Inverse Trigonometric Functions

Prove that: $\;$ $\tan^{-1} \left(\dfrac{1}{7}\right) + \tan^{-1} \left(\dfrac{1}{13}\right) = \tan^{-1} \left(\dfrac{2}{9}\right)$

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

$\begin{aligned} \therefore \; \tan^{-1} \left(\dfrac{1}{7}\right) + \tan^{-1} \left(\dfrac{1}{13}\right) & = \tan^{-1} \left(\dfrac{\dfrac{1}{7} + \dfrac{1}{13}}{1 - \dfrac{1}{7} \times \dfrac{1}{13}}\right) \\\\ & = \tan^{-1} \left(\dfrac{20}{90}\right) \\\\ & = \tan^{-1} \left(\dfrac{2}{9}\right) \end{aligned}$

Hence proved.

Inverse Trigonometric Functions

Find the value of: $\;$ $\sin^{-1} \left[\sin \left(\dfrac{5 \pi}{6}\right)\right] + \cos^{-1} \left[\cos \left(\dfrac{5 \pi}{3}\right) \right] + \tan^{-1} \left[\tan \left(\dfrac{7 \pi}{3}\right)\right]$

$\begin{aligned} \sin^{-1} \left[\sin \left(\dfrac{5 \pi}{6}\right)\right] & = \sin^{-1} \left[\sin \left(\pi - \dfrac{\pi}{6}\right)\right] & \hspace{1cm} \left\{\dfrac{5 \pi}{6} \notin \left[- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right] \right\} \\\\ & = \sin^{-1} \left[\sin \left(\dfrac{\pi}{6}\right)\right] & \\\\ & = \dfrac{\pi}{6} \hspace{1cm} & \hspace{1cm} \left\{\dfrac{\pi}{6} \in \left[- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right] \right\} \end{aligned}$

$\begin{aligned} \cos^{-1} \left[\cos \left(\dfrac{5 \pi}{3}\right)\right] & = \cos^{-1} \left[\cos \left(2 \pi - \dfrac{\pi}{3}\right)\right] & \hspace{1cm} \left\{\dfrac{5 \pi}{3} \notin \left[0, \pi\right] \right\} \\\\ & = \cos^{-1} \left[\cos \left(\dfrac{\pi}{3}\right)\right] & \\\\ & = \dfrac{\pi}{3} \hspace{1cm} & \hspace{1cm} \left\{\dfrac{\pi}{3} \in \left[0, \pi\right] \right\} \end{aligned}$

$\begin{aligned} \tan^{-1} \left[\tan \left(\dfrac{7 \pi}{3}\right)\right] & = \tan^{-1} \left[\tan \left(2 \pi + \dfrac{\pi}{3}\right)\right] & \hspace{1cm} \left\{\dfrac{7 \pi}{3} \notin \left(- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right) \right\} \\\\ & = \tan^{-1} \left[\tan \left(\dfrac{\pi}{3}\right)\right] & \\\\ & = \dfrac{\pi}{3} \hspace{1cm} & \hspace{1cm} \left\{\dfrac{\pi}{3} \in \left(- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right) \right\} \end{aligned}$

$\therefore \; \sin^{-1} \left[\sin \left(\dfrac{5 \pi}{6}\right)\right] + \cos^{-1} \left[\cos \left(\dfrac{5 \pi}{3}\right) \right] + \tan^{-1} \left[\tan \left(\dfrac{7 \pi}{3}\right)\right]$

$= \dfrac{\pi}{6} + \dfrac{\pi}{3} + \dfrac{\pi}{3}$

$= \dfrac{5 \pi}{6}$

Inverse Trigonometric Functions

Find the value of: $\;$ $\cos \left[\sin^{-1} \left(- \dfrac{4}{5}\right)\right] + \sin \left[\tan^{-1} \left(\dfrac{3}{4}\right)\right] + \cos \left[\csc^{-1} \left(\dfrac{5}{3}\right)\right]$

Let $\;$ $\sin^{-1} \left(- \dfrac{4}{5}\right) = - \sin^{-1} \left(\dfrac{4}{5}\right) = -\alpha, \hspace{1cm} -\alpha \in \left[- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$

Then, $\;$ $\sin \left(\alpha\right) = \dfrac{4}{5}$ $\;\;\; \cdots \; (1)$

Let $\;$ $\tan^{-1} \left(\dfrac{3}{4}\right) = \beta, \hspace{1cm} \beta \in \left(- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$

Then, $\;$ $\tan \left(\beta\right) = \dfrac{3}{4}$ $\;\;\; \cdots \; (2)$

Let $\;$ $\csc^{-1} \left(\dfrac{5}{3}\right) = \gamma, \hspace{1cm} \gamma \in \left[- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right] - \left\{0\right\}$

Then, $\;$ $\csc \left(\gamma\right) = \dfrac{5}{3}$ $\implies$ $\sin \left(\gamma\right) = \dfrac{3}{5}$ $\;\;\; \cdots \; (3)$

From equation $(1)$, $\;$ $\cos \left(\alpha\right) = \sqrt{1 - \sin^2 \left(\alpha\right)} = \sqrt{1 - \dfrac{16}{25}} = \sqrt{\dfrac{9}{25}} = \dfrac{3}{5}$ $\;\; \cdots \; (4a)$

From equation $(2)$, $\;$ $\sin \left(\beta\right) = \dfrac{3}{\sqrt{3^2 + 4^2}} = \dfrac{3}{\sqrt{25}} = \dfrac{3}{5}$ $\;\;\; \cdots \; (4b)$

From equation $(3)$, $\;$ $\cos \left(\gamma\right) = \sqrt{1 - \sin^2\left(\gamma\right)} = \sqrt{1 - \dfrac{9}{25}} = \sqrt{\dfrac{16}{25}} = \dfrac{4}{5}$ $\;\;\; \cdots \; (4c)$

$\therefore \;$ $\cos \left[\sin^{-1} \left(- \dfrac{4}{5}\right)\right] + \sin \left[\tan^{-1} \left(\dfrac{3}{4}\right)\right] + \cos \left[\csc^{-1} \left(\dfrac{5}{3}\right)\right]$

$= \cos \left(- \alpha\right) + \sin \left(\beta\right) + \cos \left(\gamma\right)$

$= \cos \left(\alpha\right) + \sin \left(\beta\right) + \cos \left(\gamma\right)$ $\hspace{1cm} \left[\because \; \cos \left(- \theta\right) = \cos \left(\theta\right)\right]$

$= \dfrac{3}{5} + \dfrac{3}{5} + \dfrac{4}{5}$ $\hspace{1cm}$ [from equations $(4a), \; (4b), \; (4c)$]

$= \dfrac{10}{5} = 2$

Inverse Trigonometric Functions

Find the value of: $\;$ $\cot^{-1} \left(1\right) + 3 \sin^{-1} \left(\dfrac{1}{2}\right) - \csc^{-1} \left(- 2\right) - 3 \tan^{-1} \left(\dfrac{1}{\sqrt{3}}\right)$

$\cot^{-1} \left(1\right) = \cot^{-1} \left[\cot \left(\dfrac{\pi}{4}\right)\right] = \dfrac{\pi}{4}, \hspace{1cm} \dfrac{\pi}{4} \in \left(0, \pi\right)$

$\sin^{-1} \left(\dfrac{1}{2}\right) = \sin^{-1} \left[\sin \left(\dfrac{\pi}{6}\right)\right] = \dfrac{\pi}{6}, \hspace{1cm} \dfrac{\pi}{6} \in \left[- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$

$\csc^{-1} \left(- 2\right) = \csc^{-1} \left[- \csc \left(\dfrac{\pi}{6}\right)\right] = \csc^{-1} \left[\csc \left(- \dfrac{\pi}{6}\right)\right] = - \dfrac{\pi}{6}$,

where $\;$ $- \dfrac{\pi}{6} \in \left[- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right] - \left\{0\right\}$

$\tan^{-1} \left(\dfrac{1}{\sqrt{3}}\right) = \tan^{-1} \left[\tan \left(\dfrac{\pi}{6}\right)\right] = \dfrac{\pi}{6}, \hspace{1cm} \dfrac{\pi}{6} \in \left(- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$

$\therefore \;$ $\cot^{-1} \left(1\right) + 3 \sin^{-1} \left(\dfrac{1}{2}\right) - \csc^{-1} \left(- 2\right) - 3 \tan^{-1} \left(\dfrac{1}{\sqrt{3}}\right)$

$= \dfrac{\pi}{4} + 3 \times \dfrac{\pi}{6} + \dfrac{\pi}{6} - 3 \times \dfrac{\pi}{6}$

$= \dfrac{\pi}{4} + \dfrac{\pi}{6} = \dfrac{5 \pi}{12}$

Inverse Trigonometric Functions

Find the value of: $\;$ $\sin^{-1} \left(\dfrac{\sqrt{3}}{2}\right) - \cos^{-1} \left(- \dfrac{\sqrt{3}}{2}\right) + 2 \tan^{-1} \left(1\right)$

$\sin^{-1} \left(\dfrac{\sqrt{3}}{2}\right) = \sin^{-1} \left[\sin \left(\dfrac{\pi}{3}\right)\right] = \dfrac{\pi}{3}, \hspace{1cm} \dfrac{\pi}{3} \in \left[- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$

$\cos^{-1} \left(- \dfrac{\sqrt{3}}{2}\right) = \pi - \cos^{-1} \left(\dfrac{\sqrt{3}}{2}\right) = \pi - \dfrac{\pi}{6} = \dfrac{5 \pi}{6}, \hspace{1cm} \dfrac{5 \pi}{6} \in \left[0, \pi\right]$

$\tan^{-1} \left(1\right) = \tan^{-1} \left[\tan \left(\dfrac{\pi}{4}\right)\right] = \dfrac{\pi}{4}, \hspace{1cm} \dfrac{\pi}{4} \in \left(- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$

$\begin{aligned} \therefore \; \sin^{-1} \left(\dfrac{\sqrt{3}}{2}\right) - \cos^{-1} \left(- \dfrac{\sqrt{3}}{2}\right) + 2 \tan^{-1} \left(1\right) & = \dfrac{\pi}{3} - \dfrac{5 \pi}{6} + 2 \times \dfrac{\pi}{4} \\\\ & = \dfrac{\pi}{3} - \dfrac{5 \pi}{6} + \dfrac{\pi}{2} \\\\ & = \dfrac{2 \pi - 5 \pi + 3 \pi}{6} \\\\ & = 0 \end{aligned}$

