Trigonometry - Solution of Trigonometric Equations

Solve the equation: $\;$ $\left(2 \sin x - \cos x\right) \left(1 + \cos x\right) = \sin^2 x$

Given equation: $\;\;\;$ $\left(2 \sin x - \cos x\right) \left(1 + \cos x\right) = \sin^2 x$

i.e. $\;$ $2 \sin x + 2 \sin x \cos x - \cos x - \cos^2 x = \sin^2 x$

i.e. $\;$ $2 \sin x + 2 \sin x \cos x - \cos x = \sin^2 x + \cos^2 x$

i.e. $\;$ $2 \sin x + 2 \sin x \cos x - \cos x = 1$

i.e. $\;$ $2 \sin x \left(1 + \cos x\right) - \left(1 + \cos x\right) = 0$

i.e. $\;$ $\left(2 \sin x - 1\right) \left(1 + \cos x\right) = 0$

i.e. $\;$ $2 \sin x - 1 = 0$ $\;\;\;$ or $\;\;\;$ $1 + \cos x = 0$

When $\;\;$ $2 \sin x - 1 = 0$

i.e. $\;$ $\sin x = \dfrac{1}{2}$

$\implies$ $x = \left(-1\right)^n \sin^{-1} \left(\dfrac{1}{2}\right) + n \pi, \;\;\; n \in Z$

i.e. $\;$ $x = \left(-1\right)^n \times \dfrac{\pi}{6} + n \pi, \;\;\; n \in Z$

Or when $\;\;$ $1 + \cos x = 0$

i.e. $\;$ $\cos x = -1$

$\implies$ $x = \pm \cos^{-1} \left(-1\right) + 2 m \pi, \;\;\; m \in Z$

i.e. $\;$ $x = 2 m \pi \pm \pi, \;\;\; m \in Z$

Trigonometry - Solution of Trigonometric Equations

Solve the equation: $\;$ $2 \sin x + \tan x = 0$

Given equation: $\;\;\;$ $2 \sin x + \tan x = 0$

i.e. $\;$ $2 \sin x + \dfrac{\sin x}{\cos x} = 0$

i.e. $\;$ $2 \sin x \cos x + \sin x = 0$

i.e. $\;$ $\sin x \left(2 \cos x + 1\right) = 0$

$\implies$ $\sin x = 0$ $\;\;\;$ or $\;\;\;$ $2 \cos x + 1 = 0$

When $\;\;$ $\sin x = 0$

$\implies$ $x = n \pi, \;\;\; n \in I$

When $\;\;$ $2 \cos x + 1 = 0$

$\implies$ $\cos x = \dfrac{-1}{2} = \cos \left(\pi - \dfrac{\pi}{3}\right) = \cos \left(\dfrac{2 \pi}{3}\right)$

i.e. $\;$ $x = 2 m \pi \pm \dfrac{2 \pi}{3}, \;\;\; m \in I$

Trigonometry - Solution of Trigonometric Equations

Solve the equation: $\;$ $\sqrt{2} \cos^2 \left(7x\right) - \cos \left(7x\right) = 0$

Given equation: $\;\;\;$ $\sqrt{2} \cos^2 \left(7x\right) - \cos \left(7x\right) = 0$

i.e. $\;$ $\cos \left(7x\right) \left[\sqrt{2} \cos \left(7x\right) - 1\right] = 0$

$\implies$ $\cos \left(7x\right) = 0$ $\;\;$ or $\;\;$ $\sqrt{2} \cos \left(7x\right) - 1 = 0$

