<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] Here’s a “fun” way.
Recall the standard formulas for $\cos(A+B)$ and $\sin(A)\sin(B)$. We can find:
\begin{align*}
\cos(nx+x) &= \cos(nx) \cos(x) - \sin(nx) \sin(x) \\
\cos(nx+x) &= \cos(nx) \cos(x) - \tfrac 12 ( \cos(nx-x) - \cos(nx+x) ) \\
2 \cos(nx+x) &= 2 \cos(nx) \cos(x) - \cos(nx-x) + \cos(nx+x) \\
\color{blue}{\cos((n+1)x)} &= 2 \color{blue}{\cos(nx)} \cos(x) - \color{blue}{\cos((n-1)x)}
\end{align*}
This is cool: if we know $\cos((n-1)x)$ and $\cos(nx)$, we can easily compute $\cos((n+1)x)$ using this formula. We can write down $\cos(0x)=1$ and $\cos(1x)=\cos(x)$, and then repeatedly apply the formula above to get the next term in the sequence $a_n = \cos(nx)$.
If we now substitute $t=\cos(x)$, then our sequence is just:
\begin{align*}
a_0&=1 \\
a_1&=t \\
a_n&=2t \cdot a_{n-1} - a_{n-2}
\end{align*}
Calculate up to $a_5$ and you have your answer.
…What’s that? You’re allergic to polynomial multiplication? Say no more. I got you covered.
This is a sequence of polynomials (Chebyshev polynomials of the first kind) over $t = \cos(x)$.
A simple application of induction shows us that the $n$-th polynomial has degree $n$. Another shows us that each $a_{2k}$ is even, and each $a_{2k+1}$ is odd. We conclude that
$$\cos(5x) = a \cos^5 x + b \cos^3 x + c \cos x.$$
Let's sample both sides thrice:
\begin{align*}
1 = \cos(5 \cdot 0) &= a \cos^5 0 + b \cos^3 0 + c \cos 0 = a + b + c \\
-2^{-1/2} = \cos(5 \cdot \tfrac \pi 4) &= a \cos^5 \tfrac \pi 4 + b \cos^3 \tfrac \pi 4 + c \cos \tfrac \pi 4 = 2^{-1/2} \left( a/4 + b/2 + c \right) \\
\tfrac 12 = \cos(5 \cdot \tfrac \pi 3) &= a \cos^5 \tfrac \pi 3 + b \cos^3 \tfrac \pi 3 + c \cos \tfrac \pi 3 = a/32 + b/8 + c/2
\end{align*}
In other words:
\begin{cases}
a + b + c &=1 \\
 a + 2b + 4c &=-4 \\
 a + 4b + 16c &=16
\end{cases}
Solving the system, we find $(a,b,c)=(16,-20,5)$.
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