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October 27th, 2016, 09:51 AM   #1
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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?
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October 27th, 2016, 09:54 AM   #2
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I assume the 90 is in degrees ...

$\sin^{-1}[\sin(90^\circ)] = \sin^{-1}(1) = 90^\circ$
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October 27th, 2016, 10:18 AM   #3
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If $x=\sin^{-1} \big( \sin 90^\circ \big)$ then $\sin x=\sin 90^\circ$
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October 27th, 2016, 10:25 AM   #4
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What if it were sin^-1(sin(180)) or cos^-1(sin(330))?
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October 27th, 2016, 11:39 AM   #5
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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 $[0^\circ , 180^\circ]$
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