When $\;\;$ $\cos \left(7x\right) = 0$

$\implies$ $7x = \dfrac{\left(2n + 1\right) \pi}{2}, \;\;\; n \in I$

i.e. $\;$ $x = \dfrac{\pi}{14} \left(2n + 1\right), \;\;\; n \in I$

When $\;\;$ $\sqrt{2} \cos \left(7x\right) - 1 = 0$

i.e. $\;$ $\cos \left(7x\right) = \dfrac{1}{\sqrt{2}}$

i.e. $\;$ $7x = \pm \cos^{-1} \left(\dfrac{1}{\sqrt{2}}\right) + 2 m \pi, \;\;\; m \in I$

i.e. $\;$ $7x = \pm \dfrac{\pi}{4} + 2 m \pi, \;\;\; m \in I$

i.e. $\;$ $x = \pm \dfrac{\pi}{28} + \dfrac{2}{7} m \pi, \;\;\; m \in I$

i.e. $\;$ $x = \dfrac{\pi}{28} \left(8m \pm 1\right), \;\;\; m \in I$

Trigonometry - Solution of Trigonometric Equations

Solve the equation: $\;$ $\sin \left(3x - 2\right) = -1$

Given equation: $\;\;\;$ $\sin \left(3x - 2\right) = -1$

i.e. $\;$ $3x - 2 = \sin^{-1} \left(-1\right)$

i.e. $\;$ $3x - 2 = - \sin^{-1} \left(1\right)$ $\;\;\;$ $\left\{\because \;\; \sin^{-1} \left(-x\right) = - \sin^{-1} x \;\; \forall \; x \in \left[-1, 1\right]\right\}$

i.e. $\;$ $3x - 2 = \dfrac{- \pi}{2} + 2 n \pi \;\;\; n \in Z$

i.e. $\;$ $3x = \dfrac{4 n \pi - \pi}{2} + 2 \;\;\; n \in Z$

i.e. $\;$ $x = \dfrac{4n \pi - \pi}{6} + \dfrac{2}{3} \;\;\; n \in Z$

i.e. $\;$ $x = \dfrac{\pi \left(4n - 1\right) + 4}{6} \;\;\; n \in Z$

Trigonometry - Inverse Trigonometric Functions

Check the given equality: $\;$ $\sin^{-1} \left(\dfrac{4}{5}\right) - \cos^{-1} \left(\dfrac{2}{\sqrt{5}}\right) = \tan^{-1} \left(\dfrac{1}{2}\right)$

Formulas

$\sin^{-1} x = \tan^{-1} \left(\dfrac{x}{\sqrt{1 - x^2}}\right), \;\;\; -1 < x < 1$

$\cos^{-1} x = \tan^{-1} \left(\dfrac{\sqrt{1 - x^2}}{x}\right), \;\;\; -1 < x < 0, \; 0 < x < 1$

$\tan^{-1} x - \tan^{-1} y = \tan^{-1} \left(\dfrac{x - y}{1 + x \cdot y}\right), \;\;\; x \cdot y < 1$

$\begin{aligned} LHS & = \sin^{-1} \left(\dfrac{4}{5}\right) - \cos^{-1} \left(\dfrac{2}{\sqrt{5}}\right) \\\\ & = \tan^{-1} \left(\dfrac{\dfrac{4}{5}}{\sqrt{1 - \dfrac{16}{25}}}\right) - \tan^{-1} \left(\dfrac{\sqrt{1 - \dfrac{4}{5}}}{\dfrac{2}{\sqrt{5}}}\right) \\\\ & = \tan^{-1} \left(\dfrac{4}{3}\right) - \tan^{-1} \left(\dfrac{1}{2}\right) \\\\ & = \tan^{-1} \left(\dfrac{\dfrac{4}{3} - \dfrac{1}{2}}{1 + \dfrac{4}{3} \times \dfrac{1}{2}}\right) \\\\ & = \tan^{-1} \left(\dfrac{5}{10}\right) \\\\ & = \tan^{-1} \left(\dfrac{1}{2}\right) = RHS \end{aligned}$

Trigonometry - Inverse Trigonometric Functions

Check the given equality: $\;$ $\sin^{-1} \left(\dfrac{4}{5}\right) + \sin^{-1} \left(\dfrac{5}{13}\right) + \sin^{-1} \left(\dfrac{16}{65}\right) = \dfrac{\pi}{2}$

