Inverse Trigonometric Functions: Difficult Problems with Solutions

Solution:
The sine is odd and has period [tex]2\pi[/tex]:
[tex]\sin \left(-\frac{17\pi}{6}\right)=-\sin \frac{17\pi}{6}=-\sin \frac{5\pi}{6}=-\frac{1}{2}[/tex]
The values of [tex]\arcsin [/tex] lie in [tex]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right][/tex], so the answer is [tex]\arcsin \left(-\frac{1}{2}\right)=-\frac{\pi}{6}[/tex]. The number [tex]-\frac{17\pi}{6}[/tex] is not in this interval.

Solution:
Let [tex]\alpha=\arcsin \frac{1}{3}\in\left(0, \frac{\pi}{2}\right)[/tex]. Then
[tex]\cos\alpha=\sqrt{1-\frac{1}{9}}=\frac{2\sqrt{2}}{3}\;\Rightarrow\;\arccos\frac{2\sqrt{2}}{3}=\alpha[/tex]
By [tex]\arccos(-a)=\pi-\arccos a[/tex] the sum equals [tex]\alpha+\pi-\alpha=\pi[/tex].

Solution:
Let [tex]\alpha=\arctan x\in\left(-\frac{\pi}{2}, 0\right)[/tex]. The angle [tex]\beta=-\frac{\pi}{2}-\alpha[/tex] is also in [tex]\left(-\frac{\pi}{2}, 0\right)[/tex] and
[tex]\tan \beta=\tan \left(-\frac{\pi}{2}-\alpha\right)=\cot \alpha=\frac{1}{x}[/tex]
So [tex]\arctan \frac{1}{x}=-\frac{\pi}{2}-\alpha[/tex] and the sum is [tex]-\frac{\pi}{2}[/tex]. Check: [tex]x=-1[/tex] gives [tex]-\frac{\pi}{4}-\frac{\pi}{4}=-\frac{\pi}{2}[/tex].

Solution:
Apply [tex]\tan (a+b)=\frac{\tan a+\tan b}{1-\tan a\tan b}[/tex] twice:
[tex]\frac{\frac{1}{2}+\frac{1}{5}}{1-\frac{1}{10}}=\frac{7}{9}, \quad\frac{\frac{7}{9}+\frac{1}{8}}{1-\frac{7}{72}}=\frac{\frac{65}{72}}{\frac{65}{72}}=1[/tex]
Each term is in [tex]\left(0, \frac{\pi}{4}\right)[/tex], so the sum is in [tex]\left(0, \frac{3\pi}{4}\right)[/tex]. The only angle there with tangent [tex]1[/tex] is [tex]\frac{\pi}{4}[/tex].

Solution:
Let [tex]\alpha=\arctan \frac{1}{7}[/tex] and [tex]\beta=\arctan \frac{1}{3}[/tex]. Then
[tex]\cos 2\alpha=\frac{1-\tan ^2\alpha}{1+\tan ^2\alpha}=\frac{1-\frac{1}{49}}{1+\frac{1}{49}}=\frac{24}{25}[/tex]
[tex]\tan 2\beta=\frac{2\times \frac{1}{3}}{1-\frac{1}{9}}=\frac{3}{4}, \quad\sin 4\beta=\frac{2\tan 2\beta}{1+\tan ^2 2\beta}=\frac{\frac{3}{2}}{\frac{25}{16}}=\frac{24}{25}[/tex]
So the difference is [tex]\frac{24}{25}-\frac{24}{25}=0[/tex].

Solution:
We have [tex]\arcsin \left(-\frac{1}{2}\right)=-\frac{\pi}{6}[/tex]. From [tex]\operatorname{arccot} (-a)=\pi-\operatorname{arccot} a[/tex] and [tex]\operatorname{arccot} a=\frac{\pi}{2}-\arctan a[/tex]:
[tex]\operatorname{arccot} (-\pi)=\pi-\frac{\pi}{2}+\arctan \pi=\frac{\pi}{2}+\arctan \pi[/tex]
The expression equals
[tex]\arctan \pi-\frac{\pi}{6}-\frac{\pi}{2}-\arctan \pi=-\frac{2\pi}{3}[/tex]

Solution:
We need
[tex]-1\le\frac{4x\arctan x}{\pi}\le 1[/tex]
The numbers [tex]x[/tex] and [tex]\arctan x[/tex] have the same sign, so [tex]g(x)=x\arctan x\ge 0[/tex], and [tex]g[/tex] is even. So it is enough that [tex]x\arctan x\le\frac{\pi}{4}[/tex]. For [tex]x\ge 0[/tex] the function [tex]g[/tex] increases (a product of two nonnegative increasing functions) and [tex]g(1)=\frac{\pi}{4}[/tex]. So [tex]0\le x\le 1[/tex], and by evenness the domain is [tex]\left[-1, 1\right][/tex].