Question

Помогите решить №3, 4 и 5. Ответы : №3: -2 №4: 0,2 №5 : -3 нужен процесс решения.

Answer 1

$3)\; \; \sqrt{32}\cdot sin\frac{33\pi}{4}\cdot cos\frac{34\pi}{3}=\sqrt{2^5}\cdot sin(8\pi +\frac{\pi}{4})\cdot cos(11\pi +\frac{\pi}{3})=\\\\=2^2\sqrt2\cdot sin\frac{\pi}{4}\cdot (-cos\frac{\pi}{3})=-4\sqrt2\cdot \frac{\sqrt2}{2}\cdot \frac{1}{2}=-2\\\\4)\; \; \frac{1-2cos^2111}{10cos^2114\cdot tg66} = \frac{1-(1+cos222)}{10\cdot cos^2(90+24)\cdot tg(90-24)} = \frac{-cos(180+42)}{10\cdot sin^224\cdot ctg24} =$

$= \frac{-(-cos42)}{10\cdot sin^224\cdot \frac{cos42}{sin42}} = \frac{cos42}{10sin24\cdot cos24}=\frac{cos42}{5sin48} =$

$\frac{cos(90-48)}{5sin48} = \frac{sin48}{5sin48} = \frac{1}{5}=0,2$

$5)\; \; \sqrt6\cdot \frac{sin20\cdot cos40+sin110\cdot sin40}{sin10\cdot sin35-sin100\cdot cos35} =\\\\=\sqrt6\cdot \frac{sin20\cdot cos40+sin(90+20)\cdot sin40}{sin10\cdot sin35-sin(90+10)\cdot cos35} =\sqrt6\cdot \frac{sin20\cdot cos40+cos20\cdot sin40}{sin10\cdot sin35-cos10\cdot cos35} =\\\\=\sqrt6\cdot \frac{sin(20+40)}{-cos(10+35)} =-\sqrt6\cdot \frac{sin60}{cos45} =-\sqrt6\cdot \frac{\sqrt3/2}{\sqrt2/2} =\\\\=-\frac{\sqrt6\cdot \sqrt3}{\sqrt2}=-\frac{\sqrt{2\cdot 3}\cdot \sqrt3}{\sqrt2} =$

$=-\frac{\sqrt2\cdot \sqrt3\cdot \sqrt3}{\sqrt2}=-\sqrt3\cdot \sqrt3=-3$