Solve: $\;$ $6 \sin^{-1} \left(x^2 - 6x + 8.5\right) = \pi$
Given equation: $\;$ $6 \sin^{-1} \left(x^2 - 6x + 8.5\right) = \pi$
i.e. $\;$ $\sin^{-1} \left(x^2 - 6x + 8.5\right) = \dfrac{\pi}{6}$
i.e. $\;$ $x^2 - 6x + 8.5 = \sin \left(\dfrac{\pi}{6}\right)$
i.e. $\;$ $x^2 - 6x + 8.5 = 0.5$
i.e. $\;$ $x^2 - 6x + 8 = 0$
i.e. $\;$ $\left(x - 4\right) \left(x - 2\right) = 0$
i.e. $\;$ $x = 4$ $\;$ or $\;$ $x = 2$
Calculate: $\;$ $\sin \left[2 \tan^{-1} \left(\dfrac{1}{3}\right) \right] + \cos \left[\tan^{-1} \left(2 \sqrt{2}\right)\right]$
Given Expression (GE): $\;$ $\sin \left[2 \tan^{-1} \left(\dfrac{1}{3}\right) \right] + \cos \left[\tan^{-1} \left(2 \sqrt{2}\right)\right]$
$= 2 \sin \left[\tan^{-1} \left(\dfrac{1}{3}\right)\right] \cos \left[\tan^{-1} \left(\dfrac{1}{3}\right)\right] + \cos \left[\tan^{-1} \left(2 \sqrt{2}\right)\right]$
$= 2 \times \dfrac{\dfrac{1}{3}}{\sqrt{1 + \dfrac{1}{9}}} \times \dfrac{1}{\sqrt{1 + \dfrac{1}{9}}} + \dfrac{1}{\sqrt{1 + 8}}$
$= \dfrac{\dfrac{2}{3}}{\dfrac{\sqrt{10}}{3}} \times \dfrac{1}{\dfrac{\sqrt{10}}{3}} + \dfrac{1}{3}$
$= \dfrac{2}{\sqrt{10}} \times \dfrac{3}{\sqrt{10}} + \dfrac{1}{3}$
$= \dfrac{3}{5} + \dfrac{1}{3} = \dfrac{14}{15}$
Formulas used:
$\sin 2 \theta = 2 \sin \theta \cos \theta$
$\cos \left(\tan^{-1} x\right) = \dfrac{1}{\sqrt{1 + x^2}}, \;\; 0 \leq x \leq 1$
$\sin \left(\tan^{-1} x\right) = \dfrac{x}{\sqrt{1 + x^2}}, \;\; x > 0$
Calculate: $\;$ $\cos \left[2 \tan^{-1} \left(2\right)\right]$
Given Expression (GE): $\;$ $\cos \left[2 \tan^{-1} \left(2\right)\right]$
$= 2 \times \cos^2 \left(\tan^{-1} 2\right) - 1$
$= 2 \left[\cos \left(\tan^{-1} 2\right)\right]^2 - 1$
$= 2 \times \left(\dfrac{1}{\sqrt{1 + 2^2}}\right) - 1$
$= 2 \times \left(\dfrac{1}{\sqrt{5}}\right)^2 - 1 = \dfrac{2}{5} - 1 = \dfrac{-3}{5}$
Formulas used:
$\cos 2 \theta = 2 \cos^2 \theta - 1$
$\cos \left(\tan^{-1} x\right) = \dfrac{1}{\sqrt{1 + x^2}}, \;\; 0 \leq x \leq 1$
Calculate: $\;$ $\sin \left[2 \tan^{-1} \left(3\right)\right]$
Given Expression (GE): $\;$ $\sin \left[2 \tan^{-1} \left(3\right)\right]$
$= 2 \times \sin \left[\tan^{-1} \left(3\right)\right] \cos \left[\tan^{-1} \left(3\right)\right]$
$= 2 \times \sin \left[\sin^{-1} \left(\dfrac{3}{\sqrt{1 + 3^2}}\right)\right] \cos \left[\cos^{-1} \left(\dfrac{1}{\sqrt{1 + 3^2}}\right)\right]$
$= 2 \times \dfrac{3}{\sqrt{10}} \times \dfrac{1}{\sqrt{10}}$
$= \dfrac{3}{5}$
Formulas used:
$\sin 2 \theta = 2 \sin \theta \cos \theta$
$\tan^{-1} x = \sin^{-1} \left(\dfrac{x}{\sqrt{1 + x^2}}\right), \;\; x > 0$
$\tan^{-1} x = \cos^{-1} \left(\dfrac{1}{\sqrt{1 + x^2}}\right), \;\; x > 0$
$\sin \left(\sin^{-1} x\right) = x, \;\; \forall \; x \in \left[-1, 1\right]$
