Prove that $\cos (5A) = 16 \cos^5 (A) - 20 \cos^3 (A) + 5 \cos (A)$ 
Prove the given trigonometric identity
$$\cos (5A) = 16 \cos^5 (A) - 20 \cos^3 (A) + 5 \cos (A)$$


My attempt
L.H.S.$=\cos5A$
$$\cos(A+4A)$$
$$\cos A\cos4A-\sin A\sin4A$$
Now how should I move further?

Good job. Just keep applying double angle formula.
Hint:
$$\cos 4A= \cos(2(2A))=2\cos^2(2A)-1$$
$$\cos 2A = 2 \cos^2A-1$$
$$\sin 4A = 2 \sin 2A \cos 2A$$
$$\sin 2A = 2\sin A \cos A$$
$$ \sin^2A +\cos^2A=1$$
Edit:
\begin{align}
\cos(5A)&=\cos A \cos 4A-\sin A \sin 4A \\
&= \cos A (2 \cos^2(2A)-1)-\sin A (2 \sin 2A \cos 2A) \\
&= \cos A (2 (2 \cos^2 A-1)^2)-\cos A-2 \sin A (2 \sin A \cos A) (2 \cos^2 A -1) \\
&= 2 \cos A (4 \cos^4A-4 \cos^2A+1)-\cos A-4\sin^2A \cos A (2 \cos^2A-1)\\
&= 8\cos^5A-8 \cos^3A+\cos A-4(1-\cos^2A)  (2 \cos^3A-\cos A)
\end{align}
Are you able to complete it?
