# Trigonometric factorization method

I'm looking at some textbook question and there is a part where it writes this: $$[ 16\cos^4\theta-10\cos^2\theta+1]$$ as this: $$\left( 2\cos^2 \theta-1 \right) \left( 8\cos^2 \theta-1 \right)$$

So that it can be solved out for $\cos\theta$.

I know what the correct answer is but I'm trying to find out how to factorize similar cases. Did they use any trigonometric identity, or formula, that can help me?

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Edited sorry it was 10 instead of 12, I tried factorizing previously but I was letting it equal as $\cos\theta$ instead of squared and I got confused, you may want to post as an answer so that I can accept –  Panayiotis Apr 28 '12 at 21:14
This basically comes down to factoring the polynomial $16x^4-10x^2+1$, or if you prefer $16y^2-10y+1$. You can apply the quadratic formula in the latter case to find out the possible values of $\cos^2\theta$ –  Brett Frankel Apr 28 '12 at 21:14
Set $z=\cos^2\theta$. Then the first expression is $16z^2-10z+1$. Factor this and put back in terms of $\cos\theta$... –  David Mitra Apr 28 '12 at 21:15
Writing 16\,{\cos}^{4}\theta instead of 16\cos^4\theta is weird. That spacing character wouldn't be needed if those pointless braces weren't there. Proper spacing is built in to the software, but that purpose is defeated by the braces, so then it has to be done by hand. –  Michael Hardy Apr 29 '12 at 3:07
I didn't write any of this myself, I just exported the latex from maple that's why –  Panayiotis Apr 29 '12 at 7:49
Note the powers of $\cos$ in the first expression are $4$ and $2$. This would suggest that you have a quadratic expression in the variable $\cos^2\theta$. And indeed you do: set $z=\cos^2\theta$ if you like and write $16\cos^4\theta-10\cos^2\theta +1$ as $$16(\cos^2\theta)^2 -10(\cos^2\theta)+1=16z^2-10z+1.$$ Now you can factor $$16z^2 -10z+1=(2z-1)(8z-1).$$ Writing the above expression in terms of $\cos^2\theta$ gives you your second expression.
In general, you can recognize $az^4+bz^2+c$ as a quadratic expression in the variable $z^2$, or $az^6+bz^3+c$ as a quadratic expression in the variable $z^3$, etc.