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Question

Evaluate:sin2π10+sin22π5+sin23π5+sin29π10

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Solution

Given, sin2π10+sin22π5+sin23π5+sin29π10

=sin2π10+sin22π5+sin2(π2π5)+sin2(ππ10)

=sin2π10+sin22π5+sin22π5+sin2π10 since sin(πθ)=sinθ

=2[sin2π10+sin22π5]

=2[cos2(π2π10)+sin22π5]

=2[cos25ππ10+sin22π5]

=2[cos24π10+sin22π5]

=2[cos22π5+sin22π5]

=2×1 since sin2θ+cos2θ=1

=2

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