<s>[INST] 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?
 [/INST] One approach is to use De Moivre's Theorem along with the Binomial Theorem.
Consider $(\cos A+i\sin A)^5$. By De Moivre's Theorem, we get $\cos(5A)+i\sin(5A)$. Notice that $\cos(5A)$ is the real portion of the complex expression.
On the other hand, we can use the Binomial Theorem to expand $(\cos A+i\sin A)^5$.
$$(\cos A+i\sin A)^5=(\cos A)^5+5(\cos A)^4(i\sin A)+10(\cos A)^3(i\sin A)^2+10(\cos A)^2(i\sin A)^3+5(\cos A)(i\sin A)^4+(i\sin A)^5$$
Selecting only the real terms from above and equating with $\cos(5A)$, we have
$$\cos(5A)=\cos^5A-10\cos^3A\sin^2A+5\cos A\sin^4A\ .$$
Using the Pythagorean Identity $\sin^2A=1-\cos^2A$, we get
\begin{align}
\cos(5A)&=\cos^5A-10\cos^3A(1-\cos^2A)+5\cos A(1-\cos^2A)^2\\
&=\cos^5A-10\cos^3A+10\cos^5A+5\cos A-10\cos^3A+5\cos^5A\\
&=16\cos^5A-20\cos^3A+5\cos A\ .
\end{align}
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