Trigonometric Identity - All that Good Stuff!

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November 8th 2009, 09:35 AM
saintv

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! (Happy)
November 8th 2009, 09:41 AM
adkinsjr

Quote:

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! (Happy)

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:

$\frac{1-\frac{sin^2(x)}{cos^2(x)}}{1+\frac{sin^2(x)}{cos^2 (x)}}=\frac{\frac{cos^2(x)-sin^2(x)}{cos^2(x)}}{\frac{cos^2(x)+sin^2(x)}{cos^ 2(x)}}=\frac{cos^2(x)-sin^2(x)}{cos^2(x)+sin^2(x)}=cos^2(x)-sin^2(x)$
November 8th 2009, 09:43 AM
e^(i*pi)

Quote:

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! (Happy)

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

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

$= cos^2(x)-sin^2(x) = LHS$
November 8th 2009, 10:00 AM
saintv

Thank you so much, both of you! I really understand it now! :)

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