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# .if x+270[cos2Π/7+cos4Π/7+cos6Π/7=3746,then find the value of x. Grade:11

## 2 Answers

5 years ago
try to simplify [cos2Π/7+cos4Π/7+cos6Π/7].
Multiply on top and bottom by sin(Π/7),
[cos2Π/7+cos4Π/7+cos6Π/7] = [ sin(3Π/7) – sin(Π/7) + sin(5Π/7) – sin(3Π/7) + sin(7Π/7) – sin(5Π/7)] / (2*sin(Π/7)),
most of terms will get cancelled out and [cos2Π/7+cos4Π/7+cos6Π/7] = ½.
2 years ago

Multiply both sides with sin(π7) hence we have that

sin(π7)cos(2π7)+sin(π7)cos(4π7)+sin(π7)cos(6π7)=12sin(π7)
Then because sinAcosB=12[sin(AB)+sin(A+B)]we rewrite this as

(12)[sin(π72π7)+sin(π7+2π7)]+(12)[sin(π74π7)+sin(π7+4π7)]+(12)[sin(π76π7)+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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