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!
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 $\displaystyle tan(x)=\frac{sin(x)}{cos(x)}$ on the right side of the equation. If you simplify the fraction, you'll obtain the left side:
$\displaystyle \frac{1-\frac{sin^2(x)}{cos^2(x)}}{1+\frac{sin^2(x)}{cos^2 (x)}}=\frac{\frac{cos^2(x)-sin^2(x)}{cos^2(x)}}{\frac{cos^2(x)+sin^2(x)}{cos^ 2(x)}}=\frac{cos^2(x)-sin^2(x)}{cos^2(x)+sin^2(x)}=cos^2(x)-sin^2(x)$
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 $