Here, you will learn formulas for inverse trigonometric functions, equation and inequations involving inverse trigonometric function.
Letโs begin โ
Simplified Inverse Trigonometric Functions
(a)ย y = f(x) = \(sin^{-1}({2x\over {1+x^2}})\) = \(\begin{cases} 2tan^{-1}x, & \text{if}\ |x| \le 1 \\ \pi โ 2tan^{-1}x, & \text{if}\ x > 1 \\ -(\pi + 2tan^{-1}x), & \text{if}\ x < -1 \end{cases}\)ย
(b)ย y = f(x) = \(cos^{-1}({{1-x^2}\over {1+x^2}})\) = \(\begin{cases} 2tan^{-1}x, & \text{if}\ |x| \ge 0 \\ โ 2tan^{-1}x, & \text{if}\ x < 0 \end{cases}\)
(c)ย y = f(x) = \(tan^{-1}({2x\over {1-x^2}})\) = \(\begin{cases} 2tan^{-1}x, & \text{if}\ |x| < 1 \\ \pi + 2tan^{-1}x, & \text{if}\ x < -1 \\ -(\pi โ 2tan^{-1}x), & \text{if}\ x > 1 \end{cases}\)
(d)ย y = f(x) = \(sin^{-1}({3x โ 4x^3})\) = \(\begin{cases} -(\pi + 3sin^{-1}x), & \text{if}\ -1 \le x \le {-1\over 2} \\ 3sin^{-1}x, & \text{if}\ {-1\over 2} \le x \le {1\over 2} \\ \pi โ 3sin^{-1}x, & \text{if}\ {1\over 2} \le x \le 1 \end{cases}\)
(e)ย y = f(x) = \(cos^{-1}({4x^3 โ 3x})\) = \(\begin{cases} 3cos^{-1}x โ 2\pi, & \text{if}\ -1 \le x \le {-1\over 2} \\ 2\pi โ 3cos^{-1}x, & \text{if}\ {-1\over 2} \le x \le {1\over 2} \\ 3cos^{-1}x, & \text{if}\ {1\over 2} \le x \le 1 \end{cases}\)
(f)ย y = f(x) = \(sin^{-1}({2x\sqrt{1-x^2}})\) = \(\begin{cases} -(\pi + 2sin^{-1}x), & \text{if}\ -1 \le x \le {-1\over \sqrt{2}} \\ 2sin^{-1}x, & \text{if}\ {-1\over \sqrt{2}} \le x \le {1\over \sqrt{2}} \\ \pi โ 2sin^{-1}x, & \text{if}\ {1\over \sqrt{2}} \le x \le 1 \end{cases}\)
(g)ย y = f(x) = \(cos^{-1}({2x^2-1})\) = \(\begin{cases} 2cos^{-1}x, & \text{if}\ 0 \le x \le 1 \\ 2\pi โ 2cos^{-1}x, & \text{if}\ -1 \le x \le 0 \end{cases}\)
Example : Prove that : \(2tan^{-1}{1\over 2}\) + \(tan^{-1}{1\over 7}\) = \(tan^{-1}{31\over 17}\)
Solution : We have, \(2tan^{-1}{1\over 2}\) + \(tan^{-1}{1\over 7}\)
= \(2tan^{-1}({{2\times {1\over 2}\over {1-({1\over 2})^2}}})\) + \(tan^{-1}{1\over 7}\) ย ย ย [\(\because\) \(2tan^{-1}x\) = \(tan^{-1}{2x\over {1-x^2}}\)]
\(tan^{-1}{4\over 3}\) + \(tan^{-1}{1\over 7}\) = \(tan^{-1}[{{4\over 3} + {1\over 7}\over {1 โ {4\over 3}\times {1\over 7}}}]\) = \(tan^{-1}{31\over 17}\)
Equations involving Inverse trigonometric functions
Example : Prove that the equation \(2cos^{-1}x\) + \(sin^{-1}x\) = \(11\pi\over 6\) has no solution.
Solution : Given equation is \(2cos^{-1}x\) + \(sin^{-1}x\) = \(11\pi\over 6\)
\(\implies\) \(cos^{-1}x\) + (\(cos^{-1}x\) + \(sin^{-1}x\)) = \(11\pi\over 6\)
\(\implies\) \(cos^{-1}x\) + \(\pi\over 2\) = \(11\pi\over 6\)
\(\implies\) \(cos^{-1}x\) = \(4\pi\over 3\)
which is not possible as \(cos^{-1}x\) \(\in\) [0, \(\pi\)]. Hence no solution.
Inequations involving Inverse trigonometric functions
Example : Find the complete solution set of \(sin^{-1}(sin5)\) > \(x^2\) โ 4x.
Solution : \(sin^{-1}(sin5)\) > \(x^2\) โ 4x ย \(\implies\) ย \(sin^{-1}[sin(5-2\pi)]\) > \(x^2\) โ 4x
\(\implies\) \(x^2\) โ 4x < 5 โ 2\(\pi\) ย \(\implies\) ย \(x^2\) โ 4x + 2\(\pi\) โ 5< 0
\(\implies\) 2 โ \(\sqrt{9-2\pi}\) < x < 2 + \(\sqrt{9-2\pi}\) ย \(\implies\) ย x \(\in\) (2 โ \(\sqrt{9-2\pi}\), 2 + \(\sqrt{9-2\pi}\))
Hope you learnt formulas for inverse trigonometric functions, equation and inequations involving inverse trigonometric function, learn more concepts of inverse trigonometric functions and practice more questions to get ahead in competition. Good Luck!
