Question

Sin2π/6+sin²3π/6+sin²5π/6+sin²7π/6

Answer 1

sin²π/6+sin²3π/6+sin²5π/6+sin²7π/6

=sin²π/6+sin²3π/6+sin²(π/2+π/6)+sin²(π/2+3π/6)

=sin²π/6+sin²3π/6+cos²π/6+cos²3π/6

=(sin²π/6+cos²π/6)+(sin²3π/6+cos²3π/6) [∵, sin²Ф+cos²Ф=1]

=1+1

=2 Ans.