# Finding a polynomial for this trigonometric equality

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?

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Maybe this is of help to you: Power-reduction formula... –  draks ... Jun 25 '12 at 15:47
Since I teach some maths, students come and ask strage questions like this. Some question can be soft but some of them like this one is not. –  B. S. Jun 25 '12 at 15:48
That formula is wrong, but it is right that $\cos(n\theta)=p_n(\tan(\theta))\cos^n(\theta)$. The difference is that $\tan(\theta)$ rather than $\tan(n\theta)$ –  Thomas Andrews Jun 25 '12 at 15:49
See here for an example with $n=10$. –  draks ... Jun 25 '12 at 15:50
@draks: What wonderful relations they are, I didn't know them before. –  B. S. Jun 25 '12 at 15:51

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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Dear Thomas, I edited the question as others and you noted, but how could I guess this polynomial. Guessing is a bit hard. :) –  B. S. Jun 25 '12 at 16:01
Well, the general guideline is that if you want to associate polynomials about trig functions in $\theta$ with some trig function on $n\theta$ is that the fundamental such formula is:$$(\cos\theta + i\sin\theta)^n = \cos n\theta+i\sin n\theta$$ –  Thomas Andrews Jun 25 '12 at 16:05
Thanks Thomas for the answer and hint. –  B. S. Jun 25 '12 at 16:10