$\begin{aligned} LHS & = \sin^{-1} \left(\dfrac{4}{5}\right) + \sin^{-1} \left(\dfrac{5}{13}\right) + \sin^{-1} \left(\dfrac{16}{65}\right) \\\\ & = \sin^{-1} \left(\dfrac{4}{5} \times \sqrt{1 - \dfrac{25}{169}} + \dfrac{5}{13} \times \sqrt{1 - \dfrac{16}{25}}\right) + \sin^{-1} \left(\dfrac{16}{65}\right) \\ & \left\{\because \;\; \sin^{-1} x + \sin^{-1} y = \sin^{-1} \left(x \sqrt{1 - y^2} + y \sqrt{1 - x^2}\right), \; 0 < x, y < 1\right\} \\\\ & = \sin^{-1} \left(\dfrac{4}{5} \times \dfrac{12}{13} + \dfrac{5}{13} \times \dfrac{3}{5}\right) + \sin^{-1} \left(\dfrac{16}{65}\right) \\\\ & = \sin^{-1} \left(\dfrac{63}{65}\right) + \sin^{-1} \left(\dfrac{16}{65}\right) \\\\ & = \sin^{-1} \left(\dfrac{63}{65} \times \sqrt{1 - \dfrac{256}{4225}} + \dfrac{16}{65} \times \sqrt{1 - \dfrac{3969}{4225}}\right) \\\\ & = \sin^{-1} \left(\dfrac{63}{65} \times \sqrt{\dfrac{3969}{4225}} + \dfrac{16}{65} \times \sqrt{\dfrac{256}{4225}}\right) \\\\ & = \sin^{-1} \left(\dfrac{63}{65} \times \dfrac{63}{65} + \dfrac{16}{65} \times \dfrac{16}{65}\right) \\\\ & = \sin^{-1} \left(\dfrac{4225}{4225}\right) \\\\ & = \sin^{-1} \left(1\right) \\\\ & = \dfrac{\pi}{2} = RHS \end{aligned}$

Trigonometry - Inverse Trigonometric Functions

Check the given equality: $\;$ $\tan^{-1} \left(\dfrac{\sqrt{2}}{2}\right) + \sin^{-1} \left(\dfrac{\sqrt{2}}{2}\right) = \tan^{-1} \left(3 + 2 \sqrt{2}\right)$

$\begin{aligned} LHS & = \tan^{-1} \left(\dfrac{\sqrt{2}}{2}\right) + \sin^{-1} \left(\dfrac{\sqrt{2}}{2}\right) \\\\ & = \tan^{-1} \left(\dfrac{1}{\sqrt{2}}\right) + \sin^{-1} \left(\dfrac{1}{\sqrt{2}}\right) \\\\ & = \tan^{-1} \left(\dfrac{1}{\sqrt{2}}\right) + \tan^{-1} \left(\dfrac{\dfrac{1}{\sqrt{2}}}{\sqrt{1 - \dfrac{1}{2}}}\right) \\ & \left\{\because \;\; \sin^{-1} x = \tan^{-1} \left(\dfrac{x}{\sqrt{1 - x^2}}\right), \; 0 < x < 1\right\} \\\\ & = \tan^{-1} \left(\dfrac{1}{\sqrt{2}}\right) + \tan^{-1} \left(1\right) \\\\ & = \tan^{-1} \left[\dfrac{\dfrac{1}{\sqrt{2}} + 1}{1 - \dfrac{1}{\sqrt{2}} \times 1}\right] \\ & \left\{\because \;\; \tan^{-1} x + \tan^{-1} y = \tan^{-1} \left(\dfrac{x + y}{1 - xy}\right), \; x \cdot y < 1\right\} \\\\ & = \tan^{-1} \left[\dfrac{\sqrt{2} + 1}{\sqrt{2} - 1}\right] \\\\ & = \tan^{-1} \left[\dfrac{\left(\sqrt{2} + 1\right)^2}{\left(\sqrt{2} - 1 \right) \left(\sqrt{2} + 1\right)}\right] \\\\ & = \tan^{-1} \left[\dfrac{2 + 1 + 2 \sqrt{2}}{2 - 1}\right] \\\\ & = \tan^{-1} \left[3 + 2 \sqrt{2}\right] = RHS \end{aligned}$

