Is $\frac1p(\sin^px+\cos^px)-\frac1q(\sin^qx+\cos^qx)$ constant for some reals $p$ and $q$. We know for
$$f(x)=\dfrac14(\sin^4x+\cos^4x)~~~;~~~g(x)=\dfrac16(\sin^6x+\cos^6x)$$
have $f(x)-g(x)=\dfrac{1}{12}$. My question is 

Are there other real $p$ and $q$ such that 
  $$f(x)=\dfrac1p(\sin^px+\cos^px)~~~;~~~g(x)=\dfrac1q(\sin^qx+\cos^qx)$$
  give us $f(x)-g(x)=C$ for a real constant $C$?

I had some idea but they were not useful. Thanks.
 A: Let $k \geqslant 1$ be a natural number and let
$$
f_k(x)=\frac 1k \left( \cos^k x+\sin^k x\right).
$$
It is easy to see that if $k \geqslant 5,$ then
$$
f_k^{(4)}(0)=3k-2
$$
(a math software can help a little here). Further,
$$
f_1^{(4)}(0)=1, f_2^{(4)}(0)=0, f_3^{(4)}(0)=7, f_4^{(4)}(0)=16.
$$
Thus we see that whenever $k < m$ are natural numbers, then
$$
f_k^{(4)}(0)=f_m^{(4)}(0) \iff (k,m)=(4,6).
$$
A: Yes, $p=q=2$, since then:
$$f(x)=\tfrac12(\sin^2x+\cos^2x)=\tfrac12=g(x)$$
and so $f(x)-g(x)=0$.
A: If we consider the derivative of $f(x)-g(x)$ then in order to be constant the result must be equal to 0. So after doing some operations
$$sin^{q-2}(x)[1-sin^{p-q}(x)]=cos^{q-2}(x)[1-cos^{p-q}(x)]$$
Therefore if it is equal to 0 within any value, taking the sine and cosine of $30$ degrees and calling $q-2=a$ and $p-q=b$ after some operations:
$$2^b=\frac{\sqrt{3}^{a+b}-1}{\sqrt{3}^{a}-1}$$
Since both sides must be integers for positive integer values of a and b then $a=2n$ and $b=2m$ studying the result the only possible integer solution of that is $n=1$ and $m=1$ therefore $q=4$ and $p=6$ which we know it works.
A: Let $p,q$ be integers. If $p=q$, one has nothing to do. Suppose $p<q$. Let
$$F(x)=\dfrac1p(\sin^px+\cos^px)-\dfrac1q(\sin^qx+\cos^qx)$$
and then
\begin{eqnarray}
F'(x)&=&\sin^{p-1}x\cos x-\cos^{p-1}x\sin x-\sin^{q-1}x\cos x+\cos^{q-1}x\sin x\\
&=&\sin x\cos x(\sin^{p-2}x-\cos^{p-1}x-\sin^{q-2}x-\cos^{q-1}x).
\end{eqnarray}
If $f(x)-g(x)$ is constant, then $F'(x)\equiv0$ and hence
$$ \sin^{p-2}x-\cos^{p-1}x-\sin^{q-2}x-\cos^{q-1}x\equiv0.$$
Let
$$ h(x)=\sin^{p-2}x-\cos^{p-1}x-\sin^{q-2}x-\cos^{q-1}x $$
and then
\begin{eqnarray}
h'(x)&=&(p-2)\sin^{p-3}x\cos x+(p-2)\cos^{p-3}x\sin x-(q-2)\sin^{q-3}x\cos x-(q-2)\cos^{q-3}x\sin x\\
&=&\sin x\cos x\big\{(p-2)\big[\sin^{p-4}x+\cos^{p-4}x\big]-(q-2)\big[\sin^{q-4}x+\cos^{q-4}x\big]\big].
\end{eqnarray}
Since $h(x)\equiv0$, $h'(x)\equiv0$. Clearly if $p=2$, then $q=2$. Suppose $p>2$. Let $x=\frac{\pi}{4}$ and then one has
$$ (\frac{\sqrt{2}}{2})^{p}=\frac{q-2}{p-2}(\frac{\sqrt{2}}{2})^{q}$$
or
$$ 2^{\frac12(q-p)}=\frac{q-2}{p-2}. $$
Let $p-2=2m,q-2=2n$ ($m<n$) and then one has
$$ 2^{n-m}=\frac{n}{m} $$
from which one has $n=m2^r$ ($r>0$). So
$$ 2^{m(2^r-1)}=2^r$$
or $$ m(2^r-1)=r. $$
Noting if $r>1$, $2^r-1>r$ and hence $m(2^r-1)>r$, one must have $r=1$ and hence $m=1,n=2$. Thus $p=4,q=6$.
