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in Trigonometry by (52.1k points)

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

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Let us consider the 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)

As we know that when n is odd, sin → cos.

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

Let us rearrange this

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

As we know that sin2 + cos2x = 1.

Therefore,

1 + 1

2 = RHS

∴ LHS = RHS
Thus, proved.

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