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December 22nd, 2016, 03:53 PM   #1
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find the values of sin inverse sqrt 5

Find the values of sin inverse sqrt 5
I'm stuck with this one; please someone help me.

Last edited by skipjack; December 24th, 2016 at 06:34 AM.
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December 22nd, 2016, 04:21 PM   #2
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what is the range of the sine function?

what is the approximate value of $\sqrt{5}$ ?

do you see a problem here?

unless you mean $\sin\left(\dfrac{1}{\sqrt{5}}\right)$

is that what you mean?
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December 22nd, 2016, 06:52 PM   #3
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$\displaystyle y=\sin^{-1}(\sqrt5)$

$\displaystyle \sin(y)=\sqrt5$

$\displaystyle e^{iy}-e^{-iy}=2i\sqrt5$

$\displaystyle \frac{e^{2iy}-1}{e^{iy}}=2i\sqrt5\Rightarrow e^{2iy}-2i\sqrt5e^{iy}-1=0\implies y=\log(i(\sqrt5\pm2))/i$

$\displaystyle \sin^{-1}(\sqrt5)=\log(i(\sqrt5\pm2))/i+2k\pi,k\in\mathbf{Z}$
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December 23rd, 2016, 11:18 AM   #4
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ty so much
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