Prove that $16\cos^5A-20\cos^3A+5\cos A=\cos5A$ 
Prove that 
  $$16\cos^5A-20\cos^3A+5\cos A=\cos5A$$

My solution begins here;
$$
\begin{align}
\text{RHS} & =\cos5A \\
& =\cos(A+4A) \\
& =\cos A\cos4A-\sin A\sin4A \\
& =\cos A(2\cos^2 2A-1)-\sin A(2\sin2A\cos2A) \\
& =2\cos A\cos^2 2A-\cos A-2\sin A\sin2A\cos2A \\
& =2\cos A\cos^2 2A-\cos A-2\sin A(2\sin A\cos A)\cos2A
\end{align}
$$
Now how do I move on?
 A: Now use double angle formulae to write $\cos 2A$ in terms of $\cos A$, then turn any $\sin^2 A$ into $1-\cos^2 A$
A: Use
$$
2\cos x=e^{ix}+e^{-ix}
$$
Then
$$
32\cos^5x=e^{5ix}+5e^{3ix}+10e^{ix}+10e^{-ix}+5e^{-3ix}+e^{-5ix}=
2\cos5x+10\cos3x+20\cos x
$$
Moreover
$$
8\cos^3x=e^{3ix}+3e^{ix}+3e^{-ix}+e^{-3ix}=
2\cos3x+6\cos x
$$
Therefore
\begin{align}
16\cos^5x-20\cos^3x+5\cos x
&=(\cos5x+5\cos3x+10\cos x)-5(\cos3x+3\cos x)+5\cos x\\
&=\cos5x
\end{align}
This is actually backwards, so here's a different proof:
\begin{align}
\cos 5x+i\sin 5x
&=(\cos x+i\sin x)^5\\
&=\cos^5x+5i\cos^4x\sin x-10\cos^3x\sin^2x\\
&\qquad-10i\cos^2x\sin^3x+5\cos x\sin^4x
+i\sin^5x
\end{align}
Taking the real part
$$
\cos5x=\cos^5x-10\cos^3x(1-\cos^2x)+5\cos x(1-\cos^2x)^2
$$
and you can finish.
A: You'll need to replace each instance of $\sin^2 A$ with $1-\cos^2A$ before doing the last simplifications.
A: Calculating the Chebyschev polynomial of the first kind by the recurrence
$$T_0(A)=1\\T_1(A)=A\\T_{n+1}(A)=2AT_n(A)-T_{n-1}(A)$$ you have the known result $$T_n(A)=\cos nA$$ You can verify at once for $n=5$ your formule.
Comment for laamuser user: Chebyschev polynomial of the first kind is some "magic" to find $\cos nx$. For $n=5$ you have successively
$$T_2(A)=2A\cdot A-1=2A^2-1$$  $$T_3(A)=2A(2A^2-1)-A=4A^3-3A$$  $$T_4(A)=2A(4A^3-3A)-(2A^2-1)=8A^4-8A^2+1$$ $$T_5(A)=2A(8A^4-8A^2+1)-(4A^3-3A)$$
i.e.
$$T_5(A)=16A^5-20A^3+5A$$ and you have $$T_5(\cos x)=\cos 5x$$ 