Inverse Trigonometric Functions

Evaluate: $\;$ $\sin \left[3 \sin^{-1} \left(\dfrac{1}{3}\right)\right]$

Let $\sin^{-1} \left(\dfrac{1}{3}\right) = \theta \;\;\; \text{ where } \theta \in \left[- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$

Then, $\sin \left(\theta\right) = \dfrac{1}{3}$

$\begin{aligned} \therefore \; \sin \left[3 \sin^{-1} \left(\dfrac{1}{3}\right)\right] & = \sin \left(3 \theta\right) \\\\ & = 3 \sin \theta - 4 \sin^3 \theta \\\\ & = 3 \times \dfrac{1}{3} - 4 \times \left(\dfrac{1}{3}\right)^3 \\\\ & = 1 - \dfrac{4}{27} \\\\ & = \dfrac{23}{27} \end{aligned}$

Inverse Trigonometric Functions

Evaluate: $\;$ $\tan^2 \left[\dfrac{1}{2} \sin^{-1} \left(\dfrac{2}{3}\right)\right]$

Let $\sin^{-1} \left(\dfrac{2}{3}\right) = \theta \;\;\; \text{ where } \theta \in \left[- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$

Then, $\sin \left(\theta\right) = \dfrac{2}{3}$

$ \therefore \; \tan^2 \left[\dfrac{1}{2} \sin^{-1} \left(\dfrac{2}{3}\right)\right] = \tan^2 \left(\dfrac{\theta}{2}\right)$

Now, $\;$ $\sin \left(\theta\right) = \dfrac{2 \tan \left(\dfrac{\theta}{2}\right)}{1 + \tan^2 \left(\dfrac{\theta}{2}\right)}$

i.e. $\;$ $\dfrac{2 \tan \left(\dfrac{\theta}{2}\right)}{1 + \tan^2 \left(\dfrac{\theta}{2}\right)} = \dfrac{2}{3}$

i.e. $\;$ $\tan^2 \left(\dfrac{\theta}{2}\right) - 3 \tan \left(\dfrac{\theta}{2}\right) + 1 = 0$

i.e. $\;$ $\tan \left(\dfrac{\theta}{2}\right) = \dfrac{3 \pm \sqrt{9 - 4}}{2}$

i.e. $\;$ $\tan \left(\dfrac{\theta}{2}\right) = \dfrac{3 + \sqrt{5}}{2}$ $\;$ OR $\;$ $\tan \left(\dfrac{\theta}{2}\right) = \dfrac{3 - \sqrt{5}}{2}$

$\therefore \;$ $\tan^2 \left(\dfrac{\theta}{2}\right) = \left(\dfrac{3 + \sqrt{5}}{2}\right)^2$ $\;$ OR $\;$ $\tan^2 \left(\dfrac{\theta}{2}\right) = \left(\dfrac{3 - \sqrt{5}}{2}\right)^2$

i.e. $\;$ $\tan^2 \left(\dfrac{\theta}{2}\right) = \dfrac{14 + 6 \sqrt{5}}{4}$ $\;$ OR $\;$ $\tan^2 \left(\dfrac{\theta}{2}\right) = \dfrac{14 - 6 \sqrt{5}}{4}$

i.e. $\;$ $\tan^2 \left(\dfrac{\theta}{2}\right) = \dfrac{7 \pm 3 \sqrt{5}}{2}$

$\because \;$ $\dfrac{7 + 3 \sqrt{5}}{2} \; \notin \; \left[-1, 1\right]$ $\;$ and $\;$ $\dfrac{7 - 3 \sqrt{5}}{2} \; \in \; \left[-1, 1\right]$,

$\therefore \;$ $\tan^2 \left(\dfrac{\theta}{2}\right) = \dfrac{7 - 3 \sqrt{5}}{2}$

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

Inverse Trigonometric Functions

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

Let $\tan^{-1} \left(\dfrac{2}{5}\right) = \theta \;\;\; \text{ where } \theta \in \left(- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$

Then, $\tan \left(\theta\right) = \dfrac{2}{5}$

$\begin{aligned} \therefore \; \sin \left[2 \tan^{-1} \left(\dfrac{2}{5}\right)\right] & = \sin \left(2 \theta\right) \\\\ & = \dfrac{2 \tan \theta}{1 + \tan^2 \theta} \\\\ & = \dfrac{2 \times \dfrac{2}{5}}{1 + \dfrac{4}{25}} \\\\ & = \dfrac{4 / 5}{29 / 25} \\\\ & = \dfrac{20}{29} \end{aligned}$

Inverse Trigonometric Functions

Evaluate: $\;$ $\tan^{-1} \left[\tan \left(\dfrac{5 \pi}{4}\right)\right]$

$\because \;$ $\dfrac{5 \pi}{4} \notin \left(- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$

$\begin{aligned} \therefore \; \tan^{-1} \left[\tan \left(\dfrac{5 \pi}{4}\right)\right] & = \tan^{-1} \left[\tan \left(\pi - \dfrac{5 \pi}{4}\right)\right] \\\\ & = \tan^{-1} \left[- \tan \left(- \dfrac{\pi}{4}\right)\right] \hspace{1cm} \left[\text{Note: } \tan \left(\pi - \theta\right) = - \tan \theta\right] \\\\ & = \tan^{-1} \left[\tan \left(\dfrac{\pi}{4}\right)\right] \hspace{1cm} \left[\text{Note: } \tan \left(- \theta\right) = - \tan \theta\right] \\\\ & = \dfrac{\pi}{4} \hspace{1cm} \left[\dfrac{\pi}{4} \in \left(- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right)\right] \end{aligned}$

Inverse Trigonometric Functions

Evaluate: $\;$ $\cos^{-1} \left[\sin \left(\dfrac{\pi}{7}\right)\right]$

$\begin{aligned} \cos^{-1} \left[\sin \left(\dfrac{\pi}{7}\right)\right] & = \cos^{-1} \left[\cos \left(\dfrac{\pi}{2} - \dfrac{\pi}{7}\right)\right] \hspace{1cm} \left[\text{Note: } \cos \left(\dfrac{\pi}{2} - \theta\right) = \sin \theta\right] \\\\ & = \cos^{-1} \left[\cos \left(\dfrac{5 \pi}{14}\right)\right] \\\\ & = \dfrac{5 \pi}{14} \hspace{1cm} \left(\dfrac{5 \pi}{14} \in \left[0, \pi\right]\right) \end{aligned}$

Inverse Trigonometric Functions

Prove that $\;$ $\cos^{-1} \left(x\right) + \cos^{-1} \left(y\right) = \cos^{-1} \left[x \cdot y - \sqrt{1 - x^2} \cdot \sqrt{1 - y^2}\right]$

Let $\;$ $\cos^{-1} \left(x\right) = \theta_1$, $\;$ $\cos^{-1} \left(y\right) = \theta_2$

Then, $\;$ $\cos \theta_1 = x$, $\;$ $\cos \theta_2 = y$

$\begin{aligned} \text{Now, } \cos \left(\theta_1 + \theta_2\right) & = \cos \theta_1 \cos \cdot \theta_2 - \sin \theta_1 \cdot \sin \theta_2 \\\\ & = \cos \theta_1 \cdot \cos \theta_2 - \sqrt{1 - \cos^2 \theta_1} \cdot \sqrt{1 - \cos^2 \theta_2} \\\\ & = x \cdot y - \sqrt{1 - x^2} \cdot \sqrt{1 - y^2} \end{aligned}$

$\therefore \;$ $\theta_1 + \theta_2 = \cos^{-1} \left[x \cdot y - \sqrt{1 - x^2} \cdot \sqrt{1 - y^2}\right]$

i.e. $\;$ $\cos^{-1} \left(x\right) + \cos^{-1} \left(y\right) = \cos^{-1} \left[x \cdot y - \sqrt{1 - x^2} \cdot \sqrt{1 - y^2}\right]$

Hence proved.

Inverse Trigonometric Functions

Solve: $\;$ $\tan^{-1} \left(\dfrac{2 x}{1 - x^2}\right) + \cot^{-1} \left(\dfrac{1 - x^2}{2x}\right) = \dfrac{\pi}{3}, \;$ where $\;$ $x > 0$

$\begin{aligned} \tan^{-1} \left(\dfrac{2 x}{1 - x^2}\right) + \cot^{-1} \left(\dfrac{1 - x^2}{2x}\right) & = \tan^{-1} \left(\dfrac{2 x}{1 - x^2}\right) + \cot^{-1} \left(\dfrac{1}{\dfrac{2x}{1 - x^2}}\right) \\\\ & \left[\text{Note: } \cot^{-1} \left(\dfrac{1}{x}\right) = \tan^{-1} \left(x\right)\right] \\\\ & = \tan^{-1} \left(\dfrac{2 x}{1 - x^2}\right) + \tan^{-1} \left(\dfrac{2x}{1 - x^2}\right) \\\\ & = 2 \tan^{-1} \left(\dfrac{2x}{1 - x^2}\right) \end{aligned}$

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

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

i.e. $\;$ $\tan^{-1} \left(\dfrac{2x}{1 - x^2}\right) = \dfrac{\pi}{6}$

i.e. $\;$ $\dfrac{2x}{1 - x^2} = \tan \left(\dfrac{\pi}{6}\right) = \dfrac{1}{\sqrt{3}}$

i.e. $\;$ $2 \sqrt{3} x = 1 - x^2$

i.e. $\;$ $x^2 + 2 \sqrt{3} x - 1 = 0$

i.e. $\;$ $x = \dfrac{- 2 \sqrt{3} \pm \sqrt{12 + 4}}{2}$

i.e. $\;$ $x = \dfrac{-2 \sqrt{3} \pm 4}{2}$

i.e. $\;$ $x = - \sqrt{3} + 2$ $\;$ OR $\;$ $x = - \sqrt{3} - 2$

Given: $\;$ $x > 0$

$\therefore \;$ $x = 2 - \sqrt{3}$

Inverse Trigonometric Functions

Prove that $\;$ $\tan^{-1} \left(\dfrac{m}{n}\right) - \tan^{-1} \left(\dfrac{m - n}{m + n}\right) = \dfrac{\pi}{4}$

