For all natural numbers, find the polynomials such that $\cos(n\theta)=p_n(\tan(\theta))\cos^n(\theta)$ It was suggested that taking $p_n(x)=\frac{1}{2}\{(1+ix)^n+(1-ix)^n\}$, but I don't know how?
Finding a polynomial for this trigonometric equality
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$\begingroup$
trigonometry
polynomials
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0There is even a meta post on that topic: [MathJax menu obliterates link menu when the entire link is LaTeX](http://meta.math.stackexchange.com/q/12$0$4/19341). – 2012-06-25
1 Answers
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Let $q_n(x) = (1+ix)^n$.
Then $q_n(\tan(\theta))\cos^n(\theta) = (1+i\frac{\sin\theta}{\cos \theta})^n\cos^n\theta = (\cos \theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta$
So $p_n(x) = \frac{1}{2}\left(q_n(x)+q_n(-x)\right)$
Use that $\cos(-x)=\cos x$ to show that
$p_n(\tan\theta)\cos^n\theta = \cos n\theta $
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0Thanks Thomas for the answer and hint. – 2012-06-25