Inverse Trigonometric Functions: Problems with Solutions

Solution:
From the figure
[tex]\sin \angle I=\frac{8}{10}=\frac{4}{5}\;\Rightarrow\;\angle I=\arcsin \frac{4}{5}\approx 53. 13^{\circ}[/tex]
(with a calculator).

Solution:
From the figure
[tex]\tan \angle E=\frac{4}{6}=\frac{2}{3}\;\Rightarrow\;\angle E=\arctan \frac{2}{3}\approx 33. 69^{\circ}[/tex]
(with a calculator).

Solution:
From the figure
[tex]\sin \angle B=\frac{7}{12}\;\Rightarrow\;\angle B=\arcsin \frac{7}{12}\approx 35. 69^{\circ}[/tex]
(with a calculator).

Solution:
From the figure
[tex]\cos\angle A=\frac{3}{8}\;\Rightarrow\;\angle A=\arccos\frac{3}{8}\approx 67. 98^{\circ}[/tex]
(with a calculator).

Solution:
Transform the argument:
[tex]1-8\cos^2x+8\cos^4x=1-8\cos^2x\left(1-\cos^2x\right)=1-8\sin ^2x\cos^2x=1-2\sin ^2 2x=\cos 4x[/tex]
So the equality means [tex]\arccos(\cos 4x)=4x[/tex]. This holds exactly when [tex]4x\in\left[0, \pi\right][/tex], i.e. for [tex]x\in\left(0, \frac{\pi}{4}\right][/tex]. For example, at [tex]x=\frac{\pi}{2}[/tex] the left side is [tex]\frac{\arccos 1}{\frac{\pi}{2}}=0\ne 4[/tex].

Solution:
The functions [tex]\arcsin [/tex] and [tex]\arccos[/tex] are defined only on [tex]\left[-1, 1\right][/tex]. Since [tex]\frac{\pi}{2}\approx 1. 57\gt 1[/tex], neither [tex]\arcsin \frac{\pi}{2}[/tex] nor [tex]\arccos\frac{\pi}{2}[/tex] exists. The formula [tex]\arcsin x+\arccos x=\frac{\pi}{2}[/tex] holds only for [tex]x\in\left[-1, 1\right][/tex].

Solution:
The function [tex]\arccos[/tex] is strictly decreasing on [tex]\left[-1, 1\right][/tex], so a smaller value of [tex]\arccos[/tex] corresponds to a larger argument: [tex]a\gt b[/tex]. For example, [tex]\arccos 1=0\lt\arccos 0=\frac{\pi}{2}[/tex] and [tex]1\gt 0[/tex].

Solution:
[tex]\arcsin \frac{\sqrt{2}}{2}=\frac{\pi}{4}, \quad\arccos\left(-\frac{1}{2}\right)=\pi-\frac{\pi}{3}=\frac{2\pi}{3}, \quad\arctan 1=\frac{\pi}{4}[/tex]
[tex]P=\frac{\pi}{4}+\frac{2\pi}{3}+\frac{\pi}{4}=\frac{7\pi}{6}[/tex]

Solution:
Let [tex]\alpha=\arctan \frac{\sqrt{6}}{2}[/tex] and [tex]\beta=\arccos\frac{1}{5}[/tex]; both angles are in [tex]\left(0, \frac{\pi}{2}\right)[/tex].
[tex]\cos\alpha=\frac{1}{\sqrt{1+\tan ^2\alpha}}=\frac{1}{\sqrt{1+\frac{6}{4}}}=\frac{2}{\sqrt{10}}, \quad\sin \alpha=\tan \alpha\cos\alpha=\frac{\sqrt{6}}{\sqrt{10}}[/tex]
[tex]\cos\beta=\frac{1}{5}, \quad\sin \beta=\sqrt{1-\frac{1}{25}}=\frac{2\sqrt{6}}{5}[/tex]
[tex]\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin \alpha\sin \beta=\frac{2}{5\sqrt{10}}+\frac{12}{5\sqrt{10}}=\frac{14}{5\sqrt{10}}=\frac{7\sqrt{10}}{25}\approx 0. 885[/tex]

