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?
 A: A general way to approach this is to use Euler’s formula and the binomial theorem, as follows.
$$\cos(5A)=\frac{e^{5Ai}+e^{-5Ai}}{2}=\frac{(e^{Ai})^5+(e^{-Ai})^5}{2}=\frac{(\cos A+i\sin A)^5+(\cos A-i\sin A )^5}{2}.$$ Let $x=\cos A$ and $y=\sin A$, and note for later that $y^2=1-x^2$. Now you can write
$$2\cos(5A)=(x+iy)^5+(x-iy )^5.$$
Expand each fifth power using the binomial theorem. No odd powers of $y$ or imaginary terms should remain, and you can replace $y^2$ with $1-x^2$ and simplify to get your result.
A: Note that $\cos(a+b)+\cos(a-b)=2\cos a \cos b$, so
$$\cos5x+\cos3x=2\cos x\cos4x\implies\cos5x=2\cos4x\cos x-\cos3x$$
$$\cos4x+\cos2x=2\cos x\cos3x\implies\cos4x=2\cos3x\cos x-\cos2x$$
$$\cos3x+\cos x=2\cos x\cos2x\implies\cos3x=2\cos2x\cos x-\cos x$$
$$\cos2x+\cos0x=2\cos x\cos x\implies\cos2x=2\cos^2x-1$$
Gradually pushing these relations upwards allows us to express $\cos nx$ in terms of just $\cos x$, culminating in the result you obtain.
A: 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}
A: 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 $(a,b,c)=(16,-20,5)$.
A: 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?
A: $\cos{5a} = \cos{(3a+2a)}$
$=\cos 3a \cos 2a -\sin 3a \sin 2a$
$=\cos{(2a+a)}\cos 2a -\sin{(2a+a)} \sin 2a$
$=(\cos 2a\cos a-\sin2a\sin a)\cos 2a -(\sin2a \cos a + \cos 2a \sin a) \sin 2a$
$=(\cos 2a\cos a-\sin2a\sin a)\cos 2a -(\sin2a \cos a + \cos 2a \sin a) \sin 2a$
$=((\cos^2a-\sin^2a)\cos a-(2\sin a \cos a)\sin a)(\cos^2a-\sin^2a) -((2\sin a \cos a) \cos a + (\cos^2a-\sin^2a) \sin a) (2\sin a\cos a)$
$=(\cos^3a-\sin^2 a\cos a -2\sin^2a\cos a)(\cos^2a-\sin^2a) -(2\sin a \cos^2a + \cos^2a \sin a - \sin^3 a) (2\sin a \cos a)$
$=(\cos^3a - \sin^2 a \cos a - 2 \sin^2a \cos a)(\cos^2a)-(\cos^3a - \sin^2 a \cos a - 2 \sin^2a \cos a)(\sin^2a) -(2\sin a \cos^2a + \cos^2a \sin a - \sin^3 a) (2\sin a \cos a)$
$=(\cos^5a - \sin^2 a \cos^3 a - 2 \sin^2a \cos^3 a)-(\cos^3a \sin^2a - \sin^4 a \cos a - 2 \sin^4a \cos a) - (4\sin^2 a \cos^3a + 2\cos^3a \sin^2 a - 2\sin^4 a \cos a)$
$=(\cos^5a - \sin^2 a \cos^3 a - 2 \sin^2a \cos^3 a)+(-\cos^3a \sin^2a + \sin^4 a \cos a + 2 \sin^4a \cos a) + (-4 \sin^2 a \cos^3a - 2\cos^3a \sin^2 a + 2\sin^4 a \cos a)$
$\bf =\cos^5a - 10 \sin^2a \cos^3 a + 5 \sin^4a \cos a$ // Another proof
$=\cos^5a - 10 (1 - \cos^2a) \cos^3 a + 5 (1- \cos^2a)^2 \cos a$
$=\cos^5a - 10 \cos^3a + 10 \cos^5a + (5 \cos a- 10 \cos^3a + 5 \cos^5a)$
$\cos{5a}=16\cos^5a - 20 \cos^3a + 5 \cos a$