$\begin{aligned} \text{LHS } = \tan^{-1} \left(\dfrac{m}{n}\right) - \tan^{-1} \left(\dfrac{m - n}{m + n}\right) & = \tan^{-1} \left[\dfrac{\dfrac{m}{n} - \left(\dfrac{m - n}{m + n}\right)}{1 + \dfrac{m}{n} \times \left(\dfrac{m - n}{m + n}\right)}\right] \\\\ & = \tan^{-1} \left[\dfrac{\dfrac{m^2 + m \cdot n - m \cdot n + n^2}{n \left(m + n\right)}}{\dfrac{m \cdot n + n^2 + m^2 - m \cdot n}{n \left(m + n\right)}}\right] \\\\ & = \tan^{-1} \left[\dfrac{m^2 + n^2}{n^2 + m^2}\right] \\\\ & = \tan^{-1} \left(1\right) \\\\ & = \dfrac{\pi}{4} = \text{ RHS} \;\;\; \text{ Hence proved} \end{aligned}$

Inverse Trigonometric Functions

Prove that $\;$ $\sin^{-1} \left(2 x \sqrt{1 - x^2}\right) = 2 \sin^{-1}x$

Let $\;$ $x = \sin \theta$ $\implies$ $\theta = \sin^{-1} \left(x\right)$ $\;\;\; \cdots \; (1)$

$\begin{aligned} \text{Now, } \sin^{-1} \left(2 x \sqrt{1 - x^2}\right) & = \sin^{-1} \left(2 \sin \theta \sqrt{1 - \sin^2 \theta}\right) \\\\ & = \sin^{-1} \left(2 \sin \theta \sqrt{\cos^2 \theta}\right) \\\\ & = \sin^{-1} \left(2 \sin \theta \cos \theta\right) \\\\ & = \sin^{-1} \left(\sin 2 \theta\right) \\\\ & = 2 \theta \\\\ & = 2 \sin^{-1} \left(x\right) \;\;\; \left[\text{in vew of equation } (1)\right] \\\\ & = \text{RHS} \;\;\; \text{Hence proved} \end{aligned}$

Inverse Trigonometric Functions

Prove that $\;$ $\tan^{-1} \left(\sqrt{\dfrac{1 - \cos x}{1 + \cos x}}\right) = \dfrac{x}{2}$

$\begin{aligned} \text{LHS } = \tan^{-1} \left[\sqrt{\dfrac{1 - \cos x}{1 + \cos x}}\right] & = \tan^{-1} \left[\sqrt{\dfrac{\left(1 - \cos x\right)^2}{1 - \cos^2 x}}\right] \\\\ & = \tan^{-1} \left[\dfrac{1 - \cos x}{\sin x}\right] \\\\ & = \tan^{-1} \left[\dfrac{2 \sin^2 \left(\dfrac{x}{2}\right)}{2 \sin \left(\dfrac{x}{2}\right) \cos \left(\dfrac{x}{2}\right)}\right] \\\\ & = \tan^{-1} \left[\dfrac{\sin \left(\dfrac{x}{2}\right)}{\cos \left(\dfrac{x}{2}\right)}\right] \\\\ & = \tan^{-1} \left[\tan \left(\dfrac{x}{2}\right)\right] \\\\ & = \dfrac{x}{2} = \text{RHS} \end{aligned}$

Inverse Trigonometric Functions

Evaluate $\;$ $\tan \left(\cos^{-1} \dfrac{8}{17}\right)$

Let $\;$ $\cos^{-1} \left(\dfrac{8}{17}\right) = \theta$ $\;\;\; \cdots \; (1)$

Then, $\;$ $\cos \theta = \dfrac{8}{17}$

and $\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - \dfrac{64}{289}} = \sqrt{\dfrac{225}{289}} = \dfrac{15}{17}$

$\therefore \;$ $\tan \theta = \dfrac{\sin \theta}{\cos \theta} = \dfrac{15}{8}$ $\;\;\; \cdots \; (2)$

Now, $\;$ $\tan \left(\cos^{-1} \dfrac{8}{17}\right) = \tan \theta = \dfrac{15}{8}$ $\;\;\;$ [By equations $(1)$ and $(2)$]

Inverse Trigonometric Functions

Evaluate $\;$ $\cos \left(\sin^{-1} \dfrac{5}{13}\right)$

Let $\;$ $\sin^{-1} \left(\dfrac{5}{13}\right) = \theta$ $\;\;\; \cdots \; (1)$

Then, $\;$ $\sin \theta = \dfrac{5}{13}$

and $\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \dfrac{25}{169}} = \sqrt{\dfrac{144}{169}} = \dfrac{12}{13}$ $\;\;\; \cdots \; (2)$

Now, $\;$ $\cos \left(\sin^{-1} \dfrac{5}{13}\right) = \cos \theta = \dfrac{12}{13}$ $\;\;\;$ [By equations $(1)$ and $(2)$]

Inverse Trigonometric Functions

Prove that $\;$ $2 \tan^{-1} \left(x\right) = \sin^{-1} \left(\dfrac{2 x}{1 + x^2}\right)$

Let $\;$ $\sin^{-1} \left(\dfrac{2 x}{1 + x^2}\right) = \theta$ $\;\;\; \cdots \; (1)$

Then, $\;$ $\sin \theta = \dfrac{2x}{1 + x^2}$

$\begin{aligned} \therefore \; \cos \theta & = \sqrt{1 - \sin^2 \theta} \\\\ & = \sqrt{1 - \dfrac{4x^2}{\left(1 + x^2\right)^2}} \\\\ & = \sqrt{\dfrac{1 + x^4 - 2 x^2}{\left(1 + x^2\right)^2}} \\\\ & = \sqrt{\left(\dfrac{1 - x^2}{1 + x^2}\right)^2} = \dfrac{1 - x^2}{1 + x^2} \\\\ \therefore \; \tan \theta & = \dfrac{\sin \theta}{\cos \theta} \\\\ & = \dfrac{\dfrac{2x}{1 + x^2}}{\dfrac{1 - x^2}{1 + x^2}} = \dfrac{2x}{1 - x^2} \end{aligned}$

$\implies$ $\theta = \tan^{-1} \left(\dfrac{2x}{1 - x^2}\right) = 2 \tan^{-1} \left(x\right)$ $\;\;\; \cdots \; (2)$

$\therefore \;$ We have from equations $(1)$ and $(2)$,

$2 \tan^{-1} \left(x\right) = \sin^{-1} \left(\dfrac{2 x}{1 + x^2}\right)$ $\;\;\;$ Hence proved

Inverse Trigonometric Functions

Prove that $\;$ $2 \tan^{-1} \left(\dfrac{1}{3}\right) = \tan^{-1} \left(\dfrac{3}{4}\right)$

$\begin{aligned} \text{LHS } & = 2 \tan^{-1} \left(\dfrac{1}{3}\right) \\\\ & = \tan^{-1} \left(\dfrac{1}{3}\right) + \tan^{-1} \left(\dfrac{1}{3}\right) \\\\ & = \tan^{-1} \left(\dfrac{\dfrac{1}{3} + \dfrac{1}{3}}{1 - \dfrac{1}{3} \times \dfrac{1}{3}}\right) \\\\ & \left[\text{Note: } \tan^{-1} \left(x\right) + \tan^{-1} \left(y\right) = \tan^{-1} \left(\dfrac{x + y}{1 - x \cdot y}\right), \;\; x \cdot y < 1\right] \\\\ & = \tan^{-1} \left(\dfrac{\dfrac{2}{3}}{\dfrac{8}{9}}\right) \\\\ & = \tan^{-1} \left(\dfrac{3}{4}\right) = \text{ RHS} \end{aligned}$

Inverse Trigonometric Functions

Find the principal value of $\tan^{-1} \left(\sqrt{3}\right)$

Let $\;$ $\tan^{-1} \left(\sqrt{3}\right) = y$, $\;$ where $\;$ $- \dfrac{\pi}{2} < y < \dfrac{\pi}{2}$

$\implies$ $\tan \left(y\right) = \sqrt{3} = \tan \left(\dfrac{\pi}{3}\right)$

$\implies$ $y = \dfrac{\pi}{3}$

$\therefore \;$ The principal value of $\tan^{-1} \left(\sqrt{3}\right)$ is $\dfrac{\pi}{3}$

Inverse Trigonometric Functions

Find the principal value of $\sec^{-1} \left(- \sqrt{2}\right)$

Let $\;$ $\sec^{-1} \left(- \sqrt{2}\right) = y$, $\;$ where $\;$ $0 < y \leq \pi; \;\; y \neq \dfrac{\pi}{2}$

$\implies$ $\sec \left(y\right) = - \sqrt{2} = \sec \left(\dfrac{3 \pi}{4}\right)$

$\implies$ $y = \dfrac{3 \pi}{4}$

$\therefore \;$ The principal value of $\sec^{-1} \left(- \sqrt{2}\right)$ is $\dfrac{3 \pi}{4}$

Inverse Trigonometric Functions

Find the principal value of $\sin^{-1} \left(\dfrac{\sqrt{3}}{2}\right)$

Let $\;$ $\sin^{-1} \left(\dfrac{\sqrt{3}}{2}\right) = y$, $\;$ where $\;$ $- \dfrac{\pi}{2} \leq y \leq \dfrac{\pi}{2}$

$\implies$ $\sin \left(y\right) = \dfrac{\sqrt{3}}{2} = \sin \left(\dfrac{\pi}{3}\right)$

$\implies$ $y = \dfrac{\pi}{3}$

$\therefore \;$ The principal value of $\sin^{-1} \left(\dfrac{\sqrt{3}}{2}\right)$ is $\dfrac{\pi}{3}$

Trigonometric Equations

Solve the equation giving the exact solutions which lie in $\left[0, 2\pi \right)$: $\;$ $\;$ $\tan x = \cos x$

$\tan x = \cos x$

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

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

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

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

i.e. $\;$ $\sin x = \dfrac{-1 \pm \sqrt{1 + 4}}{2} = \dfrac{-1 \pm \sqrt{5}}{2}$

i.e. $\;$ $x = n \pi + \left(-1\right)^n \times \sin^{-1} \left(\dfrac{-1 \pm \sqrt{5}}{2}\right)$

In the range $\left[0, 2 \pi \right)$,

when $\;$ $n = 0$, $\;$ $x = \sin^{-1} \left(\dfrac{-1 + \sqrt{5}}{2}\right)$

when $\;$ $n = 1$, $\;$ $x = \pi - \sin^{-1} \left(\dfrac{-1 + \sqrt{5}}{2}\right)$

$\therefore \;$ For the given equation, the exact solutions which lie in $\left[0, 2\pi \right)$ are:

$x = \sin^{-1} \left(\dfrac{-1 + \sqrt{5}}{2}\right), \; \pi - \sin^{-1} \left(\dfrac{-1 + \sqrt{5}}{2}\right)$

Trigonometric Equations

Solve the equation giving the exact solutions which lie in $\left[0, 2\pi \right)$: $\;$ $\;$ $\cos \left(4x\right) = \cos \left(2x\right)$

