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Multiply both sides with sin(π7) hence we have that
sin(π7)cos(2π7)+sin(π7)cos(4π7)+sin(π7)⋅cos(6π7)=−12⋅sin(π7)Then because sinA⋅cosB=12⋅[sin(A−B)+sin(A+B)]we rewrite this as
(12)[sin(π7−2π7)+sin(π7+2π7)]+(12)[sin(π7−4π7)+sin(π7+4π7)]+(12)[sin(π7−6π7)+sin(π7+6π7)]=−sin(π7)2
We'll divide by (12) both sides: −sin(π7)+sin(3π7)−sin(3π7)+sin(5π7)−sin(5π7)+sin(7π7)=−sin(π7)
Cancelling the same terms yields to −sin(π7)+sin(π)=−sin(π7)
But sin(π)=0 hence
−sin(π7)=−sin(π7)
Since we have the same results in both sides, the given identity is true.
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