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I am having trouble trying to show this. I have reams of paper on the floor and I am tired, I need some help.

I know that $1+\cos\theta-2\cos^2\theta$ equals $\cos\theta + \sin^2\theta-\cos^2\theta$

after that I keep going in circles.

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1  
If you want to show the equality in the title this is not trigonometry, you just have to multiply out. –  Julian Kuelshammer Oct 30 '12 at 11:48
    
Just realized that, $1+x-2x^2$. –  yiyi Oct 30 '12 at 11:50
3  
So you realized that this question has nothing to do with $\cos$ :) –  wj32 Oct 30 '12 at 11:52

3 Answers 3

up vote 4 down vote accepted

You are making it harder than it is. Use $$(a+b)(c+d)=ac+ad+bc+bd,$$ simplify, and you are done.

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and $\sin^2\theta = 1-\cos^2\theta$

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We know that if in the second order equation $ax^2+bx+c=0$, $a+b+c=0$ then; it can be decomposed to $(x-1)(x-\frac{c}{a})=0$ . Now consider $x$ in your eqaution as $\cos(\theta)$.

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I made that realization, see my comment –  yiyi Oct 30 '12 at 11:56
1  
@MaoYiyi: Yes. You did it right. In fact by adding the coefficients you get zero. –  B. S. Oct 30 '12 at 12:00

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