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Function $(\tan{x})^{\sin{2x}}$ has its minimun for $x=u\in(0,\frac{\pi}{2})$ and maximum for $x=v\in(0,\frac{\pi}{2})$.

Find $u+v$

Having calculated the derivative: $$\frac{d}{dx}(\tan{x})^{\sin{2x}} = (\tan{x})^{\sin{2x}}\cdot(2\cos{2x\cdot\ln{\tan{x}}+2})$$ I find it quite difficult to find roots of it (as if there was some other special way/technique to approach this problem)

Can You give me any hint? I will appreciate everything.

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Note that $\tan x>0$ for $x\in (0,\pi/2)$. Let $y=\tan^{\sin 2x} x$ then taking logarithms yields $$\ln y=\sin 2x\ln \tan x\Rightarrow \frac{y'}{y}=2\cos 2x\ln\tan x+\sin 2x\frac{(\tan x)'}{\tan x}$$ Overall using definition of $y$ and $(\tan x)'=1/\cos^2 x$ one obtains $$y'=\tan^{\sin 2x} (2\cos 2x\ln\tan x+2)$$ Therefore $y'=0$ implies $$2\cos 2x\ln\tan x+2=0\Leftrightarrow \cos 2x\ln\tan x=-1$$ since $\tan^{\sin 2x}>0$ for all $x\in(0,\pi/2)$. Note that $g(x):=\cos 2x\ln\tan x$ is strictly increasing on the interval $(0,\pi/4)$ since $$g'(x)=-2\sin 2x\ln\tan x+\frac{\cos 2x}{\cos x\sin x}>0$$ moreover $\lim_{x\to 0^+}g(x)=-\infty$ and $g(\pi/4)=0$ therefore $g(x)$ intersects the horizontal line $y=-1$ only once. On the other side $g(x)$ is strictly decreasing for $x\in (\pi/4,\pi/2)$ since $g'(x)<0$ there. Likewise $g(\pi/4)=0$ and $\lim_{x\to\pi/2^-}g(x)=-\infty$ (since $\cos 2x<0$ for $x\in(\pi/4,\pi/2)$) Hence $g(x)$ intersects the line $y=-1$ only once more on the interval $(\pi/4,\pi/2)$. On the other hand we have $$\cos 2x\ln\tan x=\cos(2(\frac{\pi}{2}-x))\ln\tan(\frac{\pi}{2}-x)$$ this implies that $x\in(0,\pi/4)$ solves $g(x)=-1$ if and only if $(\pi/2-x)$ solves the same equation. Therefore the two roots $x_1$ and $x_2$ are related by $x_2=\pi/2-x_1$. Since $x_1$ and $x_2$ are unique the sum of the roots of $g(x)=-1$ is then equal to $$x_1+x_2=\frac{\pi}{2}$$

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Since $u,\,v$ are roots of $1+\cos 2x\ln\tan x$, which is invariant under $x\mapsto \pi/2-x$, $u+v=\pi/2$.

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  • $\begingroup$ I think you mean $\pi$, good observation $\endgroup$ – King Tut Apr 2 '18 at 10:05
  • $\begingroup$ @KingTut No, I was right. $\endgroup$ – J.G. Apr 2 '18 at 12:10
  • $\begingroup$ Yes I made error $\endgroup$ – King Tut Apr 2 '18 at 13:36
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Since logarithm and exponential are increasing functions, inverse to one another, you can study instead $$ f(x)=\sin 2x \log\tan x=\sin 2x(\log\sin x-\log\cos x) $$ Its derivative is \begin{align} f'(x) &=2\cos2x\log\tan x+\sin 2x\left(\frac{\cos x}{\sin x}+\frac{\sin x}{\cos x}\right)\\[6px] &=2\cos2x\log\tan x + 2 \\[6px] &=2(1+\cos 2x\log\tan x) \end{align} which agrees with your computation (apart from my preference for $\log$ over $\ln$). Now consider $$ \cos2x=2\cos^2x-1=\frac{2}{1+\tan^2x}-1=\frac{1-\tan^2x}{1+\tan^2x} $$ Since $\tan x>0$ in your setting, you can set $t=\tan^2x$ and study instead $$ g(t)=2(1+t)+(1-t)\log t \qquad (t>0) $$ and find its zeros. We have $$ g'(t)=1-\log t+\frac{1-t}{t}=1-\log t+\frac{1}{t} $$ and $$ g''(t)=-\frac{1}{t}-\frac{1}{t^2}<0 $$ This means $g$ is concave. Moreover $$ \lim_{t\to0}g(t)=-\infty \qquad \lim_{t\to\infty}g(t)=-\infty $$ so we conclude that $g$ has a single point of maximum $t_0$, where $\log t_0=1+\frac{1}{t_0}$; now $$ g(t_0)=2+2t_0+(1-t_0)\left(1+\frac{1}{t_0}\right)=\frac{(1+t_0)^2}{t_0}>0 $$ We conclude that $g$ vanishes twice, say at $t_1$ and $t_2$, being negative over $(0,t_1)$, positive over $(t_1,t_2)$ and negative again over $(t_1,\infty)$.

The same behavior will have $f'$: negative over $(0,\arctan\sqrt{t_1})$, positive over $(\arctan\sqrt{t_1},\arctan\sqrt{t_2})$ and negative over $(\arctan\sqrt{t_2},\pi/2)$.

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Now that you know the existence and uniqueness of $u=\arctan\sqrt{t_1}$ and $v=\arctan\sqrt{t_2}$, it should be easy to exploit the symmetry and find their sum.

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