1. 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

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

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.

2. 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

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

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

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

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

So that means

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

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

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

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

3. Originally Posted by Prove It
Use the Pythagorean Identity

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

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

So that means

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

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

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

$\displaystyle = 18\cos^2{x} - 6$.
Cheers, thanks!