Question

Найти значение: 1)sin2(не квадрат)альфа,если 90градусов

Answer 1

$1)\; \; sin \alpha =\frac{3}{5}\; ,\ \; 90\ \textless \ \alpha \ \textless \ 180;(cos \alpha \ \textless \ 0)\\\\sin2 \alpha =2sin \alpha \cdot cos \alpha =2sin \alpha \cdot (-\sqrt{1-sin^2 \alpha })=\\\\=-2\cdot \frac{3}{5}\cdot \sqrt{1-\frac{9}{25}}=-\frac{6}{5}\cdot \frac{4}{5}=-\frac{24}{25}\\\\2)\; \; sin \alpha =\sqrt2-1\\\\cos2 \alpha =cos^2 \alpha -sin^2 \alpha =(1-sin^2 \alpha )-sin^2 \alpha =\\\\=1-2sin^2 \alpha =1-2\cdot (\sqrt2-1)^2=1-2\cdot (3-2\sqrt2)=\\\\=1-6+4\sqrt2=4\sqrt2-5$

$3)\; \; \frac{sin \alpha +tg \alpha }{1+cos \alpha }= \frac{sin \alpha +\frac{sin \alpha }{cos \alpha }}{1+cos \alpha } = \frac{sin \alpha \cdot cos \alpha +sin \alpha }{cos \alpha (1+cos \alpha )} =\\\\= \frac{sin \alpha \cdot (cos \alpha +1)}{cos \alpha (1+cos \alpha )} =\frac{sin \alpha }{cos \alpha }=tg \alpha$

Answer 2

1
cosa=-√(1-sin²a)=-√(1-25/169)=-√(144/169)=-12/13
sin2a=2sinacosa=2*5/13*(-12/13)=-120/169
2
cos2a=1-2sin²a=1-2(√2-1)²=1-4+4√2-2=4√2-5
3
(sina+sina/cosa):(1+cosa)=(sinacosa+sina)/cosa:(1+cosa)=
sina(cosa+1)/[cosa(1+cosa)]=sina/cosa=tga
tga=tga