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 $A$

  • $\begingroup$ Both $\sin$ and $\cos$ have a minimum value of $-1$ and a maximum value of $1$, so what can you deduce from that? $\endgroup$ – Robert Howard Oct 20 '18 at 21:55
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    $\begingroup$ @RobertHoward $A$ should be between $0$ and $2$, but how do I find the the maximum value ? $\endgroup$ – ss1729 Oct 20 '18 at 21:58
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    $\begingroup$ observe that since $\sin x$ and $\cos x$ are less or equal to one then we have $ \sin^{20} x + \cos^{48} \leq \sin^2x +\cos^2 x = 1$ $\endgroup$ – ALG Oct 20 '18 at 22:05
  • $\begingroup$ Good point; I should have looked at the question more carefully. From the equation you ended with, I would try re-expressing all the powers of $\sin^2x$ in terms of $\cos^2x$ and see if that helps. $\endgroup$ – Robert Howard Oct 20 '18 at 22:06

If $0\leq a\leq 1$ then $a^2\leq a$, so we use that repetadly

$$A=\sin^{20}x+\cos^{48}x\leq \sin^{2}x+\cos^{2}x =1$$

So $A_{\max}=1$ and it is achieved at $x=0$.

For a minimum I don't see quick solution. I would try like this: Let $t= \cos^2x$ and then search for the minumum of $$g(t) = (1-t)^{10}+t^{24}$$ where $0\leq t\leq 1$.


Note that $A = (1 - \cos^2 x)^{10} + \cos^{48} x$. Letting $t = \cos^2 x$, $t \in [0,1]$, we have $ A = (1-t)^{10} + t^{24}$. This function has maximum in 0 and 1, and $A_{max} = 1$. The minimum is found by differentiating and solving $24t^{23} - 10(1-t)^9 = 0$, which leads (numerically) to $t_{min} \simeq 0,643187$, and correspondingly to $A_{min} = (1-t_{min})^{10} + t_{min}^{24} \simeq 0,0000585751.$

  • $\begingroup$ WolframAlpha gives the same minimum, achieved at $x=.640178.$ $\endgroup$ – saulspatz Oct 20 '18 at 22:20

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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