anonymous
  • anonymous
How do I get -(1+sqrt(3))/(2 sqrt(2)) from cos(11/12 pi) using half/double angle identities?
Trigonometry
chestercat
  • chestercat
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anonymous
  • anonymous
\[\cos(\frac{11}{12}\pi) =>-\frac{1+\sqrt{3}}{2\sqrt{2}}\]
campbell_st
  • campbell_st
well look at it like this \[\cos(\frac{11\pi}{12}) = \cos(\frac{2\pi}{3} + \frac{\pi}{4})\]
campbell_st
  • campbell_st
then you need \[\cos(A +B) = \cos(A)\cos(B) - \sin(A)\sin(B)\]

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campbell_st
  • campbell_st
the sum of both angles are exact values.... just remember \[\frac{2\pi}{3} ~is~2nd~quadrant\]
anonymous
  • anonymous
Well, 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
  • campbell_st
sum 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
  • anonymous
That makes sense; a little bit of thinking ahead goes a long way.

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