$\cos \left(4x\right) = \cos \left(2x\right)$

i.e. $\;$ $\cos \left(4x\right) - \cos \left(2x\right) = 0$

i.e. $\;$ $- 2 \sin \left(\dfrac{4x + 2x}{2}\right) \; \sin \left(\dfrac{4x - 2x}{2}\right) = 0$

i.e. $\;$ $\sin \left(3 x\right) \; \sin \left(x\right) = 0$

i.e. $\;$ $\sin \left(3 x\right) = 0$ $\;$ OR $\;$ $\sin \left(x\right) = 0$

i.e. $\;$ $3x = n \pi, \;\;\; n \in Z$ $\;$ OR $\;$ $x = m \pi, \;\;\; m \in Z$

i.e. $\;$ $x = \dfrac{n \pi}{3}, \;\;\; n \in Z$ $\;$ OR $\;$ $x = m \pi, \;\;\; m \in Z$

In the range $\left[0, 2 \pi \right)$,

when $\;$ $n = 0$, $\;$ $x = 0$

when $\;$ $n = 1$, $\;$ $x = \dfrac{\pi}{3}$

when $\;$ $n = 2$, $\;$ $x = \dfrac{2 \pi}{3}$

when $\;$ $n = 3$, $\;$ $x = \pi$

when $\;$ $n = 4$, $\;$ $x = \dfrac{4 \pi}{3}$

when $\;$ $n = 5$, $\;$ $x = \dfrac{5 \pi}{3}$

when $\;$ $m = 0$, $\;$ $x = 0$

when $\;$ $m = 1$, $\;$ $x = \pi$

$\therefore \;$ For the given equation, the exact solutions which lie in $\left[0, 2\pi \right)$ are:

$x = 0, \; \dfrac{\pi}{3}, \; \dfrac{2 \pi}{3}, \; \pi, \; \dfrac{4 \pi}{3}, \; \dfrac{5 \pi}{3}$

Trigonometric Equations

Solve the equation giving the exact solutions which lie in $\left[0, 2\pi \right)$: $\;$ $\;$ $\sqrt{2} \; \cos \left(x\right) - \sqrt{2} \; \sin \left(x\right) = 1$

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

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

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

i.e. $\;$ $\cos \left(\dfrac{\pi}{4}\right) \; \cos x - \sin \left(\dfrac{\pi}{4}\right) \; \sin x = \dfrac{1}{2}$

i.e. $\;$ $\cos \left(x + \dfrac{\pi}{4}\right) = \dfrac{1}{2} = \cos \left(\dfrac{\pi}{3}\right)$

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

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

i.e. $\;$ $x = 2 n \pi + \dfrac{\pi}{12}, \;\;\; n \in Z$ $\;$ OR $\;$ $x = 2 n \pi - \dfrac{7 \pi}{12}, \;\;\; n \in Z$

In the range $\left[0, 2 \pi \right)$,

when $\;$ $n = 0$, $\;$ $x = \dfrac{\pi}{12}$ $\;$ OR $\;$ $x = - \dfrac{7 \pi}{12}$ $\;\;$ $\left[- \dfrac{7\pi}{12} \text{ is the same as } 2 \pi - \dfrac{7\pi}{12} = \dfrac{17 \pi}{12} \right]$

when $\;$ $n = 1$, $\;$ $x = 2 \pi + \dfrac{\pi}{12} = \dfrac{25 \pi}{12} \notin \left[0, 2 \pi \right)$ $\;$ OR $\;$ $x = 2 \pi - \dfrac{7 \pi}{12} = \dfrac{17 \pi}{12}$

$\therefore \;$ For the given equation, the exact solutions which lie in $\left[0, 2\pi \right)$ are:

$x = \dfrac{\pi}{12}, \; \dfrac{17 \pi}{12}$

Trigonometric Equations

Solve the equation giving the exact solutions which lie in $\left[0, 2\pi \right)$: $\;$ $\;$ $\cos \left(5x\right) \; \cos \left(3x\right) - \sin \left(5x\right) \; \sin \left(3x\right) = \dfrac{\sqrt{3}}{2}$

$\cos \left(5x\right) \; \cos \left(3x\right) - \sin \left(5x\right) \; \sin \left(3x\right) = \dfrac{\sqrt{3}}{2}$

i.e. $\;$ $\cos \left(5 x + 3 x\right) = \dfrac{\sqrt{3}}{2}$

i.e. $\;$ $\cos \left(8 x\right) = \dfrac{\sqrt{3}}{2} = \cos \left(\dfrac{\pi}{6}\right)$

i.e. $\;$ $8 x = 2 n \pi \pm \dfrac{\pi}{6}, \;\;\; n \in Z$

i.e. $\;$ $x = \dfrac{n \pi}{4} \pm \dfrac{\pi}{48}, \;\;\; n \in Z$

In the range $\left[0, 2 \pi \right)$,

when $\;$ $n = 0$, $\;$ $x = \dfrac{\pi}{48}$

when $\;$ $n = 1$, $\;$ $x = \dfrac{\pi}{4} - \dfrac{\pi}{48} = \dfrac{11 \pi}{48}$ $\;$ OR $\;$ $x = \dfrac{\pi}{4} + \dfrac{\pi}{48} = \dfrac{13 \pi}{48}$

when $\;$ $n = 2$, $\;$ $x = \dfrac{\pi}{2} - \dfrac{\pi}{48} = \dfrac{23 \pi}{48}$ $\;$ OR $\;$ $x = \dfrac{\pi}{2} + \dfrac{\pi}{48} = \dfrac{25 \pi}{48}$

when $\;$ $n = 3$, $\;$ $x = \dfrac{3 \pi}{4} - \dfrac{\pi}{48} = \dfrac{35 \pi}{48}$ $\;$ OR $\;$ $x = \dfrac{3 \pi}{4} + \dfrac{\pi}{48} = \dfrac{37 \pi}{48}$

when $\;$ $n = 4$, $\;$ $x = \pi - \dfrac{\pi}{48} = \dfrac{47 \pi}{48}$ $\;$ OR $\;$ $x = \pi + \dfrac{\pi}{48} = \dfrac{49 \pi}{48}$

when $\;$ $n = 5$, $\;$ $x = \dfrac{5 \pi}{4} - \dfrac{\pi}{48} = \dfrac{59 \pi}{48}$ $\;$ OR $\;$ $x = \dfrac{5 \pi}{4} + \dfrac{\pi}{48} = \dfrac{61 \pi}{48}$

when $\;$ $n = 6$, $\;$ $x = \dfrac{3\pi}{2} - \dfrac{\pi}{48} = \dfrac{71 \pi}{48}$ $\;$ OR $\;$ $x = \dfrac{3\pi}{2} + \dfrac{\pi}{48} = \dfrac{73 \pi}{48}$

when $\;$ $n = 7$, $\;$ $x = \dfrac{7 \pi}{4} - \dfrac{\pi}{48} = \dfrac{83 \pi}{48}$ $\;$ OR $\;$ $x = \dfrac{7 \pi}{4} + \dfrac{\pi}{48} = \dfrac{85 \pi}{48}$

when $\;$ $n = 8$, $\;$ $x = 2 \pi - \dfrac{\pi}{48} = \dfrac{95 \pi}{48}$

$\therefore \;$ For the given equation, the exact solutions which lie in $\left[0, 2\pi \right)$ are:

$x = \dfrac{\pi}{48}$, $\;$ $\dfrac{11\pi}{48}$, $\;$ $\dfrac{13\pi}{48}$, $\;$ $\dfrac{23\pi}{48}$, $\;$ $\dfrac{25\pi}{48}$, $\;$ $\dfrac{35\pi}{48}$, $\;$ $\dfrac{37\pi}{48}$, $\;$ $\dfrac{47\pi}{48}$, $\;$ $\dfrac{49\pi}{48}$, $\;$ $\dfrac{59\pi}{48}$, $\;$ $\dfrac{61\pi}{48}$, $\;$ $\dfrac{71\pi}{48}$, $\;$ $\dfrac{73\pi}{48}$, $\;$ $\dfrac{83\pi}{48}$, $\;$ $\dfrac{85\pi}{48}$, $\;$ $\dfrac{95\pi}{48}$

Trigonometric Equations

Solve the equation giving the exact solutions which lie in $\left[0, 2\pi \right)$: $\;$ $\;$ $\sin \left(6x\right) \; \cos x = - \cos \left(6x\right) \; \sin x$

$\sin \left(6x\right) \; \cos x = - \cos \left(6x\right) \; \sin x$

i.e. $\;$ $\sin \left(6x\right) \; \cos x + \cos \left(6x\right) \; \sin x = 0$

i.e. $\;$ $\sin \left(6x + x\right) = 0$

i.e. $\;$ $\sin \left(7 x\right) = 0$

$\implies$ $7 x = n \pi, \;\;\; n \in Z$

i.e. $\;$ $x = \dfrac{n \; \pi}{7}, \;\;\; n \in Z$

In the range $\left[0, 2 \pi \right)$,

when $\;$ $n = 0$, $\;$ $x = 0$

when $\;$ $n = 1$, $\;$ $x = \dfrac{\pi}{7}$

when $\;$ $n = 2$, $\;$ $x = \dfrac{2 \pi}{7}$

when $\;$ $n = 3$, $\;$ $x = \dfrac{3 \pi}{7}$

when $\;$ $n = 4$, $\;$ $x = \dfrac{4 \pi}{7}$

when $\;$ $n = 5$, $\;$ $x = \dfrac{5 \pi}{7}$

when $\;$ $n = 6$, $\;$ $x = \dfrac{6 \pi}{7}$

when $\;$ $n = 7$, $\;$ $x = \pi$

when $\;$ $n = 8$, $\;$ $x = \dfrac{8 \pi}{7}$

when $\;$ $n = 9$, $\;$ $x = \dfrac{9 \pi}{7}$

when $\;$ $n = 10$, $\;$ $x = \dfrac{10 \pi}{7}$

when $\;$ $n = 11$, $\;$ $x = \dfrac{11 \pi}{7}$

when $\;$ $n = 12$, $\;$ $x = \dfrac{12 \pi}{7}$

when $\;$ $n = 13$, $\;$ $x = \dfrac{13 \pi}{7}$

$\therefore \;$ For the given equation, the exact solutions which lie in $\left[0, 2\pi \right)$ are:

$x = 0, \; \dfrac{\pi}{7}, \; \dfrac{2\pi}{7}, \; \dfrac{3\pi}{7}, \cdots, \pi, \; \cdots, \dfrac{13 \pi}{7}$

