# How to prove trig identity (sin3x/sinx)-(cos3x/cosx)=2

*print*Print*list*Cite

### 2 Answers

`sin(3x)/sinx-cos(3x)/cosx`

`= [sin(3x)cosx]/(sinxcosx)-[cos(3x)sinx]/(sinxcosx)`

`= (sin(3x)cosx-cos(3x)sinx)/(sinxcosx)`

We know that;

`sin2A = 2sinAcosA`

`sinAcosA= 1/2sin2A`

`sin(A-B) = sinAcosB-cosAsinB`

Let;

`A = 3x`

`B = x`

Then;

`sin(3x-x) = sin3xcosx-cos3xsinx`

`sin2x = sin3xcosx-cos3xsinx`

Therefore we can say;

`sin(3x)/sinx-cos(3x)/cosx`

`= (sin(3x)cosx-cos(3x)sinx)/(sinxcosx)`

`= (sin(2x))/(1/2sin2x)`

`= 2`

*So the identity is proved.*

`sin(3x)/sinx-cos(3x)/cosx = 2`

**Sources:**

`(sin3x)/sinx - (cos3x)/cosx=2`

`sin3xcosx-cos3xsinx=2sinxcosx`

`sin(3x-x)= sin2x`

`sin2x= sin 2x`