# Homework Help: Trig Identities

1. Oct 13, 2005

### cscott

I can't get anywhere with these three identities. Any tips?

$$\frac{(\sec \theta - \tan \theta)^2 + 1}{\csc \theta(\sec \theta - \tan \theta)} = 2 \tan \theta$$

$$\frac{\sin^3 \theta + \cos^3 \theta}{\sin \theta + \cos \theta} = 1 - \sin \theta\cos \theta$$

$$(2a\sin \theta\cos \theta)^2 + a^2(\cos^2 \theta - \sin^2 \theta)^2 = a^2$$

2. Oct 13, 2005

### Pyrrhus

The third one is pretty obvious a (remember the double angle identities!)
$$\sin^{2} \theta + \cos^{2} \theta = 1$$
For the second one remember
$$a^3 + b^3 = (a+b)(a^2 -ab + b^2)$$

Last edited: Oct 13, 2005
3. Oct 13, 2005

### Pyrrhus

I was trying to think a simpler way for the first one, but it all occurs to me now is to

$$\frac{(\sec \theta - \tan \theta)}{\csc \theta} + \frac{1}{\csc \theta (\sec \theta - \tan \theta)} = 2 \tan \theta$$

then work it out with sines and cosines.

4. Oct 13, 2005

### cscott

Thanks for the tips so far. I got second one.

For the first, I had simplified it to that already and tried sines and cosines but I'll try again.

For the last, is there any way to get to the answer from

$$2a^2 + a^2 \sin^4 \theta + a^2 \cos^4 \theta$$

doesn't seem so...

Last edited: Oct 13, 2005
5. Oct 13, 2005

### Pyrrhus

For the last one is basicly applying
$$\sin 2 \theta = 2 \sin \theta \cos \theta$$
$$\cos 2 \theta = \cos^{2} \theta - \sin^{2} \theta$$
$$\sin^{2} 2 \theta + \cos^{2} 2 \theta = 1$$

6. Oct 13, 2005

### cscott

I've never used those identities before, no wonder I didn't know what was going on :uhh:

7. Oct 13, 2005

### Pyrrhus

Well, here are their proof
$$\sin (a+b) = \sin a \cos b + \cos a \sin b$$
$$\sin (\theta + \theta) = \sin \theta \cos \theta + \cos \theta \sin \theta$$
$$\sin (2\theta) = 2 \sin \theta \cos \theta$$
$$\cos (a+b) = \cos a \cos b - \sin a \sin b$$
$$\cos (\theta + \theta) = \cos\theta \cos \theta - \sin \theta \sin \theta$$
$$\cos (2\theta) = \cos^{2}\theta - \sin^{2} \theta$$

8. Oct 13, 2005

### cscott

Aha! Thanks again