Trigonometric Equations

Solve the equation giving the exact solutions which lie in $\left[0, 2\pi \right)$: $\;$ $\;$ $\tan \left(2x\right) - 2 \cos x = 0$

$\tan \left(2x\right) - 2 \cos x = 0$

i.e. $\;$ $\dfrac{\sin \left(2x\right)}{\cos \left(2x\right)} - 2 \cos x = 0$

i.e. $\;$ $\dfrac{2 \sin x \; \cos x}{1 - 2 \sin^2 x} - 2 \cos x = 0$

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

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

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

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

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

i.e. $\;$ $\cos x = 0$ $\;$ OR $\;$ $\sin x = \dfrac{1}{2}$ $\;$ OR $\;$ $\sin x = - 1$

Now, $\;$ $\cos x = 0$ $\implies$ $x = 2 n \pi \pm \dfrac{\pi}{2}, \;\;\; n \in Z$

OR, $\;$ $\sin x = \dfrac{1}{2} = \sin \left(\dfrac{\pi}{6}\right)$ $\implies$ $x = m \pi + \left(-1\right)^m \times \left(\dfrac{\pi}{6}\right), \;\;\; m \in Z$

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

In the range $\left[0, 2 \pi \right)$,

when $\;$ $n = 0$, $\;$ $x = \dfrac{\pi}{2}$

when $\;$ $m = 0$, $\;$ $x = \dfrac{\pi}{6}$

when $\;$ $m = 1$, $\;$ $x = \pi - \dfrac{\pi}{6} = \dfrac{5 \pi}{6}$

when $\;$ $p = 0$, $x = - \dfrac{\pi}{2}$

$\left[- \dfrac{\pi}{2} \text{ is the same as } 2 \pi - \dfrac{\pi}{2} = \dfrac{3 \pi}{2} \right]$

$\therefore \;$ For the given equation, the exact solutions which lie in $\left[0, 2\pi \right)$ are:

$x = \dfrac{\pi}{6}, \; \dfrac{\pi}{2}, \; \dfrac{5 \pi}{6}, \; \dfrac{3 \pi}{2}$

Trigonometric Equations

Solve the equation giving the exact solutions which lie in $\left[0, 2\pi \right)$: $\;$ $\;$ $\tan^3 x = 3 \tan x$

$\tan^3 x = 3 \tan x$

i.e. $\;$ $\tan x \left(\tan^2 x - 3\right) = 0$

i.e. $\;$ $\tan x = 0$ $\;$ OR $\;$ $\tan^2 x - 3 = 0$

i.e. $\;$ $\tan x = 0$ $\;$ OR $\;$ $\tan x = \pm \sqrt{3}$

Now, $\tan x = 0$ $\implies$ $x = n \pi, \;\;\; n \in Z$

OR $\;$ $\tan x = \sqrt{3} = \tan \left(\dfrac{\pi}{3}\right)$ $\implies$ $x = m \pi + \dfrac{\pi}{3}, \;\;\; m \in Z$

OR $\;$ $\tan x = - \sqrt{3} = \tan \left(- \dfrac{\pi}{3}\right)$ $\implies$ $x = q \pi - \dfrac{\pi}{3}, \;\;\; q \in Z$

In the range $\left[0, 2 \pi \right)$,

when $n = 0$, $\;$ $x = 0$

when $n = 1$, $\;$ $x = \pi$

when $m = 0$, $\;$ $x = \dfrac{\pi}{3}$

when $m = 1$, $\;$ $x = \pi + \dfrac{\pi}{3} = \dfrac{4 \pi}{3}$

when $q = 0$, $\;$ $x = - \dfrac{\pi}{3}$

$\left[- \dfrac{\pi}{3} \text{ is the same as } 2 \pi - \dfrac{\pi}{3} = \dfrac{5 \pi}{3} \right]$

when $q = 1$, $\;$ $x = \pi - \dfrac{\pi}{3} = \dfrac{2 \pi}{3}$

$\therefore \;$ For the given equation, the exact solutions which lie in $\left[0, 2\pi \right)$ are:

$x = 0, \; \dfrac{\pi}{3}, \; \dfrac{2 \pi}{3}, \; \pi, \; \dfrac{4 \pi}{3}, \; \dfrac{5 \pi}{3}$

Trigonometric Equations

Solve the equation giving the exact solutions which lie in $\left[0, 2\pi \right)$: $\;$ $\;$ $\sin \left(2x\right) = \tan x$

$\sin \left(2x\right) = \tan x$

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

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

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

i.e. $\;$ $\sin x - 2 \sin^3 x = 0$

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

i.e. $\;$ $\sin x = 0$ $\;$ OR $\;$ $1 - 2 \sin^2 x = 0$

i.e. $\;$ $\sin x = 0$ $\;$ OR $\;$ $\sin^2 x = \dfrac{1}{2}$

i.e. $\;$ $\sin x = 0$ $\;$ OR $\;$ $\sin x = \pm \dfrac{1}{\sqrt{2}}$

Now, $\sin x = 0$ $\implies$ $x = n \pi, \;\;\; n \in Z$

OR $\;$ $\sin x = \dfrac{1}{\sqrt{2}}$ $\implies$ $x = m \pi + \left(-1\right)^m \times \left(\dfrac{\pi}{4}\right), \;\;\; m \in Z$

OR $\;$ $\sin x = - \dfrac{1}{\sqrt{2}}$ $\implies$ $x = q \pi + \left(-1\right)^q \times \left(-\dfrac{\pi}{4}\right), \;\;\; q \in Z$

In the range $\left[0, 2 \pi \right)$,

when $n = 0$, $\;$ $x = 0$

when $n = 1$, $\;$ $x = \pi$

when $m = 0$, $\;$ $x = \dfrac{\pi}{4}$

when $m = 1$, $\;$ $x = \pi - \dfrac{\pi}{4} = \dfrac{3 \pi}{4}$

when $q = 0$, $\;$ $x = - \dfrac{\pi}{4}$

$\left[- \dfrac{\pi}{4} \text{ is the same as } 2 \pi - \dfrac{\pi}{4} = \dfrac{7 \pi}{4} \right]$

when $q = 1$, $\;$ $x = \pi + \dfrac{\pi}{4} = \dfrac{5 \pi}{4}$

$\therefore \;$ For the given equation, the exact solutions which lie in $\left[0, 2\pi \right)$ are:

$x = 0, \; \dfrac{\pi}{4}, \; \dfrac{3 \pi}{4}, \; \pi, \; \dfrac{5 \pi}{4}, \; \dfrac{7 \pi}{4}$

Trigonometric Equations

Solve the equation giving the exact solutions which lie in $\left[0, 2\pi \right)$: $\;$ $\;$ $\cot^2 x = 3 \csc x - 3$

$\cot^2 x = 3 \csc x - 3$

i.e. $\;$ $\dfrac{\cos^2 x}{\sin^2 x} = \dfrac{3}{\sin x} - 3$

i.e. $\;$ $\cos^2 x = 3 \sin x - 3 \sin^2 x$

i.e. $\;$ $1 - \sin^2 x = 3 \sin x - 3 \sin^2 x$

i.e. $\;$ $2 \sin^2 x - 3 \sin x + 1 = 0$

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

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

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

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

i.e. $\;$ $\sin x = \sin \left(\dfrac{\pi}{6}\right)$ $\;$ OR $\;$ $\sin x = \sin \left(\dfrac{\pi}{2}\right)$

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

OR $\;$ $x = m \pi + \left(-1\right)^m \times \left(\dfrac{\pi}{2}\right), \;\;\; m \in Z$

In the range $\left[0, 2 \pi \right)$,

when $n = 0$, $\;$ $x = \dfrac{\pi}{6}$

when $n = 1$, $\;$ $x = \pi - \dfrac{\pi}{6} = \dfrac{5 \pi}{6}$

when $m = 0$, $\;$ $x = \dfrac{\pi}{2}$

$\therefore \;$ For the given equation, the exact solutions which lie in $\left[0, 2\pi \right)$ are:

$x = \dfrac{\pi}{6}, \; \dfrac{5 \pi}{6}, \; \dfrac{\pi}{2}$

Trigonometric Equations

Solve the equation giving the exact solutions which lie in $\left[0, 2\pi \right)$: $\;$ $\;$ $3 \cos \left(2x\right) = \sin x + 2$

$3 \cos \left(2x\right) = \sin x + 2$

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

i.e. $\;$ $3 - 6 \sin^2 x = \sin x + 2$

i.e. $\;$ $6 \sin^2 x + \sin x - 1 = 0$

i.e. $\;$ $6 \sin^2 x + 3 \sin x - 2 \sin x - 1 = 0$

i.e. $\;$ $3 \sin x \left(2 \sin x + 1\right) - 1 \left(2 \sin x + 1\right) = 0$

i.e. $\;$ $\left(3 \sin x - 1\right) \left(2 \sin x + 1\right) = 0$

i.e. $\;$ $3 \sin x - 1 = 0$ $\;$ OR $\;$ $2 \sin x + 1 = 0$

$\implies$ $\sin x = \dfrac{1}{3}$ $\;$ OR $\;$ $\sin x = - \dfrac{1}{2}$

i.e. $\;$ $\sin x = \dfrac{1}{3}$ $\;$ OR $\;$ $\sin x = \sin \left(- \dfrac{\pi}{6}\right)$

i.e. $\;$ $x = n \pi + \left(-1\right)^n \times \arcsin \left(\dfrac{1}{3}\right), \;\;\; n \in Z$

OR $\;$ $x = m \pi + \left(- 1\right)^m \times \left(- \dfrac{\pi}{6}\right), \;\;\; m \in Z$

In the range $\left[0, 2 \pi \right)$,

when $n = 0$, $\;$ $x = \arcsin \left(\dfrac{1}{3}\right)$

when $n = 1$, $\;$ $x = \pi - \arcsin \left(\dfrac{1}{3}\right)$

when $m = 0$, $\;$ $x = - \dfrac{\pi}{6}$

$- \dfrac{\pi}{6}$ $\;$ is the same as $\;$ $2 \pi - \dfrac{\pi}{6} = \dfrac{11 \pi}{6}$

when $m = 1$, $\;$ $x = \pi + \dfrac{\pi}{6} = \dfrac{7 \pi}{6}$

$\therefore \;$ For the given equation, the exact solutions which lie in $\left[0, 2\pi \right)$ are:

$x = \arcsin \left(\dfrac{1}{3}\right), \; \pi - \arcsin \left(\dfrac{1}{3}\right), \; \dfrac{7 \pi}{6}, \; \dfrac{11 \pi}{6}$

