Question

Inverse circular functions,Principal values of ${sin}^{-1}x,{cos}^{-1}x,{tan}^{-1}x$.
${tan}^{-1}x+{tan}^{-1}y={tan}^{-1}\frac{x+y}{1-xy}$, $xy<1$
$\pi +{tan}^{-1}\frac{x+y}{1-xy}$, $xy>1$.
(a) ${sin}^{-1}\left(3x/5\right)+{sin}^{-1}\left(4x/5\right)={sin}^{-1}x$
(b) ${cos}^{-1}x+{sin}^{-1}\left(\frac{1}{2}x\right)=\frac{\pi }{6}$
(c) If $a\le {tan}^{-1}\left(\frac{1-x}{1+x}\right)\le b$ where $0\le x\le 1$ then $(a,b)=$
(a) $(0,\pi )$
(b) $(0,\pi /4)$
(c) $(-\pi /4,\pi /4)$
(d) $(\pi /4,\pi /2)$
(d) If $a\le {\left({sin}^{-1}x\right)}^3+{\left({cos}^{-1}x\right)}^3\le b$ then (a,b) is equal to $(\frac{{\pi }^3}{32},\frac{7{\pi }^3}{8})$.

Answer 1

(a) Let ${sin}^{-1}\frac{3x}{5}=\theta ,{sin}^{-1}\frac{4x}{5}=\varphi ,{sin}^{-1}x=\Psi$
$\therefore$ $sin\theta =\frac{3x}{5},sin\varphi =\frac{4x}{5}.sin\Psi =x$
Now the given equation can be written as
$\theta +\varphi =\Psi$ whence $sin(\theta +\varphi )=sin\Psi$.
or $sin\theta cos\varphi +cos\theta sin\varphi =sin\Psi$
or $\frac{3x}{5}\sqrt{(1-\frac{16}{25}{x}^2)}+\frac{4x}{5}\sqrt{(1-\frac{9x^2}{25})}=x$
or $x[3\sqrt{25-16x^2}+4\sqrt{25-9x^2}]=25x$
One root is $x=0$ and the other roots are given by
$4\sqrt{25-9x^2}=25-3\sqrt{25-16x^2}$
Squaring,
$16(25-9x^2)=625-150\sqrt{25-16x^2}-9(25-16x^2)$
or $150\sqrt{25-16x^2}=450$
whence $25-16x^2=9$ or $x^2=1$
$\therefore$ $x=\pm 1$.
A check shows that $x=0,1,-1$ are all roots of the given equation.
(b) The given equation is
${cos}^{-1}x+{sin}^{-1}\left(x/2\right)=\pi /6$
Since ${sin}^{-1}\left(x/2\right)=\pi /2-{cos}^{-1}\left(x/2\right)$, the given equation is equivalent to the equation
${cos}^{-1}x+\pi /2-{cos}^{-1}\left(\pi /2\right)=\pi /6$
or ${cos}^{-1}x-{cos}^{-1}\left(\pi /2\right)=\pi /6-\pi /2=-\pi /3$
or ${cos}^{-1}x={cos}^{-1}\left(x/2\right)-\pi /3$
$={cos}^{-1}\left(x/2\right)-{cos}^{-1}\left(1/2\right)$
$=cos[\frac{x}{2},\frac{1}{2}+\sqrt{1-\frac{x^2}{4}}.\sqrt{1-\frac{1}{4}}]$
$\therefore$ $x=\frac{x}{4}+\sqrt{(1-\frac{x^2}{4})}.\frac{\sqrt{3}}{2}$
$\therefore$ $3x^2=4-x^2$ or $4x^2=4$
$\therefore$ $x=\pm 1$. But the value $x=-1$ does not satisfy the given equation.
Hence $x=1$ is the only root of the given equation.
(c) Ans. (b).
$E={tan}^{-1}\frac{1-x}{1+x}={tan}^{-1}1-{tan}^{-1}x=\frac{\pi }{4}-{tan}^{-1}x$
Now $\varsigma \le x\le 1$ $\therefore$ $0\le {tan}^{-1}x\le \frac{\pi }{4}$
Multiplying by minus sign and reversing the inequality, $0\ge -{tan}^{-1}x\ge -\pi /4$.
$\therefore$ $-\frac{\pi }{4}\le -{tan}^{-1}x\le 0$
Add $\frac{\pi }{4}$ to both sides
$\therefore$ $0\le \frac{\pi }{4}-{tan}^{-1}x\le \frac{\pi }{4}$ or $0\le E\le \frac{\pi }{4}$
(d) $E={({sin}^{-1}x+{cos}^{-1}x)}^3-3\left({sin}^{-1}x{cos}^{-1}x\right)\times ({sin}^{-1}x+{cos}^{-1}x)$
$E=\frac{{\pi }^3}{8}-\frac{3\pi }{2}\left({sin}^{-1}x{cos}^{-1}x\right)$.......(1)
Now let ${sin}^{-1}=t$ $\therefore$ ${cos}^{-1}x=\frac{\pi }{2}-t$
$E=\frac{{\pi }^3}{8}-\frac{3\pi }{2}\left[t(\frac{\pi }{2}-t)\right]$
$=\frac{{\pi }^3}{8}+\frac{3\pi }{2}[t^2-\frac{\pi }{2}t+{\left(\frac{\pi }{4}\right)}^2+{\left(\frac{\pi }{4}\right)}^2]$
$=\frac{{\pi }^3}{8}+\frac{3\pi }{2}[{(t-\frac{\pi }{4})}^2-\frac{{\pi }^2}{16}]$
$=\frac{{\pi }^3}{8}-\frac{3{\pi }^3}{32}+\frac{3\pi }{2}{(t-\frac{\pi }{4})}^2$
$=\frac{{\pi }^3}{32}+\frac{3\pi }{2}{(t-\frac{\pi }{4})}^2$ ........(2)
From above, least value is $\frac{{\pi }^3}{32}$........(3)
Again $-\frac{\pi }{2}\le {sin}^{-1}x\le \frac{\pi }{2}$ or $-\frac{\pi }{2}\le t\le \frac{\pi }{2}$
$\therefore$ $-\frac{\pi }{2}-\frac{\pi }{4}\le t-\frac{\pi }{4}\le \frac{\pi }{2}-\frac{\pi }{4}$
or $-\frac{3\pi }{4}\le (t-\frac{\pi }{4})\le \frac{\pi }{4}$
Above is a part of the double inequality
${\left(\frac{\pi }{4}\right)}^2\le {(t-\frac{\pi }{4})}^2\le \frac{{9\pi }^{}}{16}$
Hence the greatest value by (2) is
$\frac{{\pi }^3}{32}+\frac{3\pi }{2}\frac{9{\pi }^2}{16}=\frac{28{\pi }^2}{32}=\frac{7{\pi }^3}{8}$ .......(4)
$\therefore$ $(a,b)=(\frac{{\pi }^3}{32},\frac{7{\pi }^3}{8})$ by (3) and (4)