# Is there a formula for $n^{th}$ power of $\cos(ax+b)$?

Is there a closed form formula or trigonometric relation for $$\cos^{n}(ax+b)$$?

• This is already a closed formula Feb 16, 2021 at 23:34
• There are other ways to write this, but I'm afraid that they are actually more cumbersome than this. Feb 16, 2021 at 23:36
• My purpose for this question to prove that for even and odd values of $n$, the period of this function is $\frac{\pi}{a}$ and $\frac{2\pi}{a}$ respectively so I will have to find a way to get a relation for $\cos^{n}(ax+b)$. Perhaps using exponentials or taylor series? @DonThousand Feb 16, 2021 at 23:38

Let's forget $$a$$ and $$b$$, as it's not the problem here.

$$\cos^n x=\left(\frac{e^{ix}+e^{-ix}}{2}\right)^n=\frac1{2^n}\sum_{k=0}^n\binom{n}{k}e^{kix}e^{-(n-k)ix}\\=\frac1{2^n}\sum_{k=0}^n\binom{n}{k}e^{(2k-n)ix}\\=\frac1{2^n}\sum_{k=0}^n\binom{n}{k}(\cos(2k-n)x+i\sin(2k-n)x)\\ =\frac1{2^n}\sum_{k=0}^n\binom{n}{k}\cos(2k-n)x$$

• Alright that looks really nice thank you. For further inspection, how would you tackle the issue of finding the period when there's binomial? Feb 16, 2021 at 23:43
• @SofiaYaseveya The binomial plays little role here, what is important is the period of the terms. Feb 16, 2021 at 23:44
• I think recasting the original closed form as a sum makes it harder to reason about the periodicity. Feb 16, 2021 at 23:45
• @RobArthan If think so. I didn't see the comments until I wrote the answer though. Feb 16, 2021 at 23:47
• I see, perhaps it would've required a once in a lifetime computation. However, I am still satisfied with this beautiful answer. Feb 16, 2021 at 23:47

As mentioned in your comment, if you are only looking to find the period of $$cos^n(ax+b)$$, it wouldn't be necessary to write a closed form formula or expression for it. When $$n$$ is even, We see that $$(\cos(\pi - (ax+b)))^n = (\cos(\pi + (ax+b)))^n$$ and this turns the period of the function into $$\pi/a$$, whereas when $$n$$ is odd, we don't have this luxury anymore. Instead we have $$\cos^n(ax+b)$$ going its full $$2\pi/a$$ cycle.

Edit: As @Rob Arthan pointed out in his comment, given $$f(x)$$ with period $$T$$ and some function $$g(x)$$; $$h(x) = g(f(x))$$ has period $$T/k$$ for some integer $$k$$. It is also possible to observe that $$T$$ is a period of $$h$$, but often depending upon the function $$g$$ we can do better.

In our case $$f = \cos(ax+b)$$ and $$g(x) = x^n$$. When $$n$$ is odd, $$g(x)$$ is injective, and so every input to $$g$$ has a unique associated output, which means there is no room in $$[0,2\pi/a]$$ (the original period of $$f$$) for repetition of the same output values. This is why, when $$n$$ is odd, the period of $$h$$ is the same as the period of $$f$$. However, when $$g$$ is not injective, we have multiple inputs being mapped to the same output. In the case of an even $$n$$, we have exactly two inputs in a given period being mapped to the same output, which essentially means we are repeating the graph of $$f$$ twice in $$[0,2\pi/a]$$ upon composing, which means the period is halved simply giving us a period of $$\pi/a$$.

• Perhaps you should explain that you are using the fact that the function $x \mapsto x^n : [-1, 1] \to [-1, 1]$ is 1-1 when $n$ is odd and 2-1 when $n$ is even (so that the periods cannot be proper divisors of $\pi/a$ or $2\pi/a$). Feb 16, 2021 at 23:49
• I didn't quite understand the reasoning of $n$ being even implying $(\cos(\pi+(ax+b)))^{n}$. I am though, interested in your answer if you could please clarify more. Feb 16, 2021 at 23:51
• If $f$ is periodic with period $t$, and $g$ is any function (in this case $y \mapsto y^n$), then $x \mapsto g(f(x))$ is either constant or periodic with a periodic with period $t/k$ for some positive integer $k$. You should find this easy to prove from the definition of a periodic function. Feb 16, 2021 at 23:56
• I can't seem to understand your intuition :( @RobArthan Feb 16, 2021 at 23:59
• If $f(x) = f(x + t)$, then $g(f(x)) = g(f(x + t))$. Feb 17, 2021 at 0:00