Expert Answer to "Provesin⁡(5θ)=16sin5(θ)−20sin3(θ)+5sin⁡(θ)"

markush35q

Answered question

2022-03-31

Prove

\sin (5 θ) = 16 \sin^{5} (θ) - 20 \sin^{3} (θ) + 5 \sin (θ)

Karsyn Wu

Beginner2022-04-01Added 17 answers

\sin 5 x = 1 = \sin \frac{π}{2} \Rightarrow 5 x = 2 n π + \frac{π}{2} = \frac{π}{2} (4 n + 1)

where n is any integer

\Rightarrow x = \frac{π (4 n + 1)}{10}

where

0 \leq n \leq 4

For

n = 1, x = \frac{π}{2} \Rightarrow \sin x = 1

So, the rest 4 roots will be from

\frac{16 x^{5} - 20 x^{3} + 5 x - 1}{x - 1} = 0 \Rightarrow 16 x^{4} + 16 x^{3} - 4 x^{2} - 4 x + 1 = 0

Now,

16 x^{4} + 16 x^{3} - 4 x^{2} - 4 x + 1 = {(4 x^{2})}^{2} + 2 \cdot 4 x^{2} (2 x) + {(2 x)}^{2} + 2 \cdot 4 x^{2} (- 1) + 2 \cdot 2 x (- 1) + {(- 1)}^{2} = {(4 x^{2} + 2 x - 1)}^{2}

which is evident from

\frac{π}{10} (with n = 0) = π - \frac{9 π}{10} (with n = 2) \Rightarrow \sin \frac{9 π}{10} = \sin \frac{π}{10}

= \sin \frac{π}{10}

Same from

n = 3, n = 4

So,

\sin \frac{9 π}{10} = \sin \frac{π}{10}, \sin \frac{13 π}{10} = \sin \frac{17 π}{10}

are the roots of

4 x^{2} + 2 x - 1 = 0

Observe that

\sin \frac{π}{10} > 0, \sin \frac{13 π}{10} = \sin (π + \frac{3 π}{10}) = - \sin \frac{3 π}{10} < 0

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