Is there a simpler way to determine m, n, p, such that the following holds for all reals? I am given the following equation and I am asked to find $m$, $n$ and $p$, so that the equation holds for all reals:
$$ \sin^4x + \cos^4x + m(\sin^6x + \cos^6x) + n(\sin^8x + \cos^8x) + p(\sin^{10}x + \cos^{10}x) = 1, \space \forall x \in \mathbb R $$
I have managed to solve it by, in advance, calculating the following power reduction formulas, and then applying them to the equation:
$$
\sin^4x + \cos^4x = 1 - \frac12\sin^2{2x} \\
\sin^6x + \cos^6x = 1 - \frac34\sin^2{2x} \\
\sin^8x + \cos^8x = 1 - \sin^2{2x} + \frac18\sin^4{2x} \\
\sin^{10}x + \cos^{10}x = 1 - \frac54\sin^2{2x} + \frac5{16}\sin^4{2x}
$$
As a result, I managed to simplify it to the following, which is only in terms of powers of $\sin{2x}$:
$$ \left( 1+m+n+p \right) - \left( \frac12 + \frac{3m}4 + n + \frac{5p}4 \right) \sin^2{2x} + \left( \frac{n}8 + \frac{5p}{16} \right) \sin^4{2x} = 1, \space \forall x \in \mathbb R $$
I then came to the conclusion that the only possible way for which this can be true is if:
$1+m+n+p=1$, $ \frac12 + \frac{3m}4 + n + \frac{5p}4 = 0 $ and $ \frac{n}8 + \frac{5p}{16} = 0 $.
By solving the system of equations below I arrive at the solutions $m=6$, $n=-10$, $p=4$.
$$
\left\{ 
\begin{aligned}
1 + m + n + p = 1 \\ 
\frac12 + \frac{3m}4 + n + \frac{5p}4 = 0 \\ 
\frac{n}8 + \frac{5p}{16} = 0
\end{aligned}
\right. 
$$
My question is if my reasoning is correct and if there exists any simpler way to solve the problem.
 A: Consider the polynomial
$$P(t):=-1+t^2+(1-t)^2+mt^3+m(1-t)^3+nt^4+n(1-t)^4+pt^5+p(1-t)^5\,.$$
If $t=\sin^2(x)$ is a root for every $x\in\mathbb{R}$, then $P\equiv 0$ identically.  In particular,
$$0=[t^4]\,P(t)=2n+5p\,,$$
$$0=[t^2]\,P(t)=2+3m+6n+10p\,,$$
and
$$0=[t^0]\,P(t)=P(0)=m+n+p\,,$$
where $[t^k]\,P(t)$ denotes the coefficient of the term $t^k$ in $P(t)$ for every $k=0,1,2,\ldots$.  (Here, the coefficients are found via the Binomial Theorem.)  It follows immediately that $m=6$, $n=-10$, and $p=4$.
We still need to show that $P\equiv 0$ is indeed the case.  First, $\deg(P)\le4$ can be easily seen.  Therefore, we need to check that $[t]\,P(t)=0$ and $[t^3]\,P(t)=0$, but
$$[t]\,P(t)=-2-3m-4n-5p=-\big([t^2]\,P(t)\big)+\big([t^4]\,P(t)\big)=0$$
and
$$[t^3]\,P(t)=-4n-10p=-2\,\big([t^4]\,P(t)\big)=0\,.$$
Thus, $P$ is indeed the zero polynomial.
A: Your reasoning is correct. You may save a little algebraa by formulating it as follows. Define $\, f(t) := t^2 + m t^3 + n t^4 + p t^5. \,$ We want the equation
 $\, f(t) + f(1-t) - 1 = 0\,$ to be true for all $\, t = \sin^2(x). \,$ If you expand the left side, it is a fourth degreee polynomial in $\,t.\,$ All of its coefficients must be zero. This gives five linear equations in the four unknowns and the solution is as you found. The advantage of this approach is that it avoids the need to express $\, \sin^{2n}(x) + \cos^{2n}(x) \,$ in terms of even powers of $\, \sin(2x). \,$ All that is needed is to know the identity $\, \sin^2 x + \cos^2 x = 1. $
