Trigonometry proof involving sum difference and product formula How would I solve the following trig problem.
$$\cos^5x = \frac{1}{16} \left( 10 \cos x + 5 \cos 3x + \cos 5x \right)$$
I am not sure what to really I know it involves the sum and difference identity but I know not what to do. 

$$\require{cancel}
\frac1{16} [ 5(\cos 3x+\cos x)+\cos 5x+5\cos x ]\\
=\frac1{16}[10\cos x \cos 2x+ \cos 5x +5 \cos x]\\
=\frac1{16} [5\cos x(2\cos 2x+1)+\cos 5x]\\
=\frac1{16} [5\cos x(2(2\cos^2 x-1)+1)+\cos 5x]\\
=\frac1{16} [5\cos x(4\cos^2 x-1)+\cos 5x]\\
=\frac1{16} [5\cos x(4\cos^2 x-1)+\cos 5x]\\
=\frac1{16} [20\cos^3 x-5\cos x+\cos 5x]\\
=\frac1{16} [20\cos^3 x-5\cos x+\cos 5x]\\
=\frac1{16} [\cancel{20\cos^3 x}\cancel{-5\cos x}+16\cos^5 x\cancel{-20\cos^3 x}+\cancel{5\cos x}]\\
=\frac1{16} (16\cos^5 x)\\=\cos^5 x$$
