Terrie Lang

2022-01-16

If $\mathrm{cos}17x=f\left(\mathrm{cos}x\right)$, then show that $\mathrm{sin}17x=f\left(\mathrm{sin}x\right)$

Elois Puryear

Expert

$\mathrm{cos}17x=f\left(\mathrm{cos}x\right)$
$\mathrm{cos}17\left(\frac{\pi }{2}-x\right)=f\left(\mathrm{cos}\left(\frac{\pi }{2}-x\right)\right)$
$\mathrm{cos}\left(\frac{17\pi }{2}-17x\right)=f\left(\mathrm{sin}x\right)$
$\mathrm{cos}\left(8\pi +\frac{\pi }{2}-17x\right)=f\left(\mathrm{sin}x\right)$
$\mathrm{cos}\left(\frac{\pi }{2}-17x\right)=f\left(\mathrm{sin}x\right)$
$\mathrm{sin}17x=f\left(\mathrm{sin}x\right)$
The generalisation is that $\mathrm{cos}nx=f\left(\mathrm{cos}x\right)⇔\mathrm{sin}nx=f\left(\mathrm{sin}x\right)$ holds for all $n=4k+1,k\in \mathbb{Z}$

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