Question

ПОМОГИТЕ СРОЧНО
№1 найти sin a
№2 найти sin2a

Answer 1

Объяснение:

1.

$\displaystyle\\sin2\alpha =-\frac{\sqrt{3} }{2} \ \ \ \ \ \ \frac{3\pi }{4} < \alpha < \pi \ \ \ \ \ \ sin\alpha =?\\\\(sin2\alpha )^2=(-\frac{\sqrt{3} }{2})^2\\\\sin^22\alpha =\frac{3}{4} \\\\2^2*sin^2\alpha *cos^2\alpha =\frac{3}{4} \\\\4*sin^2\alpha *(1-sin^2\alpha )=\frac{3}{4} \ |*4\\\\16*sin^2\alpha -16*sin^2\alpha =3\\\\16sin^4\alpha -16sin^2\alpha +3=0$

Пусть sin²α=t        ⇒

$\displaystyle\\16t^2-16t+3=0\\\\16t^2-4t-12t+3=0\\\\4t*(4t-1)-3*(4t-1)=0\\\\(4t-1)*(4t-3)=0\\\\4t-1=0\\\\4t=1\ |:4\\\\t=sin^2\alpha =\frac{1}{4} \ \ \ \ \ \ \Rightarrow\\\\sin\alpha =б\frac{1}{2} \ \ \ \ \\\\\frac{3\pi }{4} < \alpha < \pi \ \ \ \ \ \ \Rightarrow\\\\sin\alpha =\frac{1}{2}.\\\\4t-3=0\\\\4t=3\ |:4\\\\t=sin^2\alpha =\frac{3 }{4} \\sin\alpha =б\frac{\sqrt{3} }{2} \\\frac{3\pi }{4} < \alpha < \pi \ \ \ \ \ \ \Rightarrow\\\\sin\alpha =\frac{\sqrt{3} }{2} .$

Ответ: sinα=1/2,   sinα=√3/2.

2.

$\displaystyle\\sin\alpha +cos\alpha =\frac{5}{4} \ \ \ \ \ \ sin2\alpha =? \\\\(sin\alpha +cos\alpha )^2=(\frac{5}{4})^2 \\\\sin^2\alpha +2*sin\alpha *cos\alpha +cos^2\alpha =\frac{25}{16} =1\frac{9}{16} \\\\1+sin2\alpha =1\frac{9}{16} \\\\sin2\alpha =\frac{9}{16} .$

Ответ: sin2α=9/16.