If $\cos 17x = f(\cos x)$, then show that $\sin 17 x=f(\sin x)$

Question

If $f$ denotes the function which gives $\cos(17x)$ in terms of $\cos x$, that is $\cos(17 x) = f (\cos x)$, then, prove that it is the same function $f$ which gives $\sin(17x)$ in terms of $\sin x$. Generalize this result.

I am really unable to understand how I should go about this question. Thanks a lot.

Answer 1

$\displaystyle \cos 17x = f(\cos x) \\ \displaystyle\cos 17 (\frac{\pi}2 - x) = f(\cos (\frac{\pi}2 - x)) \\ \displaystyle\cos (\frac{17\pi}{2} - 17x) =f(\sin x) \\\displaystyle \cos (8\pi + \frac{\pi}{2} - 17x) = f(\sin x) \\ \displaystyle\cos(\frac{\pi}2 - 17x) = f(\sin x) \\\displaystyle \sin 17x = f(\sin x) $

The generalisation is that $\displaystyle \cos nx = f(\cos x) \iff \sin nx = f(\sin x)$ holds for all $\displaystyle n = 4k+1, k \in \mathbb{Z}$

Answer 2

For brevity let $\cos x=C$ and $\sin x=S.$

Let $n=4m+1$ with $m\in \Bbb Z^+.$ $$\cos nx=Re[(C+iS)^n]=\sum_{j=0}^{2m}C^{4m+1-2j}(iS)^{2j}\binom {4m+1}{2j}=$$ $$=\sum_{j=0}^{2m}C^{4m+1-2j}(C^2-1)^j\binom {4m+1}{2j}.$$

$$\sin nx=Im[(iS+C)^n]=\sum_{j=0}^{2m}(1/i)\cdot(iS)^{4m+1-2j}C^{2j}\binom {4m+1}{2j}=$$ $$=\sum_{j=0}^{2m}(-1)^jS^{4m+1-2j}(1-S^2)^j\binom {4m+1}{2j}=$$ $$=\sum_{j=0}^{2m}S^{4m+1-2j}(S^2-1)^j\binom {4m+1}{2j}.$$

Answer 3

Well, since $\cos(\arccos(x))=x$ for $-1 \le x \le 1$, we have $$f(\cos(x))=\cos(17x) \implies f(x)=\cos(17\arccos(x))$$ $$\implies f(\sin(x))=\cos(17\arccos(\sin(x))$$ and since always $-1\le\sin(x) \le 1$, we no more have the domain problem, it remains to prove $$\cos(17\arccos(\sin(x))=\sin(17x)$$ Notice that $\arccos(\sin(x))=x-\frac{\pi}{2}$ for $\frac{\pi}{2} \le x \le \pi$ so the periodicity implies that the above equation is true.