# Find a,b,c in sin^5(x) = asin(x) + bsin(3x) + csin(5x)

1. Nov 1, 2009

### soopo

1. The problem statement, all variables and given/known data
Find A,B,C in sin^5(x) = Asin(x) + Bsin(3x) + Csin(5x).

3. The attempt at a solution

I get by Euler the double angle identities for $sin(3x) and sin(5x)$.
They are

$$sin(3x) = s(2x) c(x) + c(2x) s(x)$$
$$sin(5x) = s(3x) c(2x) + c(3x) s(2x)$$

I have the following expression now

$$sin^5(x) = 1 - cos^5(x) - 5c^4(x)s(x) - 10c^3(x)s^2(x) - 10 c^2(x)s^3(x) - 5 c(x)s^4(x)$$

where the trigonometric terms are $(1/2) s(2x) c^3(x), (1/4) (s(2x))^2 c(x), (1/4) (s2x)^2 s(x), (1/2) s(2x) s^3(x)$, respectively.

I get the common term $[(1/2)s(2x) = c(x)s(x)$ by the double angle identity of sine.

So you have $(-5/2) s(2x) (c^3(x) + c(x) + s(x) + s^(x)$.

By comparing the terms to the given equation, we get

$$Bsin(3x) = B (s(2x) c(x) + c(2x) s(x) )$$
$$Csin(5x) = C ( s(3x) c(2x) + c(3x) s(2x) )$$

so we have

$$B c(x) + C c(3x) = (-5/2) ( c^3(x) + c(x) + s(x) + s^3(x) )$$
which implies that $C = - \frac {-5} {2}$

The other term is in the form

$$B c(x) = (-5/2) ( c(x) + s(x) + s^3(x) )$$
where where $s(x) + s^3(x) = s(x) (1 + s^(x) = 2s(x) - c^2(x) s(x)$ by Pythagoras.

However, I do not see directly how to get the term [itex] c(x) [/tex] to the RHS of the equation.

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My first attempt seems to be useless, since the answer may be found by Fourier series too.
However, I have little experience of them and cannot see to how use them here.

How can you find A, B and C by Fourier series or by other method?

2. Nov 10, 2009

### tiny-tim

Hi soopo!

How about expanding (eix - e-ix)5 ?