# General Formula for $\sin^n(x)+\cos^n(x)$

I wanted to derive an expression for $$\sin^n(x)+\cos^n(x)$$ and thought I should start of by deriving a few basic powers myself, and finding a pattern. Apart from our usual $$\sin^2(x)+\cos^2(x)=1$$, I derived the following myself,

Subsequently, I attempted to find a general form for odd numbers. $$\sin^{2n-1}+\cos^{2n-1}=(1-\sum_{i=1}^{n-1} (\sin(x)\cos(x))^i),n\in\mathbb{N},n\neq1$$

And even powers, by induction I had assumed the general form was: $$\sin^n(x)+\cos^n(x)=1-\frac{n}{2}\sin^2(x)\cos^2(x),n\in\mathbb{N}$$

But, on checking this on Geogebra, I found these only hold for $$n\in{2,3}$$

Does anyone know a general formula for a sum of powers of sine and cosine? I've been unable to find a clear pattern.

• $x^n+y^n= (x+y)(x^{n-1}+x^{n-2}y-x^{n-3}y^2 + \dots - xy^{n-2} + y^{n-1})$ for every odd $n$ and any $x,y$. That gives a lead for odd $n$. In general, I have a feeling these identities will hold just with the extra condition $x^2+y^2=1$. Commented Apr 30 at 4:52
• Beyond that, deMoivre's formula is the natural playing field for trigonometric identities. Commented Apr 30 at 4:54
• What is your intent in this question? There's no single equality for such an expression, nor is there a canonically "simplest" equivalent expression for $n>2$. Any other expression would be chosen based on what you want to do with it. Commented Apr 30 at 4:54
• Well, there are many different identities regarding these goniometric expressions. You can for example express it only using cosines, you can express them without using any powers, you can express them as a long sum or maybe even like a long product. So it is not obvious what you really want to do. If your pattern breaks at some n, then evaluating it at that n manually may help. Commented Apr 30 at 5:00
• What would you like the expression to be in terms of? Is it powers of $\sin(x)$ and $\cos(x)$ exclusively? Commented Apr 30 at 5:54

If $$n=2^uk$$, where $$k$$ is prime, $$\sin^nx+\cos^n x=(\sin^{\frac{n}{k}}x+ \cos^{\frac{n}{k}} x)\\\sum_{r=0}^{k-1} (-1)^r\sin^{\frac{n}{k}{(k-r-1)}} x \cos ^{\frac{n}{k} r} x \\ =(\sin^{\frac{n}{k}}x+ \cos^{\frac{n}{k}} x)\\\sum_{r=0}^{k-1} (-1)^r \sin ^{\frac{n}{k}(k-1)} x \cot ^{\frac{n}{k}r} x$$ $$=(\sin^{\frac{n}{k}}x+ \cos^{\frac{n}{k}} x)\\\sin^{\frac{n}{k}(k-1)} x\sum_{r=0}^{k-1} (-1)^r \cot ^{\frac{n}{k}r} x$$ $$=\sin^n x(1 + \cot^{\frac{n}{k}x})\sum_{r=0}^{k-1} (-1)^r \cot ^{\frac{n}{k}r} x$$

just used the fact that if $$n$$ is odd $$(a^n+b^n)=(a+b)\sum_{r=0}^{n-1}(-1)^r a^{n-r-1}b^r$$

Simplifying is not possible unless $$n$$ is already provided

• I assume $\cot$ is supposed to be $\cos$? And $k$ can be any odd number, not necessarily a prime? Commented May 1 at 4:25
• @TorstenSchoeneberg To your first question, the $\cot$ is supposed to be $\cot$. The step where the cotangent appears uses $\sin^{r-s}x \cos^sx = \sin^rx\cot^sx$. Commented May 1 at 6:27

Not sure if this is going to help, but i think this works for n even: if n is even, put n=2k where k is any natural number. denote $$sin^nx+cos^nx=(sinx)^{2k}+(cosx)^{2k}=T_k$$.