$\cos \left(\cos^{-1} x\right) = x, \;\; \forall \; x \in \left[-1, 1\right]$
Calculate: $\;$ $\sin \left[2 \sin^{-1} \left(\dfrac{3}{5}\right)\right]$
Given Expression (GE): $\;$ $\sin \left[2 \sin^{-1} \left(\dfrac{3}{5}\right)\right]$
Let $\;$ $\sin^{-1} \left(\dfrac{3}{5}\right) = \theta$
Then, GE $= \sin 2 \theta = 2 \sin \theta \cos \theta$
i.e. $\;$ GE $= 2 \times \sin \left[\sin^{-1} \left(\dfrac{3}{5}\right)\right] \cos \left[\sin^{-1} \left(\dfrac{3}{5}\right)\right]$ $\;\;\;$ [substituting for $\theta$]
$= 2 \times \dfrac{3}{5} \times \cos \left[\cos^{-1} \left(\sqrt{1 - \dfrac{9}{25}}\right)\right]$
$= \dfrac{6}{5} \times \cos \left[\cos^{-1} \left(\dfrac{4}{5}\right)\right]$
$= \dfrac{6}{5} \times \dfrac{4}{5} = \dfrac{24}{25}$
Formulas used:
$\sin \left(\sin^{-1} x\right) = x, \; x \in \left[-1, 1\right]$
$\sin^{-1} x = \cos^{-1} \left(\sqrt{1 - x^2}\right), \; 0 \leq x \leq 1$
$\cos \left(\cos^{-1} x\right) = x. \; x \in \left[-1, 1\right]$
Calculate: $\;$ $\cot^{-1} \left[\tan \left(- 37^{\circ}\right)\right]$
Given Expression (GE): $\;$ $\cot^{-1} \left[\tan \left(- 37^{\circ}\right)\right]$
$= \cot^{-1} \left[- \tan 37^\circ\right]$
$= \cot^{-1} \left[- \tan \left(90^\circ - 53^\circ\right)\right]$
$= \cot^{-1} \left[- \cot 53^\circ\right]$
Let $\;$ $\cot 53^\circ = x$
Then, GE $= \cot^{- 1} \left(- x\right) = \pi - \cot^{-1} x$
$= \pi - \cot^{-1} \left(\cot 53^\circ\right)$ $\;\;\;$ [substituting for $x$]
$= \pi - 53^\circ = 127^\circ$
Formulas used:
$\tan \left(- \theta\right) = - \tan \theta$
$\cot \theta = \tan \left(\dfrac{\pi}{2} - \theta\right)$
$\cot^{-1} \left(- \theta\right) = \pi - \cot^{-1} \theta, \;\;\; \forall \; \theta \in R$
$\cot^{-1} \left(\cot \theta\right) = \theta, \;\;\; \forall \; \theta \in \left(0, \pi\right)$
Calculate: $\;$ $\sin^{-1} \left(\dfrac{-\sqrt{2}}{2}\right) + \cos^{-1} \left(\dfrac{-1}{2}\right) - \tan^{-1} \left(- \sqrt{3}\right) + \cot^{-1} \left(\dfrac{-1}{\sqrt{3}}\right)$
Given: $\;$ $\sin^{-1} \left(\dfrac{-\sqrt{2}}{2}\right) + \cos^{-1} \left(\dfrac{-1}{2}\right) - \tan^{-1} \left(- \sqrt{3}\right) + \cot^{-1} \left(\dfrac{-1}{\sqrt{3}}\right)$
$= - \sin^{-1} \left(\dfrac{\sqrt{2}}{2}\right) + \pi - \cos^{-1} \left(\dfrac{1}{2}\right) + \tan^{-1} \left(\sqrt{3}\right) + \pi - \cot^{-1} \left(\dfrac{1}{\sqrt{3}}\right)$
$= - \sin^{-1} \left(\dfrac{1}{\sqrt{2}}\right) + 2 \pi - \dfrac{\pi}{3} + \dfrac{\pi}{3} - \dfrac{\pi}{3}$
$= \dfrac{- \pi}{4} + \dfrac{5 \pi}{3} = \dfrac{17 \pi}{12}$
$\left\{\because \; \sin^{-1} \left(-x\right) = - \sin^{-1} x, \; \forall \; x \in \left[-1, 1\right]\right\}$
$\left\{\cos^{-1} \left(-x\right) = \pi - \cos^{-1} x, \; \forall \; x \in \left[-1, 1\right]\right\}$
