# Trig relationship

• May 29th 2010, 07:07 PM
Ares_D1
Trig relationship
I'm working through a problem set for a class unlrelated to trigonometry, but reached a dead end when I failed to recognize that

$2cos^2(x)- sin^2(x) = 3cos^2(x)-1$

Could somebody kindly show me why this is the case? (I've never actually taken a formal trigonometry class, so my background is fairly weak).

Thanks.
• May 29th 2010, 07:10 PM
Prove It
Quote:

Originally Posted by Ares_D1
I'm working through a problem set for a class unlrelated to trigonometry, but reached a dead end when I failed to recognize that

$6 (2cos^2(x)- sin^2(x)) = 18cos^2(x)-6$

Could somebody kindly show me why this is the case? (I've never actually taken a formal trigonometry class, so my background is fairly weak).

Thanks.

Use the Pythagorean Identity

$\sin^2{x} + \cos^2{x} = 1$

$\sin^2{x} = 1 - \cos^2{x}$.

So that means

$6(2\cos^2{x} - \sin^2{x}) = 6[2\cos^2{x} - (1 - \cos^2{x})]$

$= 6(2\cos^2{x} - 1 + \cos^2{x})$

$= 6(3\cos^2{x} - 1)$

$= 18\cos^2{x} - 6$.
• May 29th 2010, 07:39 PM
Ares_D1
Quote:

Originally Posted by Prove It
Use the Pythagorean Identity

$\sin^2{x} + \cos^2{x} = 1$

$\sin^2{x} = 1 - \cos^2{x}$.

So that means

$6(2\cos^2{x} - \sin^2{x}) = 6[2\cos^2{x} - (1 - \cos^2{x})]$

$= 6(2\cos^2{x} - 1 + \cos^2{x})$

$= 6(3\cos^2{x} - 1)$

$= 18\cos^2{x} - 6$.

Cheers, thanks!