Trigonometric Equations

Solve the equation giving the exact solutions which lie in $\left[0, 2\pi \right)$: $\;$ $\;$ $\cos \left(2x\right) = 2 - 5 \cos x$

$\cos \left(2x\right) = 2 - 5 \cos x$

i.e. $\;$ $2 \cos^2 x - 1 = 2 - 5 \cos x$

i.e. $\;$ $2 \cos^2 x + 5 \cos x - 3 = 0$

i.e. $\;$ $2 \cos^2 x + 6 \cos x - \cos x - 3 = 0$

i.e. $\;$ $\cos x \left(2 \cos x - 1\right) + 3 \left(2 \cos x - 1\right) = 0$

i.e. $\;$ $\left(\cos x + 3\right) \left(2 \cos x - 1\right) = 0$

i.e. $\;$ $\cos x + 3 = 0$ $\;$ OR $\;$ $2 \cos x - 1 = 0$

Now, $\cos x + 3 = 0$ $\implies$ $\cos x = -3$, $\;$ is not possible $\;$ $\because \; -1 \leq \cos x \leq 1$

$2 \cos x - 1 = 0$ $\implies$ $\cos x = \dfrac{1}{2}$

$\left\{\text{Principal value of } \cos x \text{ lies in } \left[0, \pi\right] \text{ i.e. in first or second quadrants} \right\}$

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

$\implies$ $x = 2 n \pi \pm \dfrac{\pi}{3}, \;\;\; n \in Z$

$\therefore \;$ In the range $\left[0, 2\pi \right)$,

when $\;$ $n = 0$, $\;$ $x = \dfrac{\pi}{3}$

when $\;$ $n = 1$ $\;$ $x = 2 \pi - \dfrac{\pi}{3} = \dfrac{5 \pi}{3}$

$\therefore \;$ For the given equation, the exact solutions which lie in $\left[0, 2\pi \right)$ are:

$x = \dfrac{\pi}{3}, \; \dfrac{5 \pi}{3}$

Trigonometric Equations

Solve the equation $\;$ $\sin \left(2x\right) = \cos x$ $\;$ giving the exact solutions which lie in $\left[0, 2\pi \right)$.

$\sin \left(2x\right) = \cos x$

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

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

$\implies$ $\cos x = 0$ $\;$ OR $\;$ $2 \sin x - 1 = 0$

Now, $\cos x = 0 = \cos \left(\dfrac{\pi}{2}\right)$

$\left\{\text{Principal value of } \cos x \text{ lies in } \left[0, \pi\right] \text{ i.e. in first or second quadrants} \right\}$

$\therefore \;$ $x = 2 n \pi \pm \dfrac{\pi}{2}, \;\;\; n \in Z$

$\therefore \;$ In the range $\left[0, 2\pi \right)$,

when $\;$ $n = 0$, $\;$ $x = \dfrac{\pi}{2}$

when $\;$ $n = 1$, $\;$ $x = 2 \pi - \dfrac{\pi}{2} = \dfrac{3 \pi}{2}$

Now, $2 \sin x - 1 = 0$ $\implies$ $\sin x = \dfrac{1}{2}$ $\;\;\;$ [i.e. $\sin x$ is positive]

$\left\{\text{Principal value of } \sin x \text{ lies in } \left[- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right] \text{ i.e. in first or fourth quadrants} \right\}$

$\therefore \;$ $\sin x = \dfrac{1}{2} = \sin \left(\dfrac{\pi}{6}\right)$

$\implies$ $x = m \pi + \left(- 1\right)^m \times \dfrac{\pi}{6}, \;\;\; m \in Z$

$\therefore \;$ In the range $\left[0, 2\pi \right)$,

when $\;$ $m = 0$, $\;$ $x = \dfrac{\pi}{6}$

when $\;$ $m = 1$, $\;$ $x = \pi - \dfrac{\pi}{6} = \dfrac{5 \pi}{6}$

$\therefore \;$ For the given equation, the exact solutions which lie in $\left[0, 2\pi \right)$ are:

$x = \dfrac{\pi}{6}, \; \dfrac{\pi}{2}, \; \dfrac{5 \pi}{6}, \; \dfrac{3 \pi}{2}$

Trigonometric Equations

Solve: $\sqrt{2} \sec \theta + \tan \theta = 1$

$\sqrt{2} \sec \theta + \tan \theta = 1$

i.e. $\;$ $\dfrac{\sqrt{2}}{\cos \theta} + \dfrac{\sin \theta}{\cos \theta} = 1$

i.e. $\;$ $\sqrt{2} + \sin \theta = \cos \theta$

i.e. $\;$ $\cos \theta - \sin \theta = \sqrt{2}$

i.e. $\;$ $\dfrac{1}{\sqrt{2}} \cos \theta - \dfrac{1}{\sqrt{2}} \sin \theta = 1$

i.e. $\;$ $\cos \theta \; \cos \left(\dfrac{\pi}{4}\right) - \sin \theta \; \sin \left(\dfrac{\pi}{4}\right) = 1$

i.e. $\;$ $\cos \left(\theta + \dfrac{\pi}{4}\right) = 1$ $\;\;\;$ [i.e. $\;$ $\cos \left(\theta + \dfrac{\pi}{4}\right)$ is positive]

$\left\{\text{Principal value of } \cos x \text{ lies in } \left[0, \pi\right] \text{ i.e. in first or second quadrants} \right\}$

$\therefore \;$ We have $\cos \left(\theta + \dfrac{\pi}{4}\right) = 1 = \cos 0$

i.e. $\;$ $\theta + \dfrac{\pi}{4} = 2 n \pi \pm 0$

i.e. $\;$ $\theta = 2 n \pi - \dfrac{\pi}{4}, \;\;\; n \in Z$

Trigonometric Equations

Solve: $\sin \theta + \cos \theta = \sqrt{2}$

$\sin \theta + \cos \theta = \sqrt{2}$

i.e. $\;$ $\dfrac{1}{\sqrt{2}} \sin \theta + \dfrac{1}{\sqrt{2}} \cos \theta = 1$ $\;\;\; \cdots \; (1)$

i.e. $\;$ $\cos \left(\dfrac{\pi}{4}\right) \sin \theta + \sin \left(\dfrac{\pi}{4}\right) \cos \theta = 1$

i.e. $\;$ $\sin \left(\dfrac{\pi}{4} + \theta\right) = 1$

[Principal value of $\sin \theta$ is in $\left[- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ $\;$ i.e. $\;$ in first or fourth quadrants.]

$\because \;$ $\sin \left(\dfrac{\pi}{4} + \theta\right) = 1$ is positive

$\implies$ $\sin \left(\dfrac{\pi}{4} + \theta\right) = \sin \left(\dfrac{\pi}{2}\right)$

$\implies$ $\dfrac{\pi}{4} + \theta = n \pi + \left(-1\right)^n \left(\dfrac{\pi}{2}\right)$

$\therefore \;$ $\theta = n \pi + \left(-1\right)^n \left(\dfrac{\pi}{2}\right) - \dfrac{\pi}{4}, \;\;\; n \in Z$

OR

Equation $(1)$ can also be written as

$\cos \left(\dfrac{\pi}{4}\right) \; \cos \theta + \sin \left(\dfrac{\pi}{4}\right) \; \sin \theta = 1$

i.e. $\;$ $\cos \left(\theta - \dfrac{\pi}{4}\right) = 1$ $\;\;\;$ [i.e. $\;$ $\cos \left(\theta - \dfrac{\pi}{4}\right)$ is positive]

$\left\{\text{Principal value of } \cos x \text{ lies in } \left[0, \pi\right] \text{ i.e. in first or second quadrants} \right\}$

$\therefore \;$ We have $\cos \left(\theta - \dfrac{\pi}{4}\right) = 1 = \cos 0$

i.e. $\;$ $\theta - \dfrac{\pi}{4} = 2 m \pi \pm 0$

i.e. $\;$ $\theta = 2 m \pi + \dfrac{\pi}{4}, \;\;\; m \in Z$

Trigonometric Equations

Solve: $\sin 2x + \sin 4x = 2 \sin 3x$

$\sin 2x + \sin 4x = 2 \sin 3x$

i.e. $\;$ $2 \sin \left(\dfrac{2x + 4x}{2}\right) \cos \left(\dfrac{4x - 2x}{2}\right) = 2 \sin 3x$

i.e. $\;$ $\sin 3x \; \cos x = \sin 3x$

i.e. $\;$ $\sin 3x \left(\cos x - 1\right) = 0$

i.e. $\;$ $\sin 3x = 0$ $\;$ or $\;$ $\cos x - 1 = 0$

Now, $\;$ $\sin 3x = 0$ $\implies$ $3x = n \pi$ $\;\;$ i.e. $\;$ $x = \dfrac{n \pi}{3}, \;\;\; n \in Z$

Principal value of $\cos x$ lies in $\left[0, \pi\right]$

Now, $\cos x - 1 = 0$ $\implies$ $\cos x = 1 = \cos 0$ $\;\;\;$ $\left(\because \;\cos x = +ve\right)$

$\implies$ $x = 2 m \pi, \;\;\; m \in Z$

Trigonometric Equations

Solve: $\sin^2 \theta - 2 \cos \theta + \dfrac{1}{4} = 0$

$\sin^2 \theta - 2 \cos \theta + \dfrac{1}{4} = 0$

i.e. $\;$ $1 - \cos ^2 \theta - 2 \cos \theta + \dfrac{1}{4} = 0$

i.e. $\;$ $\cos ^2 \theta + 2 \cos \theta - \dfrac{5}{4} = 0$

i.e. $\;$ $4 \cos^2 \theta + 8 \cos \theta - 5 = 0$

i.e. $\;$ $4 \cos^2 \theta + 10 \cos \theta - 2 \cos \theta - 5 = 0$

i.e. $\;$ $2 \cos \theta \left(2 \cos \theta - 1\right) + 5 \left(2 \cos \theta - 1\right) = 0$

i.e. $\;$ $\left(2 \cos \theta + 5\right) \left(2 \cos \theta - 1\right) = 0$

i.e. $\;$ $\cos \theta = - \dfrac{5}{2}$ $\;$ or $\;$ $\cos \theta = \dfrac{1}{2}$

$\because \;$ $-1 \leq \cos \theta \leq 1$ $\;$ $\therefore \;$ $\cos \theta = - \dfrac{5}{2}$ is not a valid solution.

Principal value of $\cos \theta$ is in $\left[0, \pi\right]$ $\;$ i.e. $\;$ in first or second quadrants.

