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Question -

Prove that: sin2 π/18 + sin2 π/9 + sin2 7π/18 + sin2 4π/9 = 2



Answer -

Let us consider LHS:

sin2 π/18 + sin2 π/9 +sin2 7π/18 + sin2 4π/9

sin2 π/18 + sin2 2π/18+ sin2 7π/18 + sin2 8π/18

sin2 π/18 + sin2 2π/18+ sin2 (π/2 – 2π/18) + sin2 (π/2 – π/18)

We know that when n is odd, sin → cos.

sin2 π/18 + sin2 2π/18+ cos2 2π/18 + cos2 2π/18

when rearranged,

sin2 π/18 + cos2 2π/18+ sin2 π/18 + cos2 2π/18

We know that sin2 + cos2x =1.

So,

1 + 1

2 = RHS

LHS = RHS

Hence proved.

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