Trigonometry - Inverse Trigonometric Functions

Check the given equality: $\;$ $\sin^{-1} \left(\dfrac{7}{25}\right) + \dfrac{1}{2} \cos^{-1} \left(\dfrac{7}{25}\right) = \cos^{-1} \left(\dfrac{3}{5}\right)$

To Check That (TCT) $\;$ $\sin^{-1} \left(\dfrac{7}{25}\right) + \dfrac{1}{2} \cos^{-1} \left(\dfrac{7}{25}\right) = \cos^{-1} \left(\dfrac{3}{5}\right)$

i.e. TCT $\;$ $\left[\sin^{-1} \left(\dfrac{7}{25}\right) + \cos^{-1} \left(\dfrac{7}{25}\right)\right] - \dfrac{1}{2} \cos^{-1} \left(\dfrac{7}{25}\right) = \cos^{-1} \left(\dfrac{3}{5}\right)$

i.e. TCT $\;$ $\dfrac{\pi}{2} = \cos^{-1} \left(\dfrac{3}{5}\right) + \dfrac{1}{2} \cos^{-1} \left(\dfrac{7}{25}\right)$
$\left\{\because \;\; \sin^{-1} x + \cos^{-1} x = \dfrac{\pi}{2}, \;\;\; \forall \; x \in \left[-1, 1\right]\right\}$

i.e. TCT $\;$ $\pi = 2 \cos^{-1} \left(\dfrac{3}{5}\right) + \cos^{-1} \left(\dfrac{7}{25}\right)$

$\begin{aligned} RHS & = 2 \cos^{-1} \left(\dfrac{3}{5}\right) + \cos^{-1} \left(\dfrac{7}{25}\right) \\\\ & = \cos^{-1} \left(2 \times \dfrac{9}{25} - 1\right) + \cos^{-1} \left(\dfrac{7}{25}\right) \\ & \left\{\because \; 2 \cos^{-1} x = \cos^{-1} \left(2 x^2 - 1\right), \;\; 0 \leq x \leq 1\right\} \\\\ & = \cos^{-1} \left(\dfrac{-7}{25}\right) + \cos^{-1} \left(\dfrac{7}{25}\right) \\\\ & = \pi - \cos^{-1} \left(\dfrac{7}{25}\right) + \cos^{-1} \left(\dfrac{7}{25}\right) \\ & \left\{\because \; \cos^{-1} \left(-x\right) = \pi - \cos^{-1} \left(x\right) \;\; \forall \;\; x \in \left[-1, 1\right]\right\} \\\\ & = \pi = LHS \end{aligned}$

Trigonometry - Inverse Trigonometric Functions

Check the given equality: $\;$ $\cot^{-1} \left(\dfrac{1}{7}\right) + 2 \cot^{-1} \left(\dfrac{1}{3}\right) = \dfrac{5 \pi}{4}$

$\begin{aligned} LHS & = \cot^{-1} \left(\dfrac{1}{7}\right) + 2 \cot^{-1} \left(\dfrac{1}{3}\right) \\\\ & = \tan^{-1} \left(7\right) + 2 \tan^{-1} \left(3\right) \;\;\; \left\{\because \;\; \tan^{-1} \left(\dfrac{1}{x}\right) = \cot^{-1}\left(x\right), \; x >0\right\} \\\\ & = \left[\tan^{-1} \left(7\right) + \tan^{-1} \left(3\right)\right] + \tan^{-1} \left(3\right) \\\\ & = \pi + \tan^{-1} \left(\dfrac{7 + 3}{1 - 7 \times 3}\right) + \tan^{-1} \left(3\right) \\\\ & \left\{\because \;\; \tan^{-1} x + \tan^{-1} y = \pi + \tan^{-1} \left(\dfrac{x + y}{1 - xy}\right), \; x > 0, \; y > 0, \; xy >1\right\} \\\\ & = \pi + \tan^{-1} \left(\dfrac{10}{-20}\right) + \tan^{-1} \left(3\right) \\\\ & = \pi + \tan^{-1} \left(\dfrac{-1}{2}\right) + \tan^{-1} \left(3\right) \\\\ & = \pi + \tan^{-1} \left(\dfrac{\dfrac{-1}{2} + 3}{1 - \left(\dfrac{-1}{2}\right) \times 3}\right) \\\\ & \left\{\because \;\; \tan^{-1} x + \tan^{-1} y = \tan^{-1} \left(\dfrac{x + y}{1 - xy}\right), \;\; x y < 1\right\} \\\\ & = \pi + \tan^{-1} \left(\dfrac{\dfrac{5}{2}}{\dfrac{5}{2}}\right) \\\\ & = \pi + \tan^{-1} \left(1\right) \\\\ & = \pi + \dfrac{\pi}{4} \\\\ & = \dfrac{5 \pi}{4} = RHS \end{aligned}$

