Question

cos^2(π/8)+cos^2(3π/8)+cos^2(5π/8)+cos^2(7π/8)=? plz solve it s urgent

Answer 1

cos^2(π/8) + cos^2(3π/8) + cos^2(5π/8) + cos^2(7π/8) 
= cos^2(π/8) + cos^2(π/2 - π/8) + cos^2(5π/8) + cos^2(π/2 - (-3π/8)) 
= cos^2(π/8) + sin^2(π/8) + cos^2(5π/8) + sin^2(-3π/8) 
= cos^2(π/8) + sin^2(π/8) + cos^2(5π/8) + sin^2(5π/8) -----> sin(-3π/8) = -sin(5π/8) but after squaring it becomes positive anyway 
= 1 + 1 
= 2

Anonymous

Answer 2

we have to find the value of cos²(π/8) + cos²(3π/8) + cos²(5π/8) + cos²(7π/8)

       we know, cos(π/2 - Ф) = sinФ

  so, cos(5π/8) = cos(π/2 + π/8) = -sin(π/8)

       cos(7π/8) = cos(π/2 + 3π/8) = -sin(3π/8)

so, cos²(π/8) + cos²(3π/8) + [-sin(π/8) ]² + [-sin(3π/8) ]²

= cos²(π/8) + cos²(3π/8) + sin²(π/8) + sin²(3π/8)

= [cos²(π/8) + sin²(π/8) ] + [cos²(3π/8) + sin²(3π/8) ]

= 1 + 1 [ we know sin²Ф + cos²Ф = 1 ]

= 2

therefore, cos²(π/8) + cos²(3π/8) + cos²(5π/8) + cos²(7π/8)