{"id":3974,"date":"2021-08-11T16:36:23","date_gmt":"2021-08-11T16:36:23","guid":{"rendered":"https:\/\/mathemerize.com\/?p=3974"},"modified":"2021-11-26T16:24:57","modified_gmt":"2021-11-26T10:54:57","slug":"inverse-trigonometric-function-formula","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/inverse-trigonometric-function-formula\/","title":{"rendered":"Formulas for Inverse Trigonometric Functions"},"content":{"rendered":"
Here, you will learn formulas for inverse trigonometric functions, equation and inequations involving inverse trigonometric function.<\/p>\n
Let’s begin –<\/p>\n
(a) y = f(x) = \\(sin^{-1}({2x\\over {1+x^2}})\\) = \\(\\begin{cases} 2tan^{-1}x, & \\text{if}\\ |x| \\le 1 \\\\ \\pi – 2tan^{-1}x, & \\text{if}\\ x > 1 \\\\ -(\\pi + 2tan^{-1}x), & \\text{if}\\ x < -1 \\end{cases}\\) <\/p>\n
(b) y = f(x) = \\(cos^{-1}({{1-x^2}\\over {1+x^2}})\\) = \\(\\begin{cases} 2tan^{-1}x, & \\text{if}\\ |x| \\ge 0 \\\\ – 2tan^{-1}x, & \\text{if}\\ x < 0 \\end{cases}\\)<\/p>\n
(c) 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 – 2tan^{-1}x), & \\text{if}\\ x > 1 \\end{cases}\\)<\/p>\n
(d) y = f(x) = \\(sin^{-1}({3x – 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 – 3sin^{-1}x, & \\text{if}\\ {1\\over 2} \\le x \\le 1 \\end{cases}\\)<\/p>\n
(e) y = f(x) = \\(cos^{-1}({4x^3 – 3x})\\) = \\(\\begin{cases} 3cos^{-1}x – 2\\pi, & \\text{if}\\ -1 \\le x \\le {-1\\over 2} \\\\ 2\\pi – 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}\\)<\/p>\n
(f) 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 – 2sin^{-1}x, & \\text{if}\\ {1\\over \\sqrt{2}} \\le x \\le 1 \\end{cases}\\)<\/p>\n
(g) y = f(x) = \\(cos^{-1}({2x^2-1})\\) = \\(\\begin{cases} 2cos^{-1}x, & \\text{if}\\ 0 \\le x \\le 1 \\\\ 2\\pi – 2cos^{-1}x, & \\text{if}\\ -1 \\le x \\le 0 \\end{cases}\\)<\/p>\n\n\n
Example : <\/span> Prove that : \\(2tan^{-1}{1\\over 2}\\) + \\(tan^{-1}{1\\over 7}\\) = \\(tan^{-1}{31\\over 17}\\)<\/p>\n Solution : <\/span>We have, \\(2tan^{-1}{1\\over 2}\\) + \\(tan^{-1}{1\\over 7}\\) Example : <\/span> Prove that the equation \\(2cos^{-1}x\\) + \\(sin^{-1}x\\) = \\(11\\pi\\over 6\\) has no solution.<\/p>\n Solution : <\/span>Given equation is \\(2cos^{-1}x\\) + \\(sin^{-1}x\\) = \\(11\\pi\\over 6\\) Example : <\/span> Find the complete solution set of \\(sin^{-1}(sin5)\\) > \\(x^2\\) – 4x.<\/p>\n Solution : <\/span>\\(sin^{-1}(sin5)\\) > \\(x^2\\) – 4x \\(\\implies\\) \\(sin^{-1}[sin(5-2\\pi)]\\) > \\(x^2\\) – 4x 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!<\/p>\n\n\n
\n = \\(2tan^{-1}({{2\\times {1\\over 2}\\over {1-({1\\over 2})^2}}})\\) + \\(tan^{-1}{1\\over 7}\\) [\\(\\because\\) \\(2tan^{-1}x\\) = \\(tan^{-1}{2x\\over {1-x^2}}\\)]
\n \\(tan^{-1}{4\\over 3}\\) + \\(tan^{-1}{1\\over 7}\\) = \\(tan^{-1}[{{4\\over 3} + {1\\over 7}\\over {1 – {4\\over 3}\\times {1\\over 7}}}]\\) = \\(tan^{-1}{31\\over 17}\\)
<\/p>\n\n\nEquations involving Inverse trigonometric functions<\/strong><\/h4>\n\n\n
\n \\(\\implies\\) \\(cos^{-1}x\\) + (\\(cos^{-1}x\\) + \\(sin^{-1}x\\)) = \\(11\\pi\\over 6\\)
\n \\(\\implies\\) \\(cos^{-1}x\\) + \\(\\pi\\over 2\\) = \\(11\\pi\\over 6\\)
\n \\(\\implies\\) \\(cos^{-1}x\\) = \\(4\\pi\\over 3\\)
\n which is not possible as \\(cos^{-1}x\\) \\(\\in\\) [0, \\(\\pi\\)]. Hence no solution.
\n <\/p>\n\n\nInequations involving Inverse trigonometric functions<\/strong><\/h4>\n\n\n
\n \\(\\implies\\) \\(x^2\\) – 4x < 5 – 2\\(\\pi\\) \\(\\implies\\) \\(x^2\\) – 4x + 2\\(\\pi\\) – 5< 0
\n \\(\\implies\\) 2 – \\(\\sqrt{9-2\\pi}\\) < x < 2 + \\(\\sqrt{9-2\\pi}\\) \\(\\implies\\) x \\(\\in\\) (2 – \\(\\sqrt{9-2\\pi}\\), 2 + \\(\\sqrt{9-2\\pi}\\))
<\/p>\n\n\n