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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.

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hint: rewrite $\cos x$ with exponentials and expand the fifth power : the binomial coefficients are here – Raymond Manzoni Aug 6 '12 at 18:43
I am not sure how I would that theorem. – Fernando Martinez Aug 6 '12 at 18:48
$\cos(x)=\frac 12 (e^{ix}+e^{-ix})$ so that $\cos(x)^5=\frac 1{2^5} (e^{ix}+e^{-ix})^5$, using binomial theorem is faster for the term at the right but it is not an obligation! – Raymond Manzoni Aug 6 '12 at 18:51
I have not studied e^ix or complex numbers unfortunately. – Fernando Martinez Aug 6 '12 at 18:53
you may try it in the other direction $\cos(5x)=\cos(4x+x)=\cos(4x)\cos(x)-\sin(4x)\sin(x)$ and so on if you know the trigonometric rules for addition and no rules for powers... – Raymond Manzoni Aug 6 '12 at 18:58

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 $\cos^5x$.

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I like this response. – Fernando Martinez Aug 7 '12 at 15:08

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

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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.

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