Question

Решите пожалуйста
Решение и ответ !!!!

Answer 1

По формуле приведения  cos(α+β)=cosαcosβ-sinαsinβ
cos(5π/2)=cos(2π+π/2)=cos(π/2)=0
sin(5π/2)=sin(2π+π/2)=sin(π/2)=1 
cos(5π/2 + α)=cos(5π/2)cos(α)-sin(5π/2)sin(α)=0*cosα-1*sinα=-sinα
sinα>0 α⊂(π/2,π)
sin²α+cos²α=1
sin²α+144/169=1
sin²α=25/169
sinα=5/13
-39cos(5π/2+α)=-39*(-5/13)=5*3=15