Prove this trigonometric identity [closed]

Question

Prove the following trigonometric identity: $16\sin ^2\left(θ\right)\cos ^3\left(θ\right)=2\cos \left(θ\right)-\cos \left(3θ\right)-\cos \left(5θ\right)\:$

Answer 1

Using

1. $2\cos(A)\cos(B)=\cos(A+B)+\cos(A-B)$, and
2. $\cos^2(A)=\frac{\cos(2A)+1}{2}$
3. $\sin^2(A)=1-\cos^2(A)$

Solution with minimum steps:

Using $(3)$

$$ 16\left(1-\cos^2(\theta)\right)\cos^3(\theta)=16\left( \cos(\theta)\cos^2(\theta) - \cos(\theta)\cos^2(\theta)\cos^2(\theta)\right) $$

using $(2)$

$$ 16\ \cos(\theta)\left(\frac {\cos(2\theta)+1}{2}\right) \left[ 1-\frac {\cos(2\theta)+1}{2} \right] $$ simplify

$$ -\frac{16}{1}\cos(\theta)\left(\frac {\cos(2\theta)+1}{2}\right) \left[ \frac {\cos(2\theta)-1}{2} \right] $$ and $$ -\frac{16}{4}\cos(\theta)\left[ \cos^2(2\theta)-1 \right] $$ using $(2)$ $$ -\frac{16}{8}\cos(\theta)\left[ \cos(4\theta)-1 \right] $$ using $(1)$ $$ 2\cos(\theta)-2\cos(\theta)\cos(4\theta) = 2\cos(\theta)-\cos(5\theta)-\cos(3\theta) $$ Q.E.D.

Answer 2

$\sin^2 \alpha=\dfrac{1-\cos 2\alpha}{2}$ and such identity for $\cos^3 \alpha=\dfrac{3\cos \alpha}{4}+\dfrac{\cos 3\alpha}{4}$

Answer 3

Using Prosthaphaeresis Formulas

$$\cos A-\cos5A=2\sin3A\sin2A\text{ and }\cos A-\cos3A=2\sin A\sin2A$$

$$\implies\cos A-\cos5A+\cos A-\cos3A=2\sin2A(\sin3A+\sin A)$$

Again using Prosthaphaeresis, $$\sin3A+\sin A=2\sin2A\cos A$$

Finally use $\displaystyle\sin2A=2\sin A\cos A$

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Alternatively,

Using Prosthaphaeresis Formulas

$$\cos3A+\cos5A=2\cos A\cos4A$$

$$\implies2\cos A-\cos3A-\cos5A=2\cos A(1-\cos4A)=2\cos A(2\sin^22A)$$