Question

Show that cos^2 pi/8 + cos^2 3pi/8 + cos^2 5 pi/8 + cos^2 7pi/8 = 2 by brainly.in

Answer 1

We have,

• sin²x + cos²x = 1

• cos((π/2) + x) = - sin x => cos²((π/2) + x) = sin²x

So,

LHS

=> cos²(π/8) + cos²(3π/8) + cos²(5π/8) + cos²(7π/8)

=> cos²(π/8) + cos²(5π/8) + cos²(3π/8) + cos^2(7π/8)

=> cos²(π/8) + cos²((4π + π)/8) + cos²(3π/8) + cos²((4π + 3π)/8)

=> cos²(π/8) + cos²((π/2) + (π/8)) + cos²(3π/8) + cos²((π/2) + (3π/8))

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

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

=> 1 + 1

=> 2

=> RHS

Hence Proved!

Answer 2

Answer:

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