ПОМОГИТЕ!!!!!!!!!!!!!!Срочно!!!!!!!!!!!!!!!!!!!!ПЛИЗ!!!!!!!!!!!!!!!!!!!

Question

Алгебра

Флорентий

6 год назад

ПОМОГИТЕ!!!!!!!!!!!!!!Срочно!!!!!!!!!!!!!!!!!!!!ПЛИЗ!!!!!!!!!!!!!!!!!!!

ОТВЕТЫ

Nikodim

Jun 30, 2019

B1. sinα lt;0 при α∈(π/2,π)
sin²α=1-cos²α=1-(-√21/5)²=1-21/25=(25-21)/25=4/25
sinα=-2/5
B2
$28 \sqrt{6}cos \frac{\pi}{6} cos \frac{5\pi}{4} =28 \sqrt{6}cos \frac{\pi}{6} (-cos \frac{\pi}{4})= -28 \sqrt{6}\frac{ \sqrt{3} }{2} \frac{ \sqrt{2} }{2}=-7*6=-42$

B3. -21√3 tg(660°)=-21√3 tg(300°)=-21√3 tg(-60°)=21√3 tg(60°)=21√3 √3 =21*3=63
B4.
$30 \sqrt{6}cos(- \frac{\pi}{4}) sin(- \frac{\pi}{3}) = 30 \sqrt{6}cos(\frac{\pi}{4}) (-sin( \frac{\pi}{3}))= \\ =-30 \sqrt{6}cos(\frac{\pi}{4}) sin( \frac{\pi}{3})=-30 \sqrt{6} \frac{ \sqrt{2} }{2} \frac{ \sqrt{3} }{2} = -15*6/2=-45$
B5.
$\frac{22(sin^272-cos^272)}{cos144}=-\frac{22(cos^272-sin^272)}{cos144}= -22\frac{cos144}{cos144}=-22$
B6.
$\frac{-51 sin385}{sin25}= \frac{-51 sin(360+25)}{sin25}=\frac{-51 sin25}{sin25}=-51$
B7.
$\frac{3sin( \alpha +\pi)-2cos( \frac{\pi}{2}+ \alpha ) }{sin( \alpha -2\pi)}=\frac{3(-sin \alpha)-2(-sin\alpha ) }{sin\alpha} =\frac{-3sin \alpha+2sin\alpha }{sin\alpha}=-1$
B8.
5sin²α +11cos²α = 9
5sin²α + 5cos²α + 6cos²α= 9
5 + 6cos²α= 9
6cos²α= 4
cos²α= 4/6
sin²α=1-cos²α=1-4/6=2/6
tg²α=sin²α/cos²α=(2/6):(4/6)=1/2

B9.
$\frac{4sin \alpha +2cos \alpha }{5sin \alpha -16cos \alpha }=1 \\ 4sin \alpha +2cos \alpha= 5sin \alpha -16cos \alpha$
4sinα+2cosα=5sinα-16cosα
sinα=18cosα
tgα=sinα/cosα=18

B10.
$4sin( \alpha +\pi)+7cos( \frac{\pi}{2}+ \alpha )=-4sin \alpha -7sin \alpha =-11 sin \alpha= \\ =-11*0,25=-2,75$

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