\frac{\sqrt2}{2})*cos(-\frac{\pi}{4})=(-\sqrt2)*\frac{\sqrt2}{2}=-1$
$c)\ \sqrt6 * \frac{sin20*cos320+sin40*sin430}{sin755*sin890+sin980*sin775}=\\= \sqrt6*\frac{sin20*cos(360-40)+sin(360+(90-20))*sin40}{sin(360*2+35)*sin(360*2+(90+80))+sin(360*2+(90-35))*sin(360*3-100)}=\\ =\sqrt6 * \frac{sin20*cos(-40)+cos(-20)*sin40}{sin35*cos80+cos(35)*sin(-100)}=\\ =\sqrt6 * \frac{sin20*cos40+cos20*sin40}{sin35*cos80+cos35*sin(-180+80)}=\\ =\sqrt6 * \frac{sin(20+40)}{sin35*cos80-cos35*sin80} =\sqrt6 * \frac{sin60}{sin(35-80)}=\sqrt6 * \frac{sin60}{-sin45}=$
$=\sqrt6 * \frac{\frac{\sqrt3}{2}}{-\frac{\sqrt2}{2}}=-\frac{\sqrt{6*3}}{\sqrt2}=-\sqrt{3*3}=-3$