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 October 27th, 2016, 09:51 AM #1 Newbie   Joined: Oct 2016 From: Rainfall Posts: 2 Thanks: 0 How to figure out the exact values? I'm wondering how to solve problems like this: Sin^-1(sin(90)) Can anyone help me with this?
 October 27th, 2016, 09:54 AM #2 Math Team     Joined: Jul 2011 From: Texas Posts: 3,016 Thanks: 1600 I assume the 90 is in degrees ... $\sin^{-1}[\sin(90^\circ)] = \sin^{-1}(1) = 90^\circ$
 October 27th, 2016, 10:18 AM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,683 Thanks: 2664 Math Focus: Mainly analysis and algebra If $x=\sin^{-1} \big( \sin 90^\circ \big)$ then $\sin x=\sin 90^\circ$
 October 27th, 2016, 10:25 AM #4 Newbie   Joined: Oct 2016 From: Rainfall Posts: 2 Thanks: 0 What if it were sin^-1(sin(180)) or cos^-1(sin(330))?
 October 27th, 2016, 11:39 AM #5 Math Team     Joined: Jul 2011 From: Texas Posts: 3,016 Thanks: 1600 I had a feeling this was going to be asked next ... $\sin^{-1}[\sin(180^\circ)] = \sin^{-1}(0) = 0$, because the range of the inverse sine function is $[-90^\circ, 90^\circ]$ $\cos^{-1}[\sin(330^\circ)] = \cos^{-1}\left(-\dfrac{1}{2}\right) = 120^\circ$ because the range of the inverse cosine function is $[0^\circ , 180^\circ]$ Thanks from topsquark

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