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. 
 A: Taking you at your word that you know (or can derive)

$\cos 3x = 4\cos^3x-3\cos x$
$\cos 5x = 16\cos^5 x-20\cos^3 x+5\cos x$

all you have to do then is substitute these equalities into $(1/16)(10\cos x + 5\cos 3x + \cos 5x)$ and you'll find that it's equal to $\cos^5x$.
A: $$\require{cancel}
\frac1{16} [ 5(\cos 3x+\cos x)+\cos 5x+5\cos x ]\\
=\frac1{16}[10\cos x \cos 2x+ \cos 5x +5 \cos x]\\
=\frac1{16} [5\cos x(2\cos 2x+1)+\cos 5x]\\
=\frac1{16} [5\cos x(2(2\cos^2 x-1)+1)+\cos 5x]\\
=\frac1{16} [5\cos x(4\cos^2 x-1)+\cos 5x]\\
=\frac1{16} [5\cos x(4\cos^2 x-1)+\cos 5x]\\
=\frac1{16} [20\cos^3 x-5\cos x+\cos 5x]\\
=\frac1{16} [20\cos^3 x-5\cos x+\cos 5x]\\
=\frac1{16} [\cancel{20\cos^3 x}\cancel{-5\cos x}+16\cos^5 x\cancel{-20\cos^3 x}+\cancel{5\cos x}]\\
=\frac1{16} (16\cos^5 x)\\=\cos^5 x$$
A: Applying the same approach as that of this, 
let $A\cos5x+B\cos3x+C\cos x=\cos^5x$
As  $\cos 3x = 4\cos^3x-3\cos x$ and 
 $\cos 5x = 16\cos^5 x-20\cos^3 x+5\cos x$
$A(16\cos^5 x-20\cos^3 x+5\cos x) + B( 4\cos^3x-3\cos x)+ C\cos x=\cos^5x$
Comparing the coefficients of different powers of cosx,
5th power=>16A=1=>$A=\frac{1}{16}$ ,
3rd power=>-20A+4B=0=>B=5A=$\frac{5}{16}$ and
1st power=>5A-3B+C=0=>C=3B-5A$=3\cdot\frac{5}{16}-5\cdot\frac{1}{16}=\frac{5}{8}$

Alternatively, observe that the 3rd power of $\cos x$ is absent in the given expression.
But $A\cos5x+B\cos3x=A(16\cos^5 x-20\cos^3 x+5\cos x) + B( 4\cos^3x-3\cos x)$
$=16A\cdot\cos^5 x+ \cos^3x\cdot4(B-5A)+\cos x(5A-3B)$
So, B must be 5A to eliminate $cos^3x$ 
$A\cos5x+B\cos3x=A(\cos5x+5\cdot\cos3x)=A(16\cdot\cos^5 x - 10\cdot\cos x)$
Putting A=1, $\cos5x+5\cdot\cos3x=16\cdot\cos^5 x - 10\cdot\cos x$
So, we just need a little rearrangement. 
