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Question

Prove that:

sin2 π8+sin23π8+sin25π8+sin27π8=2


Solution

LHS,

=sin2 π8+sin23π8+sin25π8+sin27π8=sin2 π8+sin2(π2π8)+sin25π8+sin2(ππ8)=sin2π8+cos2π8+sin2(π3π8)+sin2π8=1+sin23π8+sin2π8=1+sin2(π2π8)+sin2π8=1+cos2π8+sin2π8=1+1=2

= RHS

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