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 $$A$$

• Both $\sin$ and $\cos$ have a minimum value of $-1$ and a maximum value of $1$, so what can you deduce from that? – Robert Howard Oct 20 '18 at 21:55
• @RobertHoward $A$ should be between $0$ and $2$, but how do I find the the maximum value ? – ss1729 Oct 20 '18 at 21:58
• observe that since $\sin x$ and $\cos x$ are less or equal to one then we have $\sin^{20} x + \cos^{48} \leq \sin^2x +\cos^2 x = 1$ – ALG Oct 20 '18 at 22:05
• 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. – 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 $$A_{min} = (1-t_{min})^{10} + t_{min}^{24} \simeq 0,0000585751.$$

• WolframAlpha gives the same minimum, achieved at $x=.640178.$ – 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 $$\approx 0.000058575$$)