Solution:
Let [tex]\arccos\frac{3}{5} = x[/tex];
then [tex]\cos x = \frac{3}{5},\ 0 < x < \frac{\pi}{2}[/tex]
So [tex]\sin 2x = 2\sin x\cos x = 2\sqrt{1-\frac{9}{25}}\times\frac{3}{5} = \frac{24}{25}[/tex]

Solution:
Let [tex]\alpha = \arcsin\frac{3}{5}[/tex] and [tex]\beta = \arcsin\frac{8}{17}[/tex]. Both angles lie in [tex]\left(0, \frac{\pi}{2}\right)[/tex], so
[tex]\cos\alpha = \sqrt{1-\frac{9}{25}} = \frac{4}{5}[/tex], [tex]\cos\beta = \sqrt{1-\frac{64}{289}} = \frac{15}{17}[/tex]
[tex]\sin(\alpha+\beta) = \sin\alpha\cos\beta + \cos\alpha\,\sin\beta = \frac{3}{5}\times\frac{15}{17} + \frac{4}{5}\times\frac{8}{17} = \frac{45}{85} + \frac{32}{85} = \frac{77}{85}[/tex]

Solution:
Let [tex]\alpha = \arcsin\frac{12}{13}[/tex], [tex]\alpha \in \left(0, \frac{\pi}{2}\right)[/tex]. Then [tex]\sin\alpha = \frac{12}{13}[/tex] and
[tex]\cos\alpha = \sqrt{1-\frac{144}{169}} = \frac{5}{13}[/tex]
[tex]\tan\alpha = \frac{\sin\alpha}{\cos\alpha} = \frac{12}{13} \div \frac{5}{13} = \frac{12}{5}[/tex]
Answer: [tex]\frac{12}{5}[/tex]

Solution:
The function [tex]\arcsin [/tex] is odd, so [tex]\arcsin (-x)=-\arcsin x[/tex] and the left side equals
[tex]-\left(\arcsin x+\arccos x\right)=-\frac{\pi}{2}[/tex]
because [tex]\arcsin x+\arccos x=\frac{\pi}{2}[/tex] for every [tex]x\in\left[-1, 1\right][/tex]. So [tex]C=-\frac{\pi}{2}[/tex].

Solution:
A function has an inverse on an interval when it is strictly monotonic there.
On [tex]\left[\frac{\pi}{2}, \frac{3\pi}{2}\right][/tex] the function [tex]\sin x[/tex] strictly decreases from [tex]1[/tex] to [tex]-1[/tex], so it has an inverse there.
On [tex]\left[0, \pi\right][/tex] we have [tex]\sin \frac{\pi}{6}=\sin \frac{5\pi}{6}[/tex], so there is no inverse; the other two intervals contain [tex]\left[0, \pi\right][/tex].

Solution:
For [tex]x\in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right][/tex] we have [tex]2x\in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right][/tex], where the sine increases, so an inverse exists:
[tex]y=\sin 2x\iff 2x=\arcsin y\iff x=\frac{1}{2}\arcsin y[/tex]
Swapping the letters, the inverse function is [tex]y=\frac{1}{2}\arcsin x, \ x\in\left[-1, 1\right][/tex].

Solution:
Let [tex]\alpha=\arcsin x\in\left[0, \frac{\pi}{2}\right][/tex]; then [tex]\cos\alpha\ge 0[/tex] and
[tex]\cos\alpha=\sqrt{1-\sin ^2\alpha}=\sqrt{1-x^2}[/tex]
Since [tex]\alpha\in\left[0, \pi\right][/tex], this means [tex]\alpha=\arccos\sqrt{1-x^2}[/tex], i.e. [tex]\arccos\sqrt{1-x^2}=\arcsin x[/tex].

Solution:
Let [tex]\alpha=\arctan x, \ \alpha\in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)[/tex]; then [tex]\tan \alpha=x[/tex] and [tex]\cos\alpha\gt 0[/tex]:
[tex]\cos\alpha=\frac{1}{\sqrt{1+\tan ^2\alpha}}=\frac{1}{\sqrt{1+x^2}}, \ \sin \alpha=\tan \alpha\cos\alpha=\frac{x}{\sqrt{1+x^2}}[/tex]
Note that [tex]\sin \alpha[/tex] has the same sign as [tex]x[/tex].