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.
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.
You are making it harder than it is. Use $(a+b)(c+d)=ac+ad+bc+bd,$ simplify, and you are done.
and $\sin^2\theta = 1-\cos^2\theta$
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 $\cos(\theta)$.