SOLUTION: find the exact value for sin(Θ/2) given sin Θ=12/13; Θ is obtuse.
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Question 916840: find the exact value for sin(Θ/2) given sin Θ=12/13; Θ is obtuse.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
\ =\ \frac{12}{13})
\ =\ \frac{144}{169})
\ =\ 1\ -\ \sin^2(\theta)\ =\ 1\ -\ \frac{144}{169}\ =\ \frac{25}{169})
\ =\ \pm\frac{5}{13})
But since 
is obtuse 
is in the second quadrant where cosine is negative,
\ =\ -\frac{5}{13})
Since
\ =\ \pm\sqrt{\frac{1\ -\ \cos(\alpha)}{2}})
We have
\ =\ \pm\sqrt{\frac{1\ +\ \frac{5}{13}}{2}}\ =\ \pm\sqrt{\frac{9}{13}}\ =\ \pm\frac{3}{\sqrt{13}})
Rationalizing the denominator
\ =\ \pm\frac{3\sqrt{13}}{13})
But since is in quadrant 2, 
must be in quadrant 1 (
). And since the sine is positive in quadrant 1:
\ =\ \frac{3\sqrt{13}}{13})
John
My calculator said it, I believe it, that settles it
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