{"id":6006,"date":"2021-10-04T21:56:58","date_gmt":"2021-10-04T16:26:58","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6006"},"modified":"2021-10-04T21:58:06","modified_gmt":"2021-10-04T16:28:06","slug":"inverse-trignometric-function-examples","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/inverse-trignometric-function-examples\/","title":{"rendered":"Inverse Trignometric Function Examples"},"content":{"rendered":"

Here you will learn some inverse trignometric function examples for better understanding of inverse trigonometric function concepts<\/a><\/span>.<\/p>\n\n\n

\n
\n\n \n

Example 1 : <\/span>Find the value of \\(sin^{-1}({-\\sqrt{3}\\over 2})\\) + \\(cos^{-1}(cos({7\\pi\\over 6}))\\).<\/p>\n

Solution : <\/span>\\(sin^{-1}({-\\sqrt{3}\\over 2})\\) = – \\(sin^{-1}({\\sqrt{3}\\over 2})\\) = \\(-\\pi\\over 3\\)

\n \\(cos^{-1}(cos({7\\pi\\over 6}))\\) = \\(cos^{-1}(cos({2\\pi – {5\\pi\\over 6}}))\\) = \\(cos^{-1}(cos({5\\pi\\over 6}))\\) = \\(5\\pi\\over 6\\)

\n Hence \\(sin^{-1}({-\\sqrt{3}\\over 2})\\) + \\(cos^{-1}(cos({7\\pi\\over 6}))\\) = \\(-\\pi\\over 3\\) + \\(5\\pi\\over 6\\) = \\(\\pi\\over 2\\)<\/p>\n


\n\n \n

Example 2 : <\/span> Prove that : \\(cos^{-1}{12\\over 13}\\) + \\(sin^{-1}{3\\over 5}\\) = \\(sin^{-1}{56\\over 65}\\)<\/p>\n

Solution : <\/span>We have, L.H.S. = \\(cos^{-1}{12\\over 13}\\) + \\(sin^{-1}{3\\over 5}\\) = \\(tan^{-1}{5\\over 12}\\) + \\(tan^{-1}{3\\over 4}\\)

\n \\(\\because\\)   [ \\(cos^{-1}{12\\over 13}\\) = \\(tan^{-1}{5\\over 12}\\) & \\(sin^{-1}{3\\over 5}\\) = \\(tan^{-1}{3\\over 4}\\) ]

\n L.H.S. = \\(tan^{-1}({{{5\\over 12} + {3\\over 4}}\\over {1 – {5\\over 12}.{3\\over 4}}})\\) = \\(tan^{-1}{56\\over 33}\\)

\n R.H.S. = \\(sin^{-1}{56\\over 65}\\) = \\(tan^{-1}{56\\over 33}\\)

\n L.H.S = R.H.S.   Hence Proved.<\/p>\n


\n\n \n

Example 3 : <\/span> Evaluate \\(sin^{-1}(sin10)\\)<\/p>\n

Solution : <\/span>We know that \\(sin^{-1}(sinx)\\) = x, if \\(-\\pi\\over 2\\) \\(\\le\\) x \\(\\le\\) \\(\\pi\\over 2\\)

\n Here, x = 10 radians which does not lie between -\\(\\pi\\over 2\\) and \\(\\pi\\over 2\\)

\n But, \\(3\\pi\\) – x i.e. \\(3\\pi\\) – 10 lie between -\\(\\pi\\over 2\\) and \\(\\pi\\over 2\\)

\n Also, sin(\\(3\\pi\\) – 10) = sin 10

\n \\(\\therefore\\)   \\(sin^{-1}(sin10)\\) = \\(sin^{-1}(sin(3\\pi – 10)\\) = (\\(3\\pi\\) – 10)<\/p>\n


\n\n \n

Example 4 : <\/span> Prove that : \\(sin^{-1}{12\\over 13}\\) + \\(cot^{-1}{4\\over 3}\\) + \\(tan^{-1}{63\\over 16}\\) = \\(\\pi\\)<\/p>\n