$\therefore \;$ $\cos \theta = \dfrac{1}{2} = \cos \left(\dfrac{\pi}{3}\right)$ $\;\;\;$ $\left(\because \;\cos \theta = +ve\right)$

$\implies$ $\theta = 2 n \pi \pm \dfrac{\pi}{3}, \;\;\; n \in Z$

Trigonometric Equations

Solve: $\tan 2x = \tan x$

$\tan 2x = \tan x$

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

i.e. $\;$ $2 \tan x = \tan x - \tan^3 x$

i.e. $\;$ $\tan^3 x + \tan x = 0$

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

i.e. $\;$ $\tan x = 0$ $\;$ or $\;$ $1 + \tan^2 x = 0$

But $1 + \tan^2 x = 0$ is not possible $\because$ $\tan^2 x \neq -1$

Now, $\tan x = 0 \implies x = n \pi, \;\;\; n \in Z$

Trigonometric Equations

Solve: $\sin 3x = \sin x$

$\sin 3x = \sin x$

i.e. $\;$ $3 \sin x - 4 \sin^3 x = \sin x$

i.e. $\;$ $2 \sin x - 4 \sin^3 x = 0$

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

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

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

Now, when $\sin x = 0$ $\implies$ $x = n \pi$, $\;\;\;$ $n \in Z$;

when $\cos 2x = 0$ $\implies$ $\cos 2x = \cos \left(\dfrac{\pi}{2}\right)$

$\implies$ $2x = 2 m \pi \pm \dfrac{\pi}{2}$ $\;$ i.e. $\;$ $x = m \pi \pm \dfrac{\pi}{4}, \;\;\; m \in Z$

Trigonometric Equations

Find the general solution of the equation $\cos 3 \theta = - \dfrac{1}{\sqrt{2}}$

$\cos 3 \theta = - \dfrac{1}{\sqrt{2}}$

$\implies$ $\cos 3 \theta = \cos \left(\dfrac{3 \pi}{4}\right)$

$\implies$ $3 \theta = 2 n \pi \pm \dfrac{3 \pi}{4} \;\;\; n \in Z$

$\therefore \;$ $\theta = \dfrac{2n \pi}{3} \pm \dfrac{\pi}{4}, \;\;\; n \in Z$

Trigonometric Equations

Find the general solution of the equation $\sin 2 \theta = \dfrac{1}{2}$

$\sin 2 \theta = \dfrac{1}{2}$

$\implies$ $\sin 2 \theta = \sin \left(\dfrac{\pi}{6}\right)$

$\implies$ $2 \theta = n \pi + \left(-1\right)^n \cdot \dfrac{\pi}{6}, \;\;\; n \in Z$

$\therefore \;$ $\theta = \dfrac{n \pi}{2} + \left(-1\right)^n \cdot \dfrac{\pi}{12} \;\;\; n \in Z$

Trigonometric Equations

Find the principal value of the equation $\tan \theta = - \dfrac{1}{\sqrt{3}}$

$\tan \theta = - \dfrac{1}{\sqrt{3}} < 0$

$\therefore \;$ $\theta$ lies in the second or fourth quadrants.

$\therefore \;$ $\theta = \pi - \dfrac{\pi}{6} = \dfrac{5 \pi}{6}$ $\;$ or $\;$ $\theta = 2 \pi - \dfrac{\pi}{6} = \dfrac{11 \pi}{6}$

Principal value of $\tan \theta$ is in $\left(- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$ $\;$ i.e. $\;$ in first or fourth quadrants.

$\because \;$ $\dfrac{5 \pi}{6} \notin \left(- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$,

$\therefore \;$ Principal value of the given equation is $\dfrac{11 \pi}{6}$ $\;$ i.e. $\;$ $- \dfrac{\pi}{6}$

Trigonometric Equations

Find the principal value of the equation $\sqrt{3} \sec \theta = 2$

$\sqrt{3} \sec \theta = 2$ $\implies$ $\sec \theta = \dfrac{2}{\sqrt{3}}$ $\implies$ $\cos \theta = \dfrac{\sqrt{3}}{2} > 0$

$\therefore \;$ $\theta$ lies in the first or fourth quadrants.

Now, $\cos \theta = \dfrac{\sqrt{3}}{2} = \cos \left(\dfrac{\pi}{6}\right)$

$\implies$ $\theta = \dfrac{\pi}{6}$, $\;$ $\theta = 2\pi - \dfrac{\pi}{6} = \dfrac{11\pi}{6}$

Principal value of $\cos \theta$ is in $\left[0, \pi\right]$ $\;$ i.e. $\;$ in first or second quadrants.

$\because \;$ $\dfrac{11 \pi}{6} \notin \left[0, \pi\right]$,

$\therefore \;$ Principal value of the given equation is $\dfrac{\pi}{6}$

Trigonometric Equations

Find the principal value of the equation $\sqrt{3} \cot \theta = 1$

$\sqrt{3} \cot \theta = 1$ $\implies$ $\cot \theta = \dfrac{1}{\sqrt{3}}$ $\implies$ $\tan \theta = \sqrt{3} > 0$

$\therefore \;$ $\theta$ lies in the first or the third quadrants.

Now, $\tan \theta = \sqrt{3} = \tan \left(\dfrac{\pi}{3}\right)$

$\implies$ $\theta = \dfrac{\pi}{3}$, $\;$ $\pi + \dfrac{\pi}{3} = \dfrac{4 \pi}{3}$

Principal value of $\tan \theta$ is in $\left(- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$ $\;$ i.e. $\;$ in first or fourth quadrants.

$\because \;$ $\dfrac{4 \pi}{3} \notin \left(- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$,

$\therefore \;$ Principal value of the given equation is $\dfrac{\pi}{3}$

Trigonometric Equations

Find the principal value of the equation $\sin \theta = \dfrac{1}{\sqrt{2}}$

$\sin \theta = \dfrac{1}{\sqrt{2}} > 0$

$\therefore \;$ $\theta$ lies in the first or the second quadrants.

Now, $\sin \theta = \dfrac{1}{\sqrt{2}} = \sin \left(\dfrac{\pi}{4}\right)$

$\implies$ $\theta = \dfrac{\pi}{4}$, $\;$ $\pi - \dfrac{\pi}{4} = \dfrac{3 \pi}{4}$

Principal value of $\sin \theta$ is in $\left[- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$ $\;$ i.e. $\;$ in first or fourth quadrants.

$\because \;$ $\dfrac{3 \pi}{4} \notin \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$,

$\therefore \;$ Principal value of the given equation is $\dfrac{\pi}{4}$

Probability

Find $c$, $\mu$ and $\sigma^2$ of the normal distribution whose probability function is given by $f\left(x\right) = c e^{- x^2 + 3x}$, $-\infty < X < + \infty$

The standard probability function of a continuous random variable $X$ which follows normal distribution with mean $\mu$ and standard deviation $\sigma$ is

$f \left(x\right) = \dfrac{1}{\sigma \sqrt{2 \pi}} \exp \left[- \dfrac{1}{2} \left(\dfrac{x - \mu}{\sigma}\right)^2\right]$ $\;\;\; \cdots \; (1)$

$\begin{aligned} \text{Given: } f \left(x\right) & = c \exp \left\{- x^2 + 3x\right\} \\\\ & = c \exp \left\{- \left[\left(x^2 - 3x + \dfrac{9}{4}\right) - \dfrac{9}{4}\right] \right\} \\\\ & = c \exp \left\{- \left(x - \dfrac{3}{2}\right)^2 + \dfrac{9}{4} \right\} \\\\ & = c \exp \left\{\dfrac{9}{4}\right\} \times \exp \left\{- \left(x - \dfrac{3}{2}\right)^2 \right\} \\\\ & = c \exp \left\{\dfrac{9}{4} \right\} \times \exp \left\{- \dfrac{1}{2} \left(\dfrac{x - \dfrac{3}{2}}{\dfrac{1}{\sqrt{2}}}\right)^2 \right\} \;\;\; \cdots \; (2) \end{aligned}$

Comparing equations $(1)$ and $(2)$ we have,

$\mu = \dfrac{3}{2}$

$\sigma = \dfrac{1}{\sqrt{2}}$ $\implies$ $\sigma^2 = \dfrac{1}{2}$

$c \exp \left\{\dfrac{9}{4} \right\} = \dfrac{1}{\sigma \sqrt{2 \pi}}$

$\implies$ $c = \exp \left\{- \dfrac{9}{4} \right\} \times \dfrac{1}{\dfrac{1}{\sqrt{2}} \times \sqrt{2 \pi}}$

i.e. $\;$ $c = \dfrac{\exp \left(-9/4\right)}{\sqrt{\pi}}$

Probability

If the height of $300$ students are normally distributed with mean $64.5$ inches and standard deviation $3.3$ inches, find the height below which $99\%$ of the students lie.

Mean height of students $= \mu = 64.5$ inches

Standard deviation of the height of students $= \sigma = 3.3$ inches

Let $X$ be a normal variate.

Standard normal variate $Z = \dfrac{X - \mu}{\sigma}$ $\;\;\; \cdots \; (1)$

$99\%$ of students corresponds to an area of $0.99$ under the normal probability curve.

Area to the left of $Z = 0$ is $= 0.5$

Remaining area $= 0.99 - 0.5 = 0.49$

Area of $0.49$ corresponds to $Z = 2.33$ $\;\;\;$ [from $z$ tables]

From equation $(1)$, $X = \left(Z \times \sigma\right) + \mu$

$\therefore \;$ When $Z = 2.33$, $\;$ $X = \left(2.33 \times 3.3\right) + 64.5 = 72.189$

$\therefore \;$ Height below which $99\%$ of the students lie $= 72.189 \approx 72.19$ inches

Probability

The mean weight of $500$ students in a certain college is $151$ pounds and the standard deviation is $15$ pounds. Assuming the weights are normally distributed, find how many students weigh

1. between $120$ and $155$ pounds;
2. more than $185$ pounds.

Given: Mean $= \mu = 151$; $\;$ Standard deviation $= \sigma = 15$

Let $X$ be a normal variate.

