# Math Help - Trigonometric Identity - All that Good Stuff!

1. ## Trigonometric Identity - All that Good Stuff!

Well guys, I really would appreciate it were someone able to prove to me this:

cos^2x-sin^2x = (1-tan^2x)/(1+tan^2x)

I think the left side equals cos(2x), but I'm not even sure of that! Any help is much appreciated!

2. Originally Posted by saintv
Well guys, I really would appreciate it were someone able to prove to me this:

cos^2x-sin^2x = (1-tan^2x)/(1+tan^2x)

I think the left side equals cos(2x), but I'm not even sure of that! Any help is much appreciated!
You're correct about the left side of the equation, but that isn't going to help you any. Just use the identity $tan(x)=\frac{sin(x)}{cos(x)}$ on the right side of the equation. If you simplify the fraction, you'll obtain the left side:

$\frac{1-\frac{sin^2(x)}{cos^2(x)}}{1+\frac{sin^2(x)}{cos^2 (x)}}=\frac{\frac{cos^2(x)-sin^2(x)}{cos^2(x)}}{\frac{cos^2(x)+sin^2(x)}{cos^ 2(x)}}=\frac{cos^2(x)-sin^2(x)}{cos^2(x)+sin^2(x)}=cos^2(x)-sin^2(x)$

3. Originally Posted by saintv
Well guys, I really would appreciate it were someone able to prove to me this:

cos^2x-sin^2x = (1-tan^2x)/(1+tan^2x)

I think the left side equals cos(2x), but I'm not even sure of that! Any help is much appreciated!
Yes the left side does equal cos(2x) but I'm going to start with the right side and get the left.

note that $tan^2(x) = \frac{sin^2(x)}{cos^2(x)}$ and $1+tan^2(x) = sec^2(x)$

Rewrite in terms of sin and cos:

Spoiler:
$\left(1-\frac{sin^2(x)}{cos^2(x)}\right) \times \frac{1}{sec^2(x)}$

Get the same denominator for both terms:

Spoiler:
$\left(1 - \frac{sin^2(x)}{cos^2(x)}\right) = \frac{cos^2(x)-sin^2(x)}{cos^2(x)}$

Combine the two again

$\frac{cos^2(x)-sin^2(x)}{cos^2(x)} \times \frac{1}{sec^2(x)} = \frac{cos^2(x)-sin^2(x)}{cos^2(x)} \times cos^2(x)$

$= cos^2(x)-sin^2(x) = LHS$

4. Thank you so much, both of you! I really understand it now!

Guess I need to work on the algebraic skills a bit more.