Question

Решите пожалуйста!!!!!!!! Нужно срочно сверить ответ!!!!!!!!!!!!!

sin( 7π/2- 2α)cos( ( 6π)/2-α) sinα; α=3π/16;

Answer 1

$sin(frac{7pi}{2}-2a)cos(frac{6pi}{2}-a)sina = sin(2pi+frac{3pi}{2}-2a)cos(2pi+pi-a)sina = sin(frac{3pi}{2}-2a)cos(pi-a)sina = -cos2a*(-cosa)sina = sina*cosa*cos2a = frac{1}{2}sin2a*cos2a = frac{1}{4}sin4a$

a = 3Pi/16

$frac{1}{4}sin4a = frac{1}{4}sinfrac{3pi*4}{16} = frac{1}{4}sinfrac{3pi}{4} = frac{1}{4}sin(pi-frac{pi}{4}) = frac{1}{4}sinfrac{pi}{4} = frac{1}{4}*frac{sqrt{2}}{2}=frac{sqrt{2}}{8}$