Standard normal variate $Z = \dfrac{X - \mu}{\sigma}$

1. When $X = 120$, $\;$ $Z = \dfrac{120 - 151}{15} = -2.07$
   
   When $X = 155$, $\;$ $Z = \dfrac{155 - 151}{15} = 0.27$
   
   $\therefore \;$ Probability of students weighing between $120$ and $155$ pounds is
   
   $\begin{aligned} P \left(120 < X < 155\right) & = P \left(- 2.07 < Z < 0.27\right) \\\\ & = P \left(- 2.07 < Z \leq 0\right) + P \left(0 \leq Z < 0.27\right) \\\\ & = P \left(0 \leq Z < 2.07\right) + P \left(0 \leq Z < 0.27\right) \;\;\; \left[\text{by symmetry}\right] \\\\ & = 0.4808 + 0.1064 \;\;\; \left[\text{from Normal Distribution Table}\right] \\\\ & = 0.5872 \end{aligned}$
   
   $\therefore \;$ Number of students who weigh between $120$ and $155$ pounds is $= 500 \times 0.5872 = 293.6$
   
   i.e. $\;$ approximately $294$ students
   
   
2. When $X = 185$, $\;$ $Z = \dfrac{185 - 151}{15} = 2.27$
   
   $\therefore \;$ Probability of students weighing more than 185 pounds is
   
   $\begin{aligned} P \left(X > 185\right) & = P \left(Z > 2.27\right) \\\\ & = \left(\text{area between } Z = 0 \text{ to } Z = \infty\right) \\\\ & \hspace{1.5cm} - \left(\text{area between } Z = 0 \text{ to } Z = 2.27\right) \\\\ & = 0.5 - 0.4884 \\\\ & = 0.0116 \end{aligned}$
   
   $\therefore \;$ Number of students weighing more than $185$ pounds is $= 500 \times 0.0116 = 5.8$
   
   i.e. $\;$ approximately $6$ students

Probability

The life of shoes is normally distributed with mean $8$ months and standard deviation $2$ months. If $5000$ pairs are issued, how many pairs would be expected to need replacement within $12$ months.

Given: Mean $= \mu = 8$; $\;$ Standard deviation $= \sigma = 2$

Let $X$ be a normal variate.

To find: $P \left(X < 12\right)$

Standard normal variate $Z = \dfrac{X - \mu}{\sigma}$

When $X = 12$, $\;$ $Z = \dfrac{12 - 8}{2} = 2$

$\begin{aligned} \therefore \; P \left(X < 12\right) = P \left(Z < 2\right) \\\\ & = \left(\text{area between } Z = - \infty \text{ to } Z = 0\right) \\\\ & \hspace{1.5cm} + \left(\text{area between } Z = 0 \text{ to } Z = 2\right) \\\\ & = 0.5000 + 0.4772 \;\;\; \left[\text{from Normal Distribution Table}\right] \\\\ & = 0.9772 \end{aligned}$

$\therefore \;$ Number of pairs of shoes that would need replacement within $12$ months $= 5000 \times 0.9772 = 4886$

Probability

If $Z$ is a standard normal variate, find the value of $c$ for the following:

1. $P \left(- c < Z < c\right) = 0.40$
2. $P \left(Z > c\right) = 0.85$

1. $\because \;$ $Z = - c$ and $Z = + c$ lie at equal distance from $Z = 0$,
   
   we have $P \left(0 < Z < c\right) = \dfrac{0.40}{2} = 0.20$
   
   $Z$ value for the area $0.20$ from the Normal Distribution table is $= 0.52$
   
   $\therefore \;$ $- c = - 0.52$ $\;$ and $\;$ $+c = + 0.52$
   
   
2. $P \left(Z > c\right) = 0.85$ $\implies$ $P \left(c < Z < \infty\right) = 0.85$
   
   i.e. $\;$ $c$ lies to the right of $Z = 0$
   
   Area to the right of $Z = 0$ is $= 0.5$
   
   $\therefore \;$ We have, $P \left(0 < Z < \infty\right) - P \left(0 < Z < c\right) = 0.85$
   
   i.e. $\;$ $0.5 - P \left(0 < Z < c\right) = 0.85$
   
   i.e. $P \left(0 < Z < c\right) = 0.5 - 0.85 = - 0.35$
   
   From the Normal Distribution table, $Z$ value for $- 0.35$ is $= - 1.04$
   
   $\therefore \;$ $c = -1.04$

Probability

If $X$ is a normal variate with mean $80$ and standard deviation $10$, compute the following probabilities by standardizing:

1. $P \left(X \leq 100\right)$
2. $P \left(85 \leq X \leq 95\right)$

Given: Normal variate $= X$; mean $= \mu = 80$; standard deviation $= \sigma = 10$

Standard normal variate $Z = \dfrac{X - \mu}{\sigma}$

1. When $X = 100$, $\;$ $Z = \dfrac{100 - 80}{10} = 2$
   
   $\begin{aligned} \therefore \; P \left(X \leq 100\right) & = P \left(Z \leq 2\right) \\\\ & = \left(\text{area between } Z = - \infty \text{ to } Z = 0\right) \\\\ & \hspace{1.5cm} + \left(\text{area between } Z = 0 \text{ to } Z = 2\right) \\\\ & = 0.5000 + 0.4772 \;\;\; \left[\text{from Normal Distribution Table}\right] \\\\ & = 0.9772 \end{aligned}$
   
   
2. When $X = 85$, $\;$ $Z = \dfrac{85 - 80}{10} = 0.5$
   
   When $X = 95$, $\;$ $Z = \dfrac{95 - 80}{10} = 1.5$
   
   $\begin{aligned} \therefore \; P \left(85 \leq X \leq 95\right) & = P \left(0.5 \leq Z \leq 1.5\right) \\\\ & = \left(\text{area between } Z = 0 \text{ to } Z = 1.5\right) \\\\ & \hspace{1.5cm} - \left(\text{area between } Z = 0 \text{ to } Z = 0.5\right) \\\\ & = 0.4332 - 0.1915 \;\;\; \left[\text{from Normal Distribution Table}\right] \\\\ & = 0.2417 \end{aligned}$

Probability

A business man receives an average of $2$ e-mails per hour related to his business and $1.5$ e-mails per hour on personal matters.

Find the probability that in any randomly chosen hour

1. he receives no e-mails,
2. he receives more than 5 e-mails.

Applying Poisson distribution to determine the number of e-mails received per hour (both business and personal), we have

parameter of Poisson distribution for total e-mails $= \lambda = 2 + 1.5 = 3.5$

1. $P \left(\text{receives no e-mail}\right) = P \left(X = 0\right)$
   
   $\begin{aligned} P \left(X = 0\right) & = \dfrac{e^{- \lambda} \times \lambda^0}{0!} \\\\ & = e^{- \lambda} \\\\ & = e^{- 3.5} = 0.0302 \end{aligned}$
   
   
2. $P \left(\text{receives more than 5 e-mails}\right) = P \left(X > 5\right)$
   
   $P \left(X > 5\right) = 1 - P \left(X \leq 5\right)$
   
   $\begin{aligned} P \left(X \leq 5\right) & = \sum \limits_{x = 0}^{5} \dfrac{e^{- \lambda} \times \lambda^{x}}{x!} \\\\ & = e^{- \lambda} \times \left[\dfrac{\lambda^{0}}{0!} + \dfrac{\lambda^{1}}{1!} + \dfrac{\lambda^{2}}{2!} + \dfrac{\lambda^{3}}{3!} + \dfrac{\lambda^{4}}{4!} + \dfrac{\lambda^{5}}{5!} \right] \\\\ & = e^{- 3.5} \times \left[1 + \dfrac{3.5}{1} + \dfrac{3.5^2}{2} + \dfrac{3.5^3}{6} + \dfrac{3.5^4}{24} + \dfrac{3.5^5}{120} \right] \\\\ & = 0.0302 \times 28.4003 = 0.8577 \end{aligned}$
   
   $\therefore \;$ $P \left(X > 5\right) = 1 - 0.8577 = 0.1423$

Probability

The water in a tank is contaminated with bacteria. The bacteria are located randomly and independently. The mean number per milliliter of liquid is known to be $1.1$.

Find the probability that a sample of $1$ ml of liquid contains more than $2$ bacteria.

Five samples, each of $1$ ml of liquid, are taken.

1. Find the mean number of bacteria for the five samples, taken together.


2. Find the probability that there are in total
   
   
   1. $0$ bacteria,
   2. less than $3$ bacteria
   
   
   in the five samples.

Mean number of bacteria per milliliter of liquid $= 1.1$

$\because \;$ The bacteria are located randomly and independently, Poisson distribution can be applied to determine the number of bacteria.

parameter of Poisson distribution $= \lambda = 1.1$

$P \left(\text{Sample contains more than 2 bacteria}\right) = P \left(X > 2\right)$

$P \left(X > 2\right) = 1 - P \left(X \leq 2\right)$

$\begin{aligned} P \left(X \leq 2\right) & = \sum \limits_{x = 0}^{2} \dfrac{e^{- \lambda} \times \lambda^{x}}{x!} \\\\ & = e^{- \lambda} \times \left[\dfrac{\lambda^{0}}{0!} + \dfrac{\lambda^{1}}{1!} + \dfrac{\lambda^{2}}{2!}\right] \\\\ & = e^{- 1.1} \times \left[1 + \dfrac{1.1}{1} + \dfrac{1.1^2}{2} \right] \\\\ & = 0.3329 \times 2.705 = 0.9005 \end{aligned}$

$\therefore \;$ $P \left(X > 2\right) = 1 - 0.9005 = 0.0995$

1. $\because \;$ Poisson distribution can be used to determine the number of bacteria present,
   
   $\therefore \;$ mean number of bacteria present in $5$ samples $= 1.1 \times 5 = 5.5$
   
   
2. For $5$ samples, parameter of Poisson distribution $= \lambda = 5.5$
   
   
   1. $P\left(\text{0 bacteria present in 5 samples}\right) = P \left(X = 0\right)$
      
      $\begin{aligned} P \left(X = 0\right) & = \dfrac{e^{- \lambda} \times \lambda^0}{0!} \\\\ & = e^{- \lambda} \\\\ & = e^{- 5.5} = 0.0041 \end{aligned}$
      
      
   2. $P \left(\text{less than 3 bacteria present in 5 samples}\right) = P \left(X < 3\right)$
      
      $\begin{aligned} P \left(X < 3\right) & = \sum \limits_{0}^{2} \dfrac{e^{- \lambda} \times \lambda^{x}}{x!} \\\\ & = e^{- \lambda} \times \left[\dfrac{\lambda^{0}}{0!} + \dfrac{\lambda^{1}}{1!} + \dfrac{\lambda^{2}}{2!}\right] \\\\ & = e^{- 5.5} \times \left[1 + \dfrac{5.5}{1} + \dfrac{5.5^2}{2} \right] \\\\ & = 0.0041 \times 21.625 = 0.0887 \end{aligned}$