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! (Happy)

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- Nov 8th 2009, 09:35 AMsaintvTrigonometric 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! (Happy) - Nov 8th 2009, 09:41 AMadkinsjr
You're correct about the left side of the equation, but that isn't going to help you any. Just use the identity $\displaystyle tan(x)=\frac{sin(x)}{cos(x)}$ on the right side of the equation. If you simplify the fraction, you'll obtain the left side:

$\displaystyle \frac{1-\frac{sin^2(x)}{cos^2(x)}}{1+\frac{sin^2(x)}{cos^2 (x)}}=\frac{\frac{cos^2(x)-sin^2(x)}{cos^2(x)}}{\frac{cos^2(x)+sin^2(x)}{cos^ 2(x)}}=\frac{cos^2(x)-sin^2(x)}{cos^2(x)+sin^2(x)}=cos^2(x)-sin^2(x)$ - Nov 8th 2009, 09:43 AMe^(i*pi)
Yes the left side does equal cos(2x) but I'm going to start with the right side and get the left.

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

Rewrite in terms of sin and cos:

__Spoiler__:

Get the same denominator for both terms:

__Spoiler__:

Combine the two again

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

$\displaystyle = cos^2(x)-sin^2(x) = LHS $ - Nov 8th 2009, 10:00 AMsaintv
Thank you so much, both of you! I really understand it now! :)

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