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Question

Prove that: sin2π18+sin2π9+sin27π18+sin24π9=2


Solution

LHS=sin2π18+sin2π9+sin27π18+sin24π9

=sin2π18+sin24π9+sin2π9+sin27π18

=sin2(π24π9)+sin24π9+sin2π9+sin2(π2π9)[π18=π24π9 and 7π18=π2π9]

=cos24π9+sin24π9+sin2π9+cos2π9

(sin(π2θ)=cosθ)=1+1[sin2θ+cos2θ=1]

=2 =RHS Hence proved.


Mathematics
RD Sharma
Standard XI

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