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Mathematics 14 Online
OpenStudy (anonymous):

The exact value of cos 5pi/12 is: sqrt 2 (sqrt 3+1) / 4 sqrt 2 (sqrt 3-1) / 4 0.26 0.99

OpenStudy (freckles):

\[\cos(\frac{5\pi}{12})=\cos(\frac{1}{2} \cdot \frac{ 5\pi}{6})\] halfangle formula comes in mind

OpenStudy (anonymous):

@freckles i got\[\frac{ \sqrt{6}-\sqrt{2} }{ 4 }\] i doesnt match any of the answers that im supposed to get

OpenStudy (freckles):

woah how did you get that!

OpenStudy (freckles):

\[\cos^2(\frac{a}{2})=\frac{1+\cos(a)}{2}\] this was the half angle formula

OpenStudy (freckles):

our a here is 5pi/6

OpenStudy (freckles):

5pi/12 is in the first quadrant so when we take square root of both sides we know to use the + instead of the - since cos is positive in the first quadrant

OpenStudy (anonymous):

0.25881905 that was what i got when i put it into my calculator

OpenStudy (freckles):

\[\cos(\frac{a}{2})=\sqrt{\frac{1+\cos(a)}{2}}\]

OpenStudy (freckles):

do you know what cos(5pi/6) is?

OpenStudy (anonymous):

yes - sqrt 3/2

OpenStudy (freckles):

\[\cos(\frac{1}{2} \cdot \frac{5\pi}{6})=\sqrt{\frac{1+\frac{-\sqrt{3}}{2}}{2}}\]

OpenStudy (freckles):

now we have an ugly compound fraction inside that sqrt( )

OpenStudy (anonymous):

oh so its a?

OpenStudy (freckles):

so inside multiply top and bottom by 2

OpenStudy (freckles):

\[\cos(\frac{1}{2} \cdot \frac{5\pi}{6})=\sqrt{\frac{2}{2} \cdot \frac{1+\frac{-\sqrt{3}}{2}}{2}} \]

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