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`        2cosπ/13 cos9π/13+cos3π/13+cos5π/13= Please explain fully And in detail manner `
one year ago

```							2cos(π/13).cos(9π/13)+cos(3π/13).cos(5π/13) = 0 use the formula, 2cosC.cosD = cos(C+D)+cos(C-D)LHS = cos(π/13+9π/13)+cos(9π/13-π/13) + cos(π/13).cos(5π/13) = cos(10π/13) + cos(8π/13) + cos(3π/13).cos(5π/13) = {cos(10π/13)+cos(3π/13)}+{cos(8π/13) + cos(5π/13)}use formula, cosC + cosD = 2cos(C+D)/2.cos(C-D)/2= {2cos(π/2).cos(7π/26)}+ {2cos(π/2).cos(3π/26)} we know, cos(π/2) = 0 = 0 + 0 = RHS
```
one year ago
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