Writing the $sin$ $cos$ power sum as a sum of multiple angles.

Question

Writing the $sin$ $cos$ power sum as a sum of multiple angles.

Trying to answer the se question, which did not specify the multiple angle solutions, I started to look for a generalization and arrived at the following equations depending on even or odd values of $n$: \begin{align} \cos^n x +\sin^n x & = 2^{2-n} \sum_{k=0}^{\frac{n}{4}}{\frac{1}{2^{\left\lfloor\frac{1}{1+k}\right\rfloor}}\dbinom{n}{\frac{n}{2}+2 k} \cos (4 k x)} \;& \forall \; n\equiv 0\mod 2\\ \cos^n x + \sin^n x & = 2^{1-n} \sum_{k=1}^{\frac{n+1}{2}}{\binom{n}{\frac{n+1}{2}-k} \left( \cos ((2k-1)x)-(-1)^k\sin((2k-1)x)\right)} \; & \forall \; n\equiv 1\mod 2 \end{align}

Question 1: Can the first equation for even $n$ be written without the characteristic function$\lfloor .. \rfloor$?

Question 2: Or better , can both equations be combined into one? (There can be only one)

For $n=4,6$ we see that we get two multiple angle formulae for $\cos(4x)$ : $$\cos (4 x)=4 \left(\sin ^4x+\cos ^4x\right)-3=\frac{1}{6} \left(16 \left(\sin ^6x+\cos ^6x\right)-10\right) $$

Question 3: Are there other multiple angle formulae for which sums of powers exist?
There are many Multiple-Angle formulae listed at wiki identities and Wolfram Multiple-Angle but no sums of powers as far as I could see.

Question 4: Can you recommend any references to this subject?

Answer 1

The explanation for the $\cos(4x)$ part is simpler.

Let $A_{n}(x)=\cos^{2n}(x)+\sin^{2n}(x).$

This satisfies $A_0(x)=2, A_1(x)=1,$ and $$ \begin{align} A_{n+1}(x)&=(\sin^2(x)+\cos^2(x))A_n(x)-\sin^2(x)\cos^2(x)A_{n-1}(x)\\&=A_n(x)-\frac14\sin^2(2x)A_{n-1}(x)\\&=A_n(x)+\frac18(\cos(4x)-1)A_{n-1}(x). \end{align} $$

So what you really get is $A_n(x)$ is always a polynomial of $\cos(4x),$ with $A_2,A_3$ linear in $\cos(4x)$ because $A_0$ and $A_1$ are constants.

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The above uses a rule about sequences $c_n=ar_1^n+br_2^n,$ that $$c_{n+1}=(r_1+r_2)c_n-(r_1r_2)c_{n-1}.$$

In our case, $a=b=1,$ $r_1=\cos^2x, r_2=\sin^2 x.$

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If you write the polyn9mial as polynomials $p_n(y)$ such that $p_n\left(\frac{\cos(4x)-1}8\right)=A_n(x),$ you get $$p_0(y)=2, p_1(y)=1, p_{n+1}(y)=p_n+yp_{n-1}(y)$$

So $$ \begin{align} p_2(y)&=1+2y,\\p_3(y)&=1+3y,\\ p_4(y)&=1+4y+2y^2,\\ p_5(y)&=1+5y+5y^2 \end{align} $$

This lets us solve for $\cos(4x)$ in terms of a quadratic formula involving $A_4(x)$ or $A_5(x).$ For example,

$$\cos(4x)=-7\pm4\sqrt{2+2(\cos^8x+\sin^8x)}$$ or something like that. Obviously, it is only true for on of the $\pm$ signs. If that formula is correct, the $+$ sign is requires to bring the value in the range $[-1,1].$

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If you define $B_n(x)=\frac{\cos^{2n+1}(x)+\sin^{2n+1}(x)}{\cos(x)+\sin(x)}$ you get that $B_0(x)=1$ and $B_{-1}(x)=\frac2{\sin(2x)}$ and the recursion: $$B_{n+1}(x)=B_n(x)-\frac{\sin^2(2x)}4 B_{n-1}(x),$$ which makes, for $n\geq 0,$ $B_n(x)$ is a polynomial of $\sin(2x).$

If we let $y=\sin(2x)/2,$ then $q_0(y)=1, q_1(y)=1-y,$ and $$q_{n+1}(y)=q_n(y)-y^2q_{n-1}(y).$$