Find the exact value by using a half-angle identity. cosine (5pi/12)

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Find the exact value by using a half-angle identity. cosine (5pi/12)

Mathematics
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\[\frac{5\pi}{12}\] is half of \[\frac{5\pi}{6}\]
then, 5pi/6 = -√3/2 ?
\[\cos^2(\frac{1}{2} \theta)=\frac{1}{2}(1+\cos(\theta)) \\ \text{ you have determined } \theta=\frac{5\pi}{6} \\ \text{ and I think you are actually saying } \cos(\frac{5\pi}{6})=\frac{-\sqrt{3}}{2} \\ \text{ because } \frac{5\pi}{6} \neq \frac{-\sqrt{3}}{2}\]

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would the answer maybe be cos(5pi/12) = √2 - 2√3 /(fraction) 2 ?
what does that say
|dw:1438812995121:dw|
so you have: \[\cos^2(\frac{5\pi}{12})=\frac{1}{2}(1+\frac{-\sqrt{3}}{2})\] you just need to solve for cos(5pi/12) above notice I just replaced theta with 5pi/6 and replaced cos(5pi/6) with -sqrt(3)/2 \[\cos^2(\frac{5\pi}{12})= \frac{1}{2}(\frac{2}{2}-\frac{\sqrt{3}}{2}) \\ \cos^2(\frac{5\pi}{12})=\frac{1}{2}(\frac{2-\sqrt{3}}{2}) \\ \cos^2(\frac{5\pi}{12})=\frac{2-\sqrt{3}}{4}\] now take square root of both sides
you are going to choose the positive output because 5pi/12 is between 0 and pi/2
so i wasnt right D:
for 2 - √3 / 4 what do you mean by square root both sides?
you want to find cos(5pi/12)
not cos^2(5pi/12)
oh, oh okay so could it be cos(5pi/12) = 2 - √3 / 2
well you are missing the square root on top
\[\cos^2(\frac{5\pi}{12})= \frac{1}{2}(\frac{2}{2}-\frac{\sqrt{3}}{2}) \\ \cos^2(\frac{5\pi}{12})=\frac{1}{2}(\frac{2-\sqrt{3}}{2}) \\ \cos^2(\frac{5\pi}{12})=\frac{2-\sqrt{3}}{4} \\ \text{ take square root of both sides } \\ \cos(\frac{5\pi}{12})= \sqrt{\frac{2-\sqrt{3}}{4}} \text{ note: chose positive \because } 0<\frac{5\pi}{12}<\frac{\pi}{2} \\ \cos(\frac{5\pi}{12})=\frac{\sqrt{2- \sqrt{3}}}{\sqrt{4}}=\frac{\sqrt{2- \sqrt{3}}}{2}\]

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