$\left\{\tan^{-1} \left(-x\right) = - \tan^{-1} x, \; \forall \; x \in R\right\}$
$\left\{\cot^{-1} \left(-x\right) = \pi - \cot^{-1} x, \; \forall \; x \in R\right\}$
Calculate: $\;$ $\tan^{-1}\left(1\right) + \cos^{-1} \left(\dfrac{-1}{2}\right) + \sin^{-1} \left(\dfrac{-1}{2}\right)$
Given: $\;$ $\tan^{-1}\left(1\right) + \cos^{-1} \left(\dfrac{-1}{2}\right) + \sin^{-1} \left(\dfrac{-1}{2}\right)$
$= \dfrac{\pi}{4} + \dfrac{\pi}{2} = \dfrac{3 \pi}{4}$
$\left\{\because \; \sin^{-1} x + \cos^{-1} x = \dfrac{\pi}{2}, \; \forall \; x \in \left[-1, 1\right]\right\}$ $\;$ [Cofunction Inverse Identity]
Transform the following expression into a product:
$\tan^2 \alpha - \tan^2 \beta - \dfrac{1}{2} \sin \left(\alpha - \beta\right) \sec^2 \alpha \sec^2 \beta$
Given: $\;$ $\tan^2 \alpha - \tan^2 \beta - \dfrac{1}{2} \sin \left(\alpha - \beta\right) \sec^2 \alpha \sec^2 \beta$
$= \dfrac{\sin^2 \alpha}{\cos^2 \alpha} - \dfrac{\sin^2 \beta}{\cos^2 \beta} - \dfrac{1}{2} \sin \left(\alpha - \beta\right) \sec^2 \alpha \sec^2 \beta$
$= \dfrac{\sin^2 \alpha \cos^2 \beta - \sin^2 \beta \cos^2 \alpha}{\cos^2 \alpha \cos^2 \beta} - \dfrac{1}{2} \sin \left(\alpha - \beta\right) \sec^2 \alpha \sec^2 \beta$
$= \sec^2 \alpha \sec^2 \beta \left[\sin^2 \alpha \cos^2 \beta - \cos^2 \alpha \sin^2 \beta\right] - \dfrac{1}{2} \sin \left(\alpha - \beta\right) \sec^2 \alpha \sec^2 \beta$
$= \sec^2 \alpha \sec^2 \beta \left[\left(\sin \alpha \cos \beta + \cos \alpha \sin \beta\right) \left(\sin \alpha \cos \beta - \cos \alpha \sin \beta\right)\right]$
$\hspace{3cm} - \sec^2 \alpha \sec^2 \beta \times \dfrac{1}{2} \sin \left(\alpha - \beta\right)$
$= \sec^2 \alpha \sec^2 \beta \left[\sin \left(\alpha + \beta\right) \sin \left(\alpha - \beta\right) - \dfrac{1}{2} \sin \left(\alpha - \beta\right)\right]$
$\left\{\because \; \sin \left(A \pm B\right) = \sin A \cos B \pm \cos A \sin B\right\}$
$= \sec^2 \alpha \sec^2 \beta \left[\sin \left(\alpha + \beta\right) \sin \left(\alpha - \beta\right) - \sin \left(\dfrac{\pi}{6}\right) \sin \left(\alpha - \beta\right)\right]$
$\left\{\because \; \sin \left(\dfrac{\pi}{6}\right) = \dfrac{1}{2}\right\}$
$= \sec^2 \alpha \sec^2 \beta \sin \left(\alpha - \beta\right) \left[\sin \left(\alpha + \beta\right) - \sin \left(\dfrac{\pi}{6}\right)\right]$
$= \sec^2 \alpha \sec^2 \beta \sin \left(\alpha - \beta\right) \times 2 \sin \left(\dfrac{\alpha + \beta - \dfrac{\pi}{6}}{2}\right) \cos \left(\dfrac{\alpha + \beta + \dfrac{\pi}{6}}{2}\right)$
$\left\{\because \; \sin A - \sin B = 2 \sin \left(\dfrac{A - B}{2}\right) \cos \left(\dfrac{A + B}{2}\right)\right\}$
Transform the following expression into a product:
$\sin \alpha + \sin 60^\circ + \sin \left(\alpha + 60^\circ\right)$
