Solving Trig Identities: Tips for Getting Unstuck

[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

 

[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)

 

[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.

 

[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...

 

[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

 

[cscott]

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

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

Thanks for your help

 

[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

 

[cscott]

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

Aha! Thanks again