Trigonometry - Inverse Trigonometric Functions

Simplify: $\;$ $\cos \left(2 \tan^{-1} x\right)$

Given expression: $\;$ $\cos \left(2 \tan^{-1} x\right)$

$= 2 \cos^2 \left(\tan^{-1} x\right) - 1$ $\;\;\; \left\{\because \; \cos 2 \theta = 2 \cos^2 \theta - 1\right\}$

$= 2 \left[\cos \left(\tan^{-1} x\right)\right]^2 - 1$

$= 2 \times \left(\dfrac{1}{\sqrt{1 + x^2}}\right)^2 - 1$ $\;\;\; \left\{\because \; \cos \left(\tan^{-1} x \right) = \dfrac{1}{\sqrt{1 + x^2}}, \; x > 0\right\}$

$= \dfrac{2}{1 + x^2} - 1$

$= \dfrac{1 - x^2}{1 + x^2}$

Trigonometry - Inverse Trigonometric Functions

Compute the given expression: $\;$ $\sin \left\{\tan^{-1} \left(\dfrac{8}{15}\right) - \sin^{-1} \left(\dfrac{8}{17}\right)\right\}$

Given expression: $\;$ $\sin \left\{\tan^{-1} \left(\dfrac{8}{15}\right) - \sin^{-1} \left(\dfrac{8}{17}\right)\right\}$

$= \sin \left\{\tan^{-1} \left(\dfrac{8}{15}\right)\right\} \cos \left\{\sin^{-1} \left(\dfrac{8}{17}\right)\right\}$

$\hspace{3cm}$ $- \cos \left\{\tan^{-1} \left(\dfrac{8}{15}\right)\right\} \sin \left\{\sin^{-1} \left(\dfrac{8}{17}\right)\right\}$

$\left[\because \;\; \sin \left(A - B\right) = \sin A \cos B - \cos A \sin B\right]$

$= \dfrac{\dfrac{8}{15}}{\sqrt{1 + \left(\dfrac{8}{15}\right)^2}} \times \sqrt{1 - \left(\dfrac{8}{17}\right)^2} - \dfrac{1}{\sqrt{1 + \left(\dfrac{8}{15}\right)^2}} \times \dfrac{8}{17}$

$\left[\because \;\; \sin \left(\tan^{-1} x\right) = \dfrac{x}{\sqrt{1 + x^2}}, \;\; x > 0 \right.$
$\left. \cos \left(\sin^{-1} x\right) = \sqrt{1 - x^2}, \;\; 0 \leq x \leq 1 \right.$
$\left. \sin \left(\sin^{-1} x\right) = x, \;\; x \in \left[-1, 1\right] \right.$
$\left. \cos \left(\tan^{-1} x\right) = \dfrac{1}{\sqrt{1 + x^2}}, \;\; x >0 \right]$

$= \dfrac{8}{17} \times \dfrac{15}{17} - \dfrac{15}{17} \times \dfrac{8}{17}$

$= 0$

Trigonometry - Inverse Trigonometric Functions

Compute the given expression: $\;$ $\cot \left\{\dfrac{1}{2} \cos^{-1} \left(\dfrac{-4}{7}\right)\right\}$

