Trigonometric Identities: Proving \frac{sin3x}{sinx}-\frac{cos3x}{cosx} = 2

[GrandMaster87]

Homework Statement

Prove that
$$\frac{sin3x}{sinx}$$-$$\frac{cos3x}{cosx}$$ = 2

Homework Equations

The Attempt at a Solution

LHS:$$\frac{sin(2x+x)}{sin}-\frac{cos(2x+x)}{cosx}$$

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

 

[GrandMaster87]

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

 

[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?

 

[mathie.girl]

Homework Statement

Prove that
$$\frac{sin3x}{sinx}$$-$$\frac{cos3x}{cosx}$$ = 2

Homework Equations

The Attempt at a Solution

LHS:$$\frac{sin(2x+x)}{sin}-\frac{cos(2x+x)}{cosx}$$

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

Did you try getting a common denominator at this point?

 

[GrandMaster87]

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?

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

 

[kbaumen]

Homework Statement

Prove that
$$\frac{sin3x}{sinx}$$-$$\frac{cos3x}{cosx}$$ = 2

Homework Equations

The Attempt at a Solution

LHS:$$\frac{sin(2x+x)}{sin}-\frac{cos(2x+x)}{cosx}$$

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

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

 

[njama]

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

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

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