Given: $\;$ $\sin \alpha + \sin 60^\circ + \sin \left(\alpha + 60^\circ\right)$
$= \sin \alpha + \dfrac{\sqrt{3}}{2} + \sin \alpha \cos 60^\circ + \cos \alpha \sin 60^\circ$
$= \sin \alpha + \dfrac{\sqrt{3}}{2} + \dfrac{1}{2} \times \sin \alpha + \dfrac{\sqrt{3}}{2} \times \cos \alpha$
$= \dfrac{3}{2} \times \sin \alpha + \dfrac{\sqrt{3}}{2} \left(1 + \cos \alpha\right)$
$= \dfrac{3}{2} \times 2 \sin \left(\dfrac{\alpha}{2}\right) \cos \left(\dfrac{\alpha}{2}\right) + \dfrac{\sqrt{3}}{2} \times 2 \cos^2 \left(\dfrac{\alpha}{2}\right)$
$= \sqrt{3} \cos \left(\dfrac{\alpha}{2}\right) \left[\sqrt{3} \sin \left(\dfrac{\alpha}{2}\right) + \cos \left(\dfrac{\alpha}{2}\right)\right]$
$= 2 \sqrt{3} \cos \left(\dfrac{\alpha}{2}\right) \left[\dfrac{\sqrt{3}}{2} \sin \left(\dfrac{\alpha}{2}\right) + \dfrac{1}{2} \cos \left(\dfrac{\alpha}{2}\right)\right]$
$= 2 \sqrt{3} \cos \left(\dfrac{\alpha}{2}\right) \left[\sin \left(\dfrac{\pi}{6}\right) \cos \left(\dfrac{\alpha}{2}\right) + \cos \left(\dfrac{\pi}{6}\right) \sin \left(\dfrac{\alpha}{2}\right)\right]$
$= 2 \sqrt{3} \cos \left(\dfrac{\alpha}{2}\right) \sin \left(\dfrac{\pi}{6} + \dfrac{\alpha}{2}\right)$
Transform the following expression into a product:
$\sec \alpha - \cos \alpha + \sec 60^\circ \cos 2 \alpha \sin 3 \alpha - \sin 5 \alpha$
Given: $\;$ $\sec \alpha - \cos \alpha + \sec 60^\circ \cos 2 \alpha \sin 3 \alpha - \sin 5 \alpha$
$= \dfrac{1}{\cos \alpha - \cos \alpha + 2 \cos 2 \alpha \sin 3 \alpha - \sin 5 \alpha}$
$= \dfrac{1}{\cos \alpha} - \cos \alpha + \sin \left(3 \alpha - 2 \alpha\right) + \sin \left(3 \alpha + 2 \alpha\right) - \sin 5 \alpha$
$\left[\because \; 2 \sin A \cos B = \sin \left(A - B\right) + \sin \left(A + B\right)\right]$
$= \dfrac{1}{\cos \alpha} - \cos \alpha + \sin \alpha + \sin 5 \alpha - \sin 5 \alpha$
$= \dfrac{1}{\cos \alpha} - \cos \alpha + \sin \alpha$
$= \dfrac{1 - \cos^2 \alpha}{\cos \alpha} + \sin \alpha$
$= \dfrac{\sin^2 \alpha}{\cos \alpha} + \sin \alpha$
$= \sin \alpha \left(\dfrac{\sin \alpha}{\cos \alpha} + 1\right)$
$= \dfrac{\sin \alpha}{\cos \alpha} \left(\sin \alpha + \cos \alpha\right)$
$= \dfrac{\sin \alpha}{\cos \alpha} \times \sqrt{2} \times \left(\dfrac{1}{\sqrt{2}} \sin \alpha + \dfrac{1}{\sqrt{2}} \cos \alpha\right)$
$= \sqrt{2} \tan \alpha \left(\sin \dfrac{\pi}{4} \cos \alpha + \cos \dfrac{\pi}{4} \sin \alpha\right)$
$= \sqrt{2} \tan \alpha \sin \left(\dfrac{\pi}{4} + \alpha\right)$ $\;\; \left[\because \; \sin \left(A + B\right) = \sin A \cos B + \cos A \sin B\right]$
Calculate, without using tables $\;$ $\dfrac{\sin \alpha}{\sin^3 \alpha + 3 \cos^3 \alpha}$, if $\tan \alpha = 2$ and $\alpha$ is in the first quadrant.