Given expression: $\;$ $\cot \left\{\dfrac{1}{2} \cos^{-1} \left(\dfrac{-4}{7}\right)\right\}$

$= \cot \left\{\dfrac{1}{2} \left[\pi - \cos^{-1} \left(\dfrac{4}{7}\right)\right]\right\}$ $\;\;\;$ $\left\{\because \; \cos^{-1} \left(-x\right) = \pi - \cos^{-1} x \;\; \forall \;\; x \in \left[-1, 1\right]\right\}$

$= \cot \left\{\dfrac{\pi}{2} - \dfrac{1}{2} \cos^{-1} \left(\dfrac{4}{7}\right)\right\}$

$= \tan \left\{\dfrac{1}{2} \cos^{-1} \left(\dfrac{4}{7}\right)\right\}$ $\;\;\;$ $\left\{\because \; \cot \left(\dfrac{\pi}{2} - \theta\right) = \tan \theta\right\}$

$= \sqrt{\dfrac{1 - \cos \left\{\cos^{-1} \left(\dfrac{4}{7}\right)\right\}}{1 + \cos \left\{\cos^{-1} \left(\dfrac{4}{7}\right)\right\}}}$ $\;\;\;$ $\left\{\because \; \tan \left(\dfrac{\theta}{2}\right) = \sqrt{\dfrac{1 - \cos \theta}{1 + \cos \theta}}\right\}$

$= \sqrt{\dfrac{1 - \dfrac{4}{7}}{1 + \dfrac{4}{7}}}$ $\;\;\;$ $\left\{\because \; \cos \left(\cos^{-1} x \right) = x \;\; \forall \;\; x \in \left[-1, 1\right]\right\}$

$= \sqrt{\dfrac{3}{11}}$

Trigonometry - Inverse Trigonometric Functions

Compute the given expression: $\;$ $\tan \left[\dfrac{1}{2} \sin^{-1} \left(\dfrac{5}{13}\right)\right]$

Given expression: $\;$ $\tan \left[\dfrac{1}{2} \sin^{-1} \left(\dfrac{5}{13}\right)\right]$

$= \dfrac{1 - \cos \left[\sin^{-1} \left(\dfrac{5}{13}\right)\right]}{\sin \left[\sin^{-1} \left(\dfrac{5}{13}\right)\right]}$ $\;\;\;$ $\left\{\because \; \tan \left(\dfrac{\theta}{2}\right) = \dfrac{1 - \cos \theta}{\sin \theta}\right\}$

$= \dfrac{1 - \sqrt{1 - \left(\dfrac{5}{13}\right)^2}}{\dfrac{5}{13}}$ $\;\;\;$ $\left\{\because \; \sin \left(\sin^{-1} x\right) = x \; \forall \; x \in \left[-1,1\right], \right.$
$\hspace{4.5cm} \left. \cos \left(\sin^{-1} x\right) = \sqrt{1- x^2}, \; x \geq 0 \right\}$

$= \dfrac{1 - \dfrac{12}{13}}{\dfrac{5}{13}}$

$= \dfrac{1}{5}$

Trigonometry - Inverse Trigonometric Functions

Compute the given expression: $\;$ $\tan^{-1} \left[- \tan \left(\dfrac{13 \pi}{8}\right)\right] + \cot^{-1} \left[\cot \left(\dfrac{-19 \pi}{6}\right)\right]$

Given expression: $\;$ $\tan^{-1} \left[- \tan \left(\dfrac{13 \pi}{8}\right)\right] + \cot^{-1} \left[\cot \left(\dfrac{-19 \pi}{6}\right)\right]$

$= - \tan^{-1} \left[\tan \left(\dfrac{13 \pi}{8}\right)\right] + \cot^{-1} \left[- \cot \left(\dfrac{19 \pi}{8}\right)\right]$

