markush35q

2022-03-31

Prove
$\mathrm{sin}\left(5\theta \right)=16{\mathrm{sin}}^{5}\left(\theta \right)-20{\mathrm{sin}}^{3}\left(\theta \right)+5\mathrm{sin}\left(\theta \right)$

### Answer & Explanation

Karsyn Wu

$\mathrm{sin}5x=1=\mathrm{sin}\frac{\pi }{2}⇒5x=2n\pi +\frac{\pi }{2}=\frac{\pi }{2}\left(4n+1\right)$
where n is any integer
$⇒x=\frac{\pi \left(4n+1\right)}{10}$
where $0\le n\le 4$
For $n=1,x=\frac{\pi }{2}⇒\mathrm{sin}x=1$
So, the rest 4 roots will be from
$\frac{16{x}^{5}-20{x}^{3}+5x-1}{x-1}=0⇒16{x}^{4}+16{x}^{3}-4{x}^{2}-4x+1=0$
Now, $16{x}^{4}+16{x}^{3}-4{x}^{2}-4x+1={\left(4{x}^{2}\right)}^{2}+2\cdot 4{x}^{2}\left(2x\right)+{\left(2x\right)}^{2}+2\cdot 4{x}^{2}\left(-1\right)+2\cdot 2x\left(-1\right)+{\left(-1\right)}^{2}={\left(4{x}^{2}+2x-1\right)}^{2}$ which is evident from

$=\mathrm{sin}\frac{\pi }{10}$
Same from
So, are the roots of $4{x}^{2}+2x-1=0$
Observe that

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