Question

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

Answer 1

$3); ; sqrt{32}cdot sinfrac{33pi}{4}cdot cosfrac{34pi}{3}=sqrt{2^5}cdot sin(8pi +frac{pi}{4})cdot cos(11pi +frac{pi}{3})=\\=2^2sqrt2cdot sinfrac{pi}{4}cdot (-cosfrac{pi}{3})=-4sqrt2cdot frac{sqrt2}{2}cdot frac{1}{2}=-2\\4); ; frac{1-2cos^2111}{10cos^2114cdot tg66} = frac{1-(1+cos222)}{10cdot cos^2(90+24)cdot tg(90-24)} = frac{-cos(180+42)}{10cdot sin^224cdot ctg24} =$

$= frac{-(-cos42)}{10cdot sin^224cdot frac{cos42}{sin42}} = frac{cos42}{10sin24cdot cos24}=frac{cos42}{5sin48} =$

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

$5); ; sqrt6cdot frac{sin20cdot cos40+sin110cdot sin40}{sin10cdot sin35-sin100cdot cos35} =\\=sqrt6cdot frac{sin20cdot cos40+sin(90+20)cdot sin40}{sin10cdot sin35-sin(90+10)cdot cos35} =sqrt6cdot frac{sin20cdot cos40+cos20cdot sin40}{sin10cdot sin35-cos10cdot cos35} =\\=sqrt6cdot frac{sin(20+40)}{-cos(10+35)} =-sqrt6cdot frac{sin60}{cos45} =-sqrt6cdot frac{sqrt3/2}{sqrt2/2} =\\=-frac{sqrt6cdot sqrt3}{sqrt2}=-frac{sqrt{2cdot 3}cdot sqrt3}{sqrt2} =$

$=-frac{sqrt2cdot sqrt3cdot sqrt3}{sqrt2}=-sqrt3cdot sqrt3=-3$