$\left\{\tan^{-1} \left(-x\right) = - \tan^{-1} x \;\; \forall \;\; x \in R\right\}$

$\left\{\cot \left(-x\right) = - \cot x\right\}$

$= - \tan^{-1} \left[\tan \left(\dfrac{13 \pi}{8}\right)\right] + \pi - \cot^{-1} \left[\cot \left(\dfrac{19 \pi}{8}\right)\right]$

$\left\{\cot^{-1} \left(-x\right) = \pi - \cot^{-1}x \;\; \forall \;\; x \in R\right\}$

$= - \tan^{-1} \left[\tan \left(2 \pi - \dfrac{3 \pi}{8}\right)\right] + \pi - \cot^{-1} \left[\cot \left(2 \pi + \dfrac{3 \pi}{8}\right)\right]$

$= - \tan^{-1} \left[- \tan \left(\dfrac{3 \pi}{8}\right)\right] + \pi - \cot^{-1} \left[\cot \left(\dfrac{3 \pi}{8}\right)\right]$

$\left\{\tan \left(2 \pi - \theta\right) = - \tan \theta, \;\;\; \cot \left(2 \pi + \theta\right) = \cot \theta\right\}$

$= - \left(-\right)\tan^{-1} \left[\tan \left(\dfrac{3 \pi}{8}\right)\right] + \pi - \cot^{-1} \left[\cot \left(\dfrac{3 \pi}{8}\right)\right]$

$= \tan^{-1} \left[\tan \left(\dfrac{3 \pi}{8}\right)\right] + \pi - \cot^{-1} \left[\cot \left(\dfrac{3 \pi}{8}\right)\right]$

$= \dfrac{3 \pi}{8} + \pi - \dfrac{3 \pi}{8}$

$\left\{\tan^{-1}\left(\tan x\right) = x \;\; \forall \;\; x \in \left(\dfrac{-\pi}{2}, \dfrac{\pi}{2}\right) \right\}$

$\left\{\cot^{-1} \left(\cot x\right) = x \;\; \forall \;\; x \in \left(0, \pi\right)\right\}$

$= \pi$

Trigonometry - Inverse Trigonometric Functions

Compute the given expression: $\;$ $\sin^{-1} \left[\sin \left(\dfrac{33 \pi}{7}\right)\right] + \cos^{-1} \left[\cos \left(\dfrac{46 \pi}{7}\right)\right]$

Given expression: $\;$ $\sin^{-1} \left[\sin \left(\dfrac{33 \pi}{7}\right)\right] + \cos^{-1} \left[\cos \left(\dfrac{46 \pi}{7}\right)\right]$

$= \sin^{-1} \left[\sin \left(5 \pi - \dfrac{2 \pi}{7}\right)\right] + \cos^{-1} \left[\cos \left(7 \pi - \dfrac{3 \pi}{7}\right)\right]$

$= \sin^{-1} \left[\sin \left(\dfrac{2 \pi}{7}\right)\right] + \cos^{-1} \left[-\cos \left(\dfrac{3 \pi}{7}\right)\right]$

$= \dfrac{2 \pi}{7} + \pi - \cos^{-1} \left[\cos \left(\dfrac{3 \pi}{7}\right)\right]$

$= \dfrac{2 \pi}{7} + \pi - \dfrac{3 \pi}{7} = \dfrac{6 \pi}{7}$

Formulas:

$\sin^{-1} \left(\sin x\right) = x \;\; \forall \;\; x \in \left[\dfrac{- \pi}{2}, \dfrac{\pi}{2}\right]$

$\cos^{-1} \left(\cos x\right) = x \;\; \forall \;\; x \in \left[0, \pi\right]$

$\cos^{-1} \left(-x\right) = \pi - \cos^{-1} x \;\; \forall \;\; x \in \left[-1, 1\right]$

Trigonometry - Inverse Trigonometric Functions

Compute the given expression: $\;$ $\cos^{-1} \left[- \cos \left(\dfrac{3 \pi}{4}\right)\right]$

