Question

Решите пожалуйста, срочно. Даю 35 баллов

Answer 1

1.

1) да

2) да

3) нет

2.

$1)sin(\alpha +\beta )-sin\beta cos\alpha =sin\alpha cos\beta +cos\alpha sin\beta -sin\beta cos\alpha =sin\alpha cos\beta$

$2)\sqrt{2} cos(\frac{\pi }{4} -\alpha )-sin\alpha =\sqrt{2} cos\frac{\pi }{4} cos\alpha +\sqrt{2} sin\frac{\pi }{4} sin\alpha -sin\alpha =\\\\=\sqrt{2}\cdot \frac{\sqrt{2} }{2} cos\alpha +\sqrt{2}\cdot \frac{\sqrt{2} }{2} sin\alpha -sin\alpha =cos\alpha +sin\alpha -sin\alpha =cos\alpha$

$3)sin3xcosx-cos3xsinx=sin(3x-x)=sin2x$

3.

$1)\frac{sin2x}{sinx}+cosx= \frac{2sinx\cdot cosx}{sinx}+cosx=2cosx+cosx=3cosx$

$2)\frac{cos6\alpha }{cos3\alpha -sin3\alpha } =\frac{cos^23\alpha -sin^2\alpha}{cos3\alpha -sin3\alpha } =\frac{(cos3\alpha -sin3\alpha )(cos3\alpha +sin3\alpha) }{cos3\alpha -sin3\alpha } =cos3\alpha +sin3\alpha$

$3)\frac{cos6x-cos2x}{sin2x} =\frac{(-2sin\frac{6x+2x}{2}sin\frac{6x-2x}{2} ) }{sin2x}=\frac{(-2sin4x\cdot sin2x)}{sin2x}=-2sin4x$

4.

$sin^2\alpha +cos^2\alpha =1\Rightarrow cos^2\alpha =1-sin^2\alpha$

$sin \alpha =0,8 \Rightarrow cos^2\alpha =1-0,8^2=1-0,64=0,36 ; ; 0 < \alpha< \frac{\pi }{2} \Rightarrow cos\alpha =0,6\\\\cos\beta =-0,6; \Rightarrow sin^2\beta =1-cos^2\beta =1-(-0,6)^2=1-0,36=0,64; \\\\\frac{\pi }{2} <\beta < \pi \Rightarrow sin\beta =-0,8$

$tg\alpha =\frac{sin\alpha }{cos\alpha } =\frac{0,8}{0,6}=\frac{4}{3}\\\\tg\beta =\frac{sin\beta }{cos\beta } =\frac{(-0,8)}{(-0,6)}=\frac{4}{3} \\\\tg (\alpha -\beta )=\frac{tg\alpha -tg\beta }{1+tg\alpha \cdot tg\beta } =0$

5.

$2sin20^{o}\cdot sin100^{o}+cos100^{o}=\\\\=2\cdot \frac{cos(20^{o}-100^{o})-cos(20^{o}+100^{o})}{2}+cos100^{o}=\\\\=cos(-80^{o})-cos120^{o} +cos100^{o}=\\\\=cos80^{o}+cos100^{o}-cos120^{o} =\\\\=2cos\frac{80^{o}+100^{o}}{2} cos\frac{80^{o}-100^{0}}{2}-cos120^{0}=0-cos(180^{o}-60^{o})=cos60^{o}= \frac{1}{2}$