Denote $$sin^2xcos^2x=p$$ for brevity.

Note that $$T_k$$ satifies the recurrence relation $$T_{k+2}-T_{k+1}+pT_{k}=0$$ With boundary conditions $$T_1=1,T_2=1-2p$$ Now you can use this to recursively generate the subsequent terms as functions of p.

For even powers one can use the Chebyshev polynomials of the first kind $$T_n$$ defined by $$T_n(\cos{x})=\cos{n x}$$ to find a unique sum such that we have : $$\cos^n x +\sin^n x= \frac{1}{2^k}\sum_m a_m T_{4m}(\cos x)$$ with integer coefficients $$a_m$$ and $$k such that : $$2^k= \sum_m a_m$$ This allows us to define : \begin{align} \cos^4 x +\sin^4 x &= \frac{1}{2^2}\left(3+\cos 4 x\right)\\ \cos^6 x +\sin^6 x &= \frac{1}{2^3}\left(5+3 \cos 4 x\right)\\ \cos^8 x + \sin^8 x &= \frac{1}{2^6} \left(35 + 28 \cos 4 x + \cos 8 x\right)\\ \cos^{10} x + \sin^{10} x &= \frac{1}{2^7}\left(63 + 60 \cos 4 x + 5 \cos 8 x\right) \\ \cos^{12} x + \sin^{12} x &= \frac{1}{2^{10}}\left(462 +495 \cos 4 x +66 \cos 8 x + \cos 12 x\right) \\ \cos^{14} x + \sin^{14} x &= \frac{1}{2^{11}}\left(858 +1001 \cos 4 x +182 \cos 8 x + 7\cos 12 x\right) \\ \cos^{16} x + \sin^{16} x &= \frac{1}{2^{14}}\left(6435 +8008 \cos 4 x +1820 \cos 8 x + 120\cos 12 x + \cos 16 x\right) \\ ....\\ \end{align} To see how this pattern emerges we start for $$n=4$$ by replacing $$\sin x$$ by $$\sqrt{1-\cos^2 x}$$ and notice that for even powers the square root vanishes.

Substituting $$\cos x$$ by $$y$$ we also see the relationship to the Chebyshev polynomial $$T_n$$ : $$y^4 + \left(\sqrt{1 - y^2}\right)^4 = y^4 + (1 - y^2)^2 = \frac{1}{2^2} \left(3 + T_4(y)\right)$$ For odd powers it will be more difficult to come up with a system of equations but there is hope because we have the second kind of Chebyshev polynomials $$U_n$$ with the relationship: \begin{align} T_n(x) + U_{n-1}(x)\,\sqrt{x^2-1} = \left(x + \sqrt{x^2-1}\right)^n\\ \end{align} EDIT For the odd powers we obtain the following equations: \begin{align} \cos^3 x +\sin^3 x &= \frac{1}{2^2}\left(3 (\cos x + \sin x) + \cos 3 x - \sin 3 x\right)\\ \cos^5 x +\sin^5 x &= \frac{1}{2^4}\left(10 (\cos x + \sin x) + 5 (\cos 3 x - \sin 3 x) + \cos 5 x + \sin 5 x\right)\\ \cos^7 x +\sin^7 x &= \frac{1}{2^6}\left(35 (\cos x + \sin x) + 21 (\cos 3 x - \sin 3 x) + 7 (\cos 5 x + \sin 5 x) + \cos 7 x - \sin 7 x\right)\\ \cos^9 x + \sin^9 x &= \frac{1}{2^8} \left(126 (\cos x + \sin x) +84 (\cos 3 x - \sin 3 x) +36 (\cos 5 x + \sin 5 x) + 9 (\cos 7 x - \sin 7 x) +\cos 9 x +\sin 9 x\right)\\ ....\\ \end{align}

I don't know of any refernces for this pattern and would be happy if someone could note them in the comments.