Question

Help!! Please..
Prove that (1-tan^2(pi/4 - theta)) divided by (1 + tan^2(pi/4 - theta)) = 2sin(theta)cos(theta)

Answer 1

It certainly looks horrible! :

\(\dfrac{1-\tan^2 (\frac{\pi}{4}-\theta)}{1+\tan^2 (\frac{\pi}{4}-\theta)}=2\sin \theta \cos \theta\).

I will do the first step, then you can try the rest.

Remember, \(\tan \theta=\dfrac{\sin \theta}{\cos \theta }\), so the LHS of the identity becomes:

\(\dfrac{1-\dfrac{\sin^2 (\frac{\pi}{4}-\theta)}{\cos^2(\frac{\pi}{4}-\theta)}}{1+\dfrac{\sin^2 (\frac{\pi}{4}-\theta)}{\cos^2(\frac{\pi}{4}-\theta)}} \).

Answer 2

I agree with you that this looks even worse, but from now on it gets better.
If you multiply both the numerator and the denominator of the large fraction with \(\cos^2(\frac{\pi}{4}-\theta)\), you will see that the whole thing looks much simpler.

Go ahead and try it!

Answer 3

ok

Answer 4

After multiplying with \(\cos^2(\frac{\pi}{4}-\theta)\), this is what comes out:

\(\dfrac{\cos^2(\frac{\pi}{4}-\theta)-\sin^2(\frac{\pi}{4}-\theta)}{\cos^2(\frac{\pi}{4}-\theta)+\sin^2(\frac{\pi}{4}-\theta)}\)

Answer 5

What is the next step?

Answer 6

It simplifies to \(\dfrac{\cos 2(\frac{\pi}{4}-\theta)}{1}=\cos(\frac{\pi}{2}-2\theta)\).

Think about what can connect this to the RHS of the identity!

Answer 7

Sin2theta which =2sin(theta)cos(theta ) thanks alot

Answer 8

YW!