Trig relationship

Apr 2009
48
3
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
Aug 2008
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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

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

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

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

\(\displaystyle = 18\cos^2{x} - 6\).
 
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Apr 2009
48
3
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!