# Trignometric Identities

1. Sep 7, 2009

### GrandMaster87

1. The problem statement, all variables and given/known data
Prove that
$$\frac{sin3x}{sinx}$$-$$\frac{cos3x}{cosx}$$ = 2

2. Relevant equations

3. The attempt at a solution
LHS:$$\frac{sin(2x+x)}{sin}-\frac{cos(2x+x)}{cosx}$$

=$$\frac{sin2x.cosx + cos2x.sinx}{sin}-\frac{cos2xcosx - sin2x.sinx}{cosx}$$

2. Sep 7, 2009

### GrandMaster87

thats as far i got...im really new with trig ..caught a wake up call at school so i started working with it...

3. Sep 7, 2009

### njama

By multiplying $\frac{sin3x}{sinx}$ with $\frac{cos(x)}{cos(x)}$ and $\frac{cos(3x)}{cos(x)}$ with $\frac{sin(x)}{sin(x)}$ you got:

$$\frac{cos(x)sin(3x)-sin(x)cos(3x)}{sin(x)cos(x)}$$

What can you spot now?

4. Sep 7, 2009

### mathie.girl

Did you try getting a common denominator at this point?

5. Sep 8, 2009

### GrandMaster87

can we expand sin(3x) and cos(3x) using double angle formula?

6. Sep 8, 2009

### kbaumen

It can also be done onwards from here. Do you know how to expand cos(2x) and sin(2x)?

7. Sep 8, 2009

### njama

No.

You can write the nominator:

$$cos(x)sin(3x)-sin(x)cos(3x)$$

as

$$sin(3x-x)=sin(2x)$$

using the sum difference formula.

Also you can write the denominator

$$cos(x)sin(x)$$

as

$$\frac{sin(2x)}{2}$$