Tutorial 6: Inverse Trigonometric Functions

Inverse Trigonometric Functions

$f_1(x) = \tan^{-1}x$
Domain: $(-\infty, \infty)$ (all real numbers)
$f_2(x) = \cos^{-1}(3x)$
Domain: $-\frac{1}{3} \leq x \leq \frac{1}{3}$ (since $\cos^{-1}u$ requires $|u| \leq 1$)
$g(x) = \tan^{-1}x - \cos^{-1}(3x)$
Domain: Intersection of both domains
$$x \in \left[-\frac{1}{3}, \frac{1}{3}\right]$$

Right triangle approach: For $\theta = \tan^{-1}\left(\frac{4}{3}\right)$
$$\sin\theta = \frac{4}{5}, \quad \cos\theta = \frac{3}{5}, \quad \cot\theta = \frac{3}{4}, \quad \sec\theta = \frac{5}{3}, \quad \csc\theta = \frac{5}{4}$$
For $\theta = \sec^{-1}(2.6)$:
$\sec\theta = \frac{13}{5}$, so $\cos\theta = \frac{5}{13}$, then $\sin\theta = \frac{12}{13}$
Compound expressions: Use trigonometric identities
For $\sin\left(2\tan^{-1}\frac{2}{3}\right)$: let $\alpha = \tan^{-1}\frac{2}{3}$, then use $\sin 2\alpha = \frac{2\tan\alpha}{1+\tan^2\alpha}$
$$\sin 2\alpha = \frac{2\cdot\frac{2}{3}}{1+\left(\frac{2}{3}\right)^2} = \frac{4/3}{13/9} = \frac{12}{13}$$
Sum/difference formulas:
For $\tan\left(\frac{\pi}{4} + \cot^{-1}\frac{4}{5}\right)$: let $\beta = \cot^{-1}\frac{4}{5}$, then $\tan\beta = \frac{5}{4}$
$$\tan\left(\frac{\pi}{4} + \beta\right) = \frac{1+\tan\beta}{1-\tan\beta} = \frac{1+5/4}{1-5/4} = \frac{9/4}{-1/4} = -9$$

For $\cos\left(\sin^{-1}\frac{x-1}{x}\right)$, valid for $x \geq \frac{1}{2}$:
Let $\theta = \sin^{-1}\frac{x-1}{x}$, then $\sin\theta = \frac{x-1}{x}$
$$\cos\theta = \sqrt{1-\sin^2\theta} = \sqrt{1-\frac{(x-1)^2}{x^2}} = \sqrt{\frac{2x-1}{x^2}} = \frac{\sqrt{2x-1}}{x}$$
For $\sin\left(\cos^{-1}\frac{x+1}{\sqrt{x^2}}\right)$, valid for $x \leq -1$:
Let $\phi = \cos^{-1}\frac{x+1}{|x|}$ (since $\sqrt{x^2} = |x|$)

Camera-missile problem: For missile of length $a$, base $b$ above lens, camera at distance $x$
$$\theta = \cot^{-1}\left(\frac{x}{a+b}\right) - \cot^{-1}\left(\frac{x}{b}\right)$$
Law of cosines: $a^2 = b^2 + c^2 - 2bc\cos\theta$
With $a=4, b=2, c=3$:
$$\cos\theta = \frac{b^2+c^2-a^2}{2bc} = \frac{4+9-16}{2(2)(3)} = \frac{-3}{12} = -\frac{1}{4}$$
$$\theta = \cos^{-1}\left(-\frac{1}{4}\right) \approx 104^\circ$$