IMO Practice Test — Inverse Trigonometric Functions
6 Questions • 20 min • Olympiad level
20:00
Question 1 of 6
hard
$\cot^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)$ for $x \in (0,\pi/4)$ equals:
$x/2$
$\pi/2-x$
$x$
$\pi/4+x/2$
Explanation: $\sqrt{1\pm\sin x}=\cos(x/2)\pm\sin(x/2)$. The ratio simplifies to $\cot(x/2)$. So $\cot^{-1}(\cot(x/2))=x/2$.
Question 2 of 6
hard
If $\tan^{-1}x + \tan^{-1}y + \tan^{-1}z = \pi/2$, then $xy+yz+zx$ equals:
$0$
$1$
$xyz$
$-1$
Explanation: $\tan^{-1}x+\tan^{-1}y+\tan^{-1}z=\pi/2$ iff $xy+yz+zx=1$ (a known identity from the triple tangent formula).
Question 3 of 6
hard
The number of solutions of $\tan^{-1}(x+1)+\tan^{-1}(x-1)=\tan^{-1}(8/31)$ is:
0
1
2
3
Explanation: Solving gives $4x^2+31x-8=0$, i.e. $(4x-1)(x+8)=0$. $x=1/4$ is valid; $x=-8$ fails the domain check. One solution.
Question 4 of 6
hard
$\sin^{-1}\left(\sin 10\right)$ (radians) equals:
$10$
$10-3\pi$
$3\pi-10$
$10-2\pi$
Explanation: $10 \approx 3.18\pi$, so $10 \in (3\pi-\pi/2, 3\pi+\pi/2)$? Actually: $3\pi \approx 9.42$, $10 \in (3\pi, 3\pi+\pi/2)$. Then $\sin(10)=\sin(10-3\pi)=\sin(3\pi-10)$ only if ... Let me recalculate: $\pi/2 < 10-3\pi\approx 0.575 <\pi/2$? No: $10-3\pi\approx 0.575$, which is in $(0,\pi/2)\subset [-\pi/2,\pi/2]$. But $\sin(10)=\sin(\pi-10+3\pi)$? Better: $\sin(10)=\sin(10-4\pi+4\pi)$... Actually $10 = 3\pi + (10-3\pi)$, and $\sin(3\pi+\theta)=-\sin\theta$. So $\sin(10)=-\sin(10-3\pi)=\sin(-(10-3\pi))=\sin(3\pi-10)$. Since $3\pi-10\approx -0.575\in[-\pi/2,\pi/2]$, $\sin^{-1}(\sin 10)=3\pi-10$.
Question 5 of 6
hard
If $f(x)=\sin^{-1}x + \cos^{-1}x + \tan^{-1}x$, the range of $f$ is:
$[\pi/4, 3\pi/4]$
$\{\pi/2\}$
$[0,\pi]$
$[-\pi/2,\pi]$
Explanation: $\sin^{-1}x+\cos^{-1}x=\pi/2$ always for $x\in[-1,1]$. So $f(x)=\pi/2+\tan^{-1}x$ for $x\in[-1,1]$. $\tan^{-1}x$ ranges from $-\pi/4$ to $\pi/4$ on $[-1,1]$. So $f$ ranges from $\pi/4$ to $3\pi/4$.
Question 6 of 6
hard
$\cos^{-1}\frac{4}{5} + \cos^{-1}\frac{12}{13}$ equals:
$\cos^{-1}\frac{33}{65}$
$\cos^{-1}\frac{56}{65}$
$\cos^{-1}\frac{16}{65}$
$\pi - \cos^{-1}\frac{33}{65}$
Explanation: Let $\alpha=\cos^{-1}(4/5)$, $\beta=\cos^{-1}(12/13)$. $\cos\alpha=4/5,\sin\alpha=3/5$; $\cos\beta=12/13,\sin\beta=5/13$. $\cos(\alpha+\beta)=(4/5)(12/13)-(3/5)(5/13)=48/65-15/65=33/65$.