<s>[INST] Find the range of $A$ if $A=\sin^{20}x+\cos^{48}x$ 
Find  the range of $A$ if $$A=\sin^{20}x+\cos^{48}x$$

$$
A'=20\sin^{19}x\cos x-48\cos^{47}x\sin x=0\implies5\sin^{19}x\cos x=12\cos^{47}x\sin x\\
\implies5\sin^{18}x=12\cos^{46}x
$$
How do I proceed further and prove that $A\in(0,1]$ ?
Is it possible to find the range of $A$ without using differentiation ?
Note: $\sin^2 x,\cos^2 x\in[0,1]\implies A\in[0,2]$ but $2$ is not "the" maximum value of $A$
 [/INST] Apart from the trivial upper bound $A\le 2$, we have the stronger (and sharp - try $x=0$) bound
$$ \tag1A\le 1.$$
Consider $f(x):=(1-x)^{10}+x^{24}$ for $0\le x\le 1$. Then $f'(x)=24x^{23}-10(1-x)^9$ is strictly increasing (as each summand is) on $[0,1]$, hence has at most one root there. As $f'(0)=-10$ and $f'(1)=24$, we conclude that there is exactly one such root $\alpha$. As $f'$ goes from negative to positive, $f$ must have a local minimum there. We conclude that $f$ has its only minimum at $\alpha$ and its maximum at the boundary - in fact, at both ends of the boundary since $f(0)=f(1)$.  As $A=f(\cos^2 x)$ and $\cos^2 x$ ranges from $0$ to $1$, inclusive, we conclude that the maximal value of $A$ is also $1$ (thus proving $(1)$), and the minimum value of $A$ is $f(\alpha)$.
Using $(1-\alpha)^9=\frac{12}5\alpha^{23}$, we have 
$$ f(\alpha)=(1-\alpha)^{10}+\alpha^{24}=(1-\alpha)\cdot\frac{12}5\alpha^{23}+\alpha^{24}=\alpha^{23}\cdot \frac{12-7\alpha}5=(1-\alpha)^9\cdot \frac{12-7\alpha}{12},$$
so certainly $$\min A=\min  f>0,$$ but not by much.
From $f'(\frac 35)=24\frac{3^{23}}{5^{23}}-10\frac{2^{9}}{5^{9}}=\frac{24\cdot 3^{23}-10\cdot 2^95^{14}}{5^{23}}<0$(!), we conclude that $\alpha>\frac35$ and hence
$$ f(\alpha)=(1-\alpha)^9\cdot \frac{12-7\alpha}{12}<(1-\tfrac35)^9\cdot \frac{12-7\cdot\frac35}{12}=\frac{1664}{9765625}\approx 0.00017$$
(whereas the true minimal value is $\approx 0.000058575$)
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