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`        Show that cos2π/15. cos4π/15.cos8π/15.cos16π/15=1/16`
3 years ago

```							cos(2π/15) cos(4π/15) cos(8π/15) cos(14π/15) = cos(2π/15) cos(4π/15) cos(8π/15) cos(π - π/15) = cos(2π/15) cos(4π/15) cos(8π/15) * -cos(π/15) = -cos(π/15) cos(2π/15) cos(4π/15) cos(8π/15) = -16sin(π/15) cos(π/15) cos(2π/15) cos(4π/15) cos(8π/15) / [ 16 sin(π/15) ] = -8 * [ 2sin(π/15) cos(π/15) ] cos(2π/15) cos(4π/15) cos(8π/15) / [ 16 sin(π/15) ] = -8 sin(2π/15) cos(2π/15) cos(4π/15) cos(8π/15) / [ 16 sin(π/15) ] = -4 * [ 2 sin(2π/15) cos(2π/15) ] cos(4π/15) cos(8π/15) / [ 16 sin(π/15) ] = -4 sin(4π/15) cos(4π/15) cos(8π/15) / [ 16 sin(π/15) ] = -2 * [2 sin(4π/15) cos(4π/15) ] cos(8π/15) / [ 16 sin(π/15) ] = -2 sin(8π/15) cos(8π/15) / [ 16 sin(π/15) ] = -sin(16π/15) / [ 16 sin(π/15) ] = -sin(π + π/15) / [ 16 sin(π/15) ] = - * -sin(π/15) / [ 16 sin(π/15) ] = 1/16
```
3 years ago
```							sorry,it should be cos(16π/15),in the first line. anyway it wouldn’t make a difference because cos(π+π/15)=-cos(π/15),which is same as taking cos(14π/15).
```
3 years ago
```							Solving it easily there is a trick for solving the cos  multiple series . (ONly for cos multiple series)). When the angles are in gp or the next angle is double of previous and series  continues.... It is like this ..{Sin(2 *largest angle)}/{sin (smallest angle)*2^n}Where "n" is number of terms in the series In this case    ~~~~{Sin (32π/15)}/sin(2π/15)*2^4{Sin(2π/15)}/sin(2π/15)*161/16
```
3 years ago
```							Solving it easily there is a trick for solving the cos multiple series . (ONly for cos multiple series)). When the angles are in gp or the next angle is double of previous and series continues.... It is like this ..{Sin(2 *largest angle)}/{sin (smallest angle)*2^n}Where "n" is nuhmber of terms in the series In this case ~~~~=={Sin (32π/15)}/sin(2π/15)*2^4={Sin(2π/15)}/sin(2π/15)*16==1/16
```
3 years ago
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