## anonymous one year ago Find the exact value by using a half-angle identity. cosine (5pi/12)

1. anonymous

$\frac{5\pi}{12}$ is half of $\frac{5\pi}{6}$

2. anonymous

then, 5pi/6 = -√3/2 ?

3. freckles

$\cos^2(\frac{1}{2} \theta)=\frac{1}{2}(1+\cos(\theta)) \\ \text{ you have determined } \theta=\frac{5\pi}{6} \\ \text{ and I think you are actually saying } \cos(\frac{5\pi}{6})=\frac{-\sqrt{3}}{2} \\ \text{ because } \frac{5\pi}{6} \neq \frac{-\sqrt{3}}{2}$

4. anonymous

would the answer maybe be cos(5pi/12) = √2 - 2√3 /(fraction) 2 ?

5. freckles

what does that say

6. anonymous

|dw:1438812995121:dw|

7. freckles

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

8. freckles

you are going to choose the positive output because 5pi/12 is between 0 and pi/2

9. anonymous

so i wasnt right D:

10. anonymous

for 2 - √3 / 4 what do you mean by square root both sides?

11. freckles

you want to find cos(5pi/12)

12. freckles

not cos^2(5pi/12)

13. anonymous

oh, oh okay so could it be cos(5pi/12) = 2 - √3 / 2

14. freckles

well you are missing the square root on top

15. freckles

$\cos^2(\frac{5\pi}{12})= \frac{1}{2}(\frac{2}{2}-\frac{\sqrt{3}}{2}) \\ \cos^2(\frac{5\pi}{12})=\frac{1}{2}(\frac{2-\sqrt{3}}{2}) \\ \cos^2(\frac{5\pi}{12})=\frac{2-\sqrt{3}}{4} \\ \text{ take square root of both sides } \\ \cos(\frac{5\pi}{12})= \sqrt{\frac{2-\sqrt{3}}{4}} \text{ note: chose positive \because } 0<\frac{5\pi}{12}<\frac{\pi}{2} \\ \cos(\frac{5\pi}{12})=\frac{\sqrt{2- \sqrt{3}}}{\sqrt{4}}=\frac{\sqrt{2- \sqrt{3}}}{2}$