A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
How do I get (1+sqrt(3))/(2 sqrt(2)) from cos(11/12 pi) using half/double angle identities?
anonymous
 one year ago
How do I get (1+sqrt(3))/(2 sqrt(2)) from cos(11/12 pi) using half/double angle identities?

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\cos(\frac{11}{12}\pi) =>\frac{1+\sqrt{3}}{2\sqrt{2}}\]

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.3well look at it like this \[\cos(\frac{11\pi}{12}) = \cos(\frac{2\pi}{3} + \frac{\pi}{4})\]

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.3then you need \[\cos(A +B) = \cos(A)\cos(B)  \sin(A)\sin(B)\]

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.3the sum of both angles are exact values.... just remember \[\frac{2\pi}{3} ~is~2nd~quadrant\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, ok. I was hoping not to have to use sum/difference formulas because they are longer than:\[\cos(\frac{\theta}{2})=\sqrt{\frac{1+\cos(\theta)}{2}}\] But I guess I just have to not be lazy and evaluate more trig functions.

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.3sum and difference are the easiest for this question \[\cos(\frac{\pi}{4}) = \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}\] \[\cos(\frac{2\pi}{3}) =  \frac{1}{2}~~~~\sin(\frac{2\pi}{3}) = \frac{\sqrt{3}}{2}\] now just multiply the fractions... with is certainly easier than the half angle formula or double angle formula

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That makes sense; a little bit of thinking ahead goes a long way.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.