# Thread: Stuck on a problem in Homework (Trigonometric Identities and Formulas)

1. ## Stuck on a problem in Homework (Trigonometric Identities and Formulas)

I've been trying to figure it out for the past 30 minutes but can't seem to get anywhere.

I was able to get somewhere but I can't seem to get it balanced:

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

I took the left side and I started by combining the two terms:

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

From there I recognized that the numerator is an expansion of the sine subtraction formula so I rewrote it to:

$\displaystyle \frac {sin(3x-x)}{sinxcosx}$

which is basically

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

That's where I am stuck.

2. Originally Posted by dagbayani481
I've been trying to figure it out for the past 30 minutes but can't seem to get anywhere.

I was able to get somewhere but I can't seem to get it balanced:

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

I took the left side and I started by combining the two terms:

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

From there I recognized that the numerator is an expansion of the sine subtraction formula so I rewrote it to:

$\displaystyle \frac {sin(3x-x)}{sinxcosx}$

which is basically

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

That's where I am stuck.
Recall that $\displaystyle \sin(2x)=2\sin x\cos x$.

You should be able to finish things off now.

3. Ha! I had totally forgotten about the double angle formula. I should have known. Thank you!