Solution : <\/span>We have, A = \\(sin^{-1}{12\\over 13}\\) + \\(cot^{-1}{4\\over 3}\\) + \\(tan^{-1}{63\\over 16}\\)

\n A = \\(tan^{-1}{12\\over 5}\\) + \\(tan^{-1}{3\\over 4}\\) + \\(tan^{-1}{63\\over 16}\\)

\n \\(\\implies\\) A = \\(\\pi\\) + \\(tan^{-1}({{{12\\over 5} + {3\\over 4}}\\over {1 – {12\\over 5} \\times {3\\over 4}}})\\) + \\(tan^{-1}{63\\over 16}\\)

\n \\(\\implies\\) A = \\(\\pi\\) + \\(tan^{-1}{63\\over (-16)}\\) + \\(tan^{-1}{63\\over 16}\\)

\n = \\(\\pi\\) – \\(tan^{-1}{63\\over 16}\\) + \\(tan^{-1}{63\\over 16}\\)

\n = \\(\\pi\\) <\/p>\n


\n\n

Example 5 : <\/span>Solve the equation : 2\\(tan^{-1}({2x+1})\\) = \\(cos^{-1}x\\)<\/p>\n

Solution : <\/span>Here, 2\\(tan^{-1}({2x+1})\\) = \\(cos^{-1}x\\)

\n cos(2\\(tan^{-1}({2x+1})\\)) = x       { We Know cos2x = \\({1-tan^2x\\over {1+tan^2x}}\\)}

\n \\(\\therefore\\)     \\({{1-{(2x+1)}^2}\\over {1-{(2x+1)}^2}}\\) = x   \\(\\implies\\)   (1 – 2x – 1)(1 + 2x + 1) = x(\\(4x^2 + 4x + 2\\))

\n \\(\\implies\\)   -2x.2(x + 1) = 2x(\\(2x^2 + 2x + 1\\))   \\(\\implies\\)   2x(\\(2x^2 + 2x + 1 + 2x + 2\\)) = 0

\n \\(\\implies\\)   x = 0   or   \\(2x^2 + 4x + 3\\) = 0   { No Solution }

\n Verify       x = 0

\n \\(2tan^{-1}(1)\\) = \\(cos^{-1}(1)\\)   \\(\\implies\\)   \\(\\pi\\over 2\\) = \\(\\pi\\over 2\\)

\n \\(\\therefore\\)     x = 0 is only the solution.<\/p>\n
\n

Practice these given inverse trignometric function examples to test your knowledge on concepts of inverse trigonometric function.<\/p>\n \n <\/div>\n <\/div>\n","protected":false},"excerpt":{"rendered":"

Here you will learn some inverse trignometric function examples for better understanding of inverse trigonometric function concepts. Example 1 : Find the value of \\(sin^{-1}({-\\sqrt{3}\\over 2})\\) + \\(cos^{-1}(cos({7\\pi\\over 6}))\\). Solution : \\(sin^{-1}({-\\sqrt{3}\\over 2})\\) = – \\(sin^{-1}({\\sqrt{3}\\over 2})\\) = \\(-\\pi\\over 3\\) \\(cos^{-1}(cos({7\\pi\\over 6}))\\) = \\(cos^{-1}(cos({2\\pi – {5\\pi\\over 6}}))\\) = \\(cos^{-1}(cos({5\\pi\\over 6}))\\) = \\(5\\pi\\over 6\\) Hence \\(sin^{-1}({-\\sqrt{3}\\over …<\/p>\n

Inverse Trignometric Function Examples<\/span> Read More »<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-global-header-display":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":""},"categories":[27],"tags":[],"yoast_head":"\nInverse Trignometric Function Examples - Mathemerize<\/title>\n<meta name=\"description\" content=\"Solve these Inverse Trignometric Function examples to practice and test your knowledge of Inverse Trignometric Function concepts.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mathemerize.com\/inverse-trignometric-function-examples\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Inverse Trignometric Function Examples - 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