Given expression: $\;$ $\cos^{-1} \left[- \cos \left(\dfrac{3 \pi}{4}\right)\right]$

$= \pi - \cos^{-1} \left[\cos \left(\dfrac{3 \pi}{4}\right)\right]$ $\;\;\;$ $\left\{\because \;\; \cos^{-1} \left(- x\right) = \pi - \cos^{-1} x \;\; \forall \;\; x \in \left[-1, 1\right]\right\}$

$= \pi - \dfrac{3 \pi}{4}$ $\;\;\;$ $\left\{\because \;\; \cos^{-1} \left(\cos x\right) = x \;\; \forall \;\; x \in \left[0, \pi\right]\right\}$

$= \dfrac{\pi}{4}$

Trigonometry - Inverse Trigonometric Functions

Compute the given expression: $\;$ $\sin \left[3 \tan^{-1} \left(\sqrt{3}\right) + 2 \cos^{-1} \left(\dfrac{1}{2}\right)\right]$

Given expression: $\;$ $\sin \left[3 \tan^{-1} \left(\sqrt{3}\right) + 2 \cos^{-1} \left(\dfrac{1}{2}\right)\right]$

$= \sin \left[3 \times \dfrac{\pi}{3} + 2 \times \dfrac{\pi}{3}\right]$

$= \sin \left[\pi + \dfrac{2 \pi}{3}\right]$

$= - \sin \left[\dfrac{2 \pi}{3}\right]$

$= - \sin \left[\pi - \dfrac{\pi}{3}\right]$

$= - \sin \left[\dfrac{\pi}{3}\right]$

$= \dfrac{- \sqrt{3}}{2}$

Trigonometry - Inverse Trigonometric Functions

Compute the given expression: $\;$ $\tan \left[5 \tan^{-1} \left(\dfrac{\sqrt{3}}{3}\right) - \dfrac{1}{4} \sin^{-1} \left(\dfrac{\sqrt{3}}{2}\right)\right]$

Given expression: $\;$ $\tan \left[5 \tan^{-1} \left(\dfrac{\sqrt{3}}{3}\right) - \dfrac{1}{4} \sin^{-1} \left(\dfrac{\sqrt{3}}{2}\right)\right]$

$= \tan \left[5 \tan^{-1} \left(\dfrac{1}{\sqrt{3}}\right) - \dfrac{1}{4} \sin^{-1} \left(\dfrac{\sqrt{3}}{2}\right)\right]$

$= \tan \left[5 \times \dfrac{\pi}{6} - \dfrac{1}{4} \times \dfrac{\pi}{3}\right]$

$= \tan \left[\dfrac{9 \pi}{12}\right]$

$= \tan \left[\dfrac{3 \pi}{4}\right]$

$= \tan \left[\dfrac{\pi}{2} + \dfrac{\pi}{4}\right]$

$= - \cot \left[\dfrac{\pi}{4}\right]$

$= -1$

Trigonometry - Inverse Trigonometric Functions

Solve: $\;$ $5 \tan^{-1} x + 3 \cot^{-1} x = 2 \pi$

Given equation: $\;$ $5 \tan^{-1} x + 3 \cot^{-1} x = 2 \pi$

i.e. $\;$ $2 \tan^{-1} x + \left(3 \tan^{-1} x + 3 \cot^{-1} x\right) = 2 \pi$

i.e. $\;$ $2 \tan^{-1} x + 3 \left(\tan^{-1} x + \cot^{-1} x \right) = 2 \pi$

i.e. $\;$ $2 \tan^{-1} x + 3 \times \dfrac{\pi}{2} = 2 \pi$ $\;\;\;$ $\left[\because \; \tan^{-1} x + \cot^{-1} x = \dfrac{\pi}{2}, \;\; \forall \; x \in R\right]$

i.e. $\;$ $2 \tan^{-1} x = \dfrac{\pi}{2}$

i.e. $\;$ $\tan^{-1} x = \dfrac{\pi}{4}$

i.e. $\;$ $x = \tan \left(\dfrac{\pi}{4}\right) = 1$