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`        .if x+270[cos2Π/7+cos4Π/7+cos6Π/7=3746,then find the value of x. `
4 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.
```
4 years ago
```							Multiply both sides with sin(π7) hence we have thatsin(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)2We'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.
```
one year ago
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