Given: $\;$ $\tan \alpha = 2$ $\implies$ $\tan^2 \alpha = 4$
$\therefore \;$ $1 + \tan^2 \alpha = 1 + 4 = 5$
i.e. $\;$ $\sec^2 \alpha = 5$
i.e. $\;$ $\dfrac{1}{\cos^2 \alpha} = 5$ $\implies$ $\cos^2 \alpha = \dfrac{1}{5}$
$\implies$ $\cos \alpha = \dfrac{\pm 1}{\sqrt{5}}$
But since $\alpha$ is in the first quadrant, $\;$ $\therefore \;$ $\cos \alpha = \dfrac{+ 1}{\sqrt{5}}$
Then, $\;$ $\sin \alpha = \sqrt{1 - \cos^2 \alpha} = \sqrt{1 - \dfrac{1}{5}} = \dfrac{\pm 2}{\sqrt{5}}$
But since $\alpha$ is in the first quadrant, $\;$ $\therefore \;$ $\sin \alpha = \dfrac{+ 2}{\sqrt{5}}$
$\therefore \;$ The given expression $\;$ $\dfrac{\sin \alpha}{\sin^3 \alpha + 3 \cos^3 \alpha}$ $\;$ becomes
$\dfrac{\dfrac{2}{\sqrt{5}}}{\left(\dfrac{2}{\sqrt{5}}\right)^3 + 3 \left(\dfrac{1}{\sqrt{5}}\right)^3}$
$= \dfrac{\dfrac{2}{\sqrt{5}}}{\dfrac{8}{5 \sqrt{5}} + \dfrac{3}{5 \sqrt{5}}}$
$= \dfrac{2}{\sqrt{5}} \times \dfrac{5 \sqrt{5}}{11} = \dfrac{10}{11}$
Without using tables, find $\;$ $\sin \left(\dfrac{\alpha}{2}\right)$, if $\sin \alpha = \dfrac{-5}{13}$ and $\pi < \alpha < \dfrac{3 \pi}{2}$
Given: $\;$ $\pi < \alpha < \dfrac{3 \pi}{2}$ $\implies$ $\dfrac{\pi}{2} < \dfrac{\alpha}{2} < \dfrac{3 \pi}{4}$
$\implies$ $\alpha$ lies in the third quadrant $\implies$ Both $\sin \alpha$ and $\cos \alpha$ are negative.
And, $\;$ $\dfrac{\alpha}{2}$ lies in the second quadrant $\implies$ $\sin \left(\dfrac{\alpha}{2}\right)$ is positive.
Given: $\;$ $\sin \alpha = \dfrac{-5}{13}$
$\therefore \;$ $\cos \alpha = \sqrt{1- \sin^2 \alpha} = \sqrt{1 - \left(\dfrac{-5}{13}\right)^2} = \dfrac{\pm 12}{13}$
$\because \;$ $\alpha$ is in the third quadrant $\implies$ $\cos \alpha = \dfrac{-5}{13}$
Now, $\;$ $\cos \alpha = 1 - 2 \sin^2 \left(\dfrac{\alpha}{2}\right)$
$\therefore \;$ $\sin^2 \left(\dfrac{\alpha}{2}\right) = \dfrac{1 - \cos \alpha}{2} = \dfrac{1 - \left(\dfrac{-12}{13}\right)}{2} = \dfrac{25}{26}$
$\implies$ $\sin \left(\dfrac{\alpha}{2}\right) = \dfrac{\pm 5}{\sqrt{26}}$
$\because \;$ $\dfrac{\alpha}{2}$ is in the second quadrant $\implies$ $\sin \left(\dfrac{\alpha}{2}\right) = \dfrac{+5}{\sqrt{26}}$