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? $\endgroup$ Oct 20, 2018 at 21:55
  • 2
    $\begingroup$ @RobertHoward $A$ should be between $0$ and $2$, but how do I find the the maximum value ? $\endgroup$
    – Sooraj S
    Oct 20, 2018 at 21:58
  • 1
    $\begingroup$ 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, 2018 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. $\endgroup$ Oct 20, 2018 at 22:06

3 Answers 3


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, 2018 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 $\approx 0.000058575$)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.