# Trig relationship

#### 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

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

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.

#### Prove It

MHF Helper
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

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

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

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

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

Ares_D1

#### Ares_D1

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!