Sin^n(theta) as product of sin^1, cos^1 (m*theta)

• mikeph
In summary, the conversation is about finding a way to decompose a function f(x + a sin(t)) into a Fourier series. The speaker mentions that they have already expanded the function into a Taylor series and are now trying to find a way to express it in terms of single powers of sines and cosines of multiples of t. They mention they have made some progress with cos^n(t) but are still looking for a way to find the bracket terms in the expression. The other person suggests using a formula for cos^n(t) in terms of cos(t), cos(2t), etc. and mentions that they will think about the problem later.
mikeph
Hello,

Somewhat urgent question, I would normally try and do this myself but I have a feeling it will take a while, and I sort of need to be working through this pretty quickly, so any help much appreciated. Plus I might end up wasting half a day trying formulas on this.

I am expanding a function f(x + a sin(t)) as a Taylor series, so I get:

f(x + a sin (t)) = f(x) + (a sin t)f'(x) + (a sin t)^2/2! *f''(x) + ...

Now I am trying to decompose this into some sort of Fourier series (not exactly clear on final goal or method yet), so I need this in terms of:

f(x + a sin (t)) = [ ... ] + [ ... ]*sin(t) + [ ... ]*sin(2t) + ... + [ ... ]*cos(t) + [ ... ]*cos(2t) + ...

I am wondering if there is any simple way to do this. I have worked out the first one, eg:
sin^2(t) = 1/2 - 1/2 *cos(t), and was wondering if there are any simple recursion formulas or anything out there to help me find an expression for sin^n(t) in terms of single powers of sines and cosines of multiples of t?

Thanks for any help,
Mike

Ok, a little progress has been made:

I am now working with cos^n (t) as that can be put into only cosine terms. I have also found the formula for this in terms of cos(t), cos(2t), etc. but it is not so helpful for finding what my bracket terms [ ... ] are, since there are many overlaps, as the expansion of f is a series of increasing powers of cos(t) each with a different weighting by a.

For example, the coefficient of cos(t) would be: [a*f'(x) + (a^3)*f''(x)/2 + ... ]

First of all note that the powers of trig function can be found from
$$\cos^n(x)=\Re\left(e^{ix}\left(\frac{e^{ix}+e^{-ix}}{2}\right)^{n-1}\right)$$
and similarly for sine.
I have to leave now, but I can think about the full problem later.

1. What is the formula for expressing sin^n(theta) as a product of sin^1 and cos^1 (m*theta)?

The formula for sin^n(theta) as a product of sin^1 and cos^1 (m*theta) is:

sin^n(theta) = (sin(theta))^n = (sin(theta))*(sin(theta))^(n-1) = (sin(theta))*(cos(m*theta))^2

2. How is this formula derived?

This formula can be derived using the trigonometric identity sin^2(theta) + cos^2(theta) = 1. By substituting sin^2(theta) = (sin(theta))*(sin(theta)) and cos^2(theta) = (cos(m*theta))^2, the formula can be rearranged to obtain the desired formula.

3. Can this formula be used for any value of n?

Yes, this formula can be used for any value of n as long as n is a positive integer.

4. How is this formula useful in solving trigonometric equations?

This formula can be useful in simplifying trigonometric equations involving higher powers of sin(theta). It can be used to express a given trigonometric equation in terms of sin^1 and cos^1 (m*theta), which may make it easier to solve.

5. Are there any limitations to using this formula?

One limitation of this formula is that it can only be used for values of n that are positive integers. It may also not be applicable for certain complex trigonometric equations that involve multiple trigonometric functions.

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