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
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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. – Mikasa Jun 25 '12 at 15:48
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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
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See here for an example with $n=10$. – draks ... Jun 25 '12 at 15:50
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@draks: What wonderful relations they are, I didn't know them before. – Mikasa Jun 25 '12 at 15:51
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@BabakSorouh you're welcome. – draks ... Jun 25 '12 at 15:52
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It's probably closely related to Chebyshev polynomials. – lhf Jun 25 '12 at 15:55
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@draks: Do you think this one can be completely carried out by what you had linked here for me, or we can use induction on $n$? – Mikasa Jun 25 '12 at 15:56
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@BabakSorouh I'm not sure, but the answer below solves it very nice... – draks ... Jun 25 '12 at 15:59
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@draks Just a stylistic recommendation. When you create a link, always include some non-math in the link code, so that people can right click on the link and open it in a new tab if they want to. Unfortunately, when you right click in MathJax code, it only pops of the MathJax menu. So, above, you should link to "example with $n=10$" rather than just the text "$n=10"$. – Thomas Andrews Jun 25 '12 at 16:02
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@ThomasAndrews Thanks for the hint, I tend to use the middle mouse click so I never realized it. – draks ... Jun 25 '12 at 19:18
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There is even a meta post on that topic: MathJax menu obliterates link menu when the entire link is LaTeX. – draks ... Jun 25 '12 at 19:18
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
$$p_n(\tan\theta)\cos^n\theta = \cos n\theta $$
Thomas Andrews
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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. :) – Mikasa Jun 25 '12 at 16:01
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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
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