1. ## half angle trouble

I am trying to use the half angle formula to solve these problems:
Sin(-7pi/12)
Sin(7pi/12)
cos(13pi/12)
cos(3pi/8)
here is what I did cos(3pi/8) = 1/2(1+cis13pi/4)= sqrt (2)+sqrt(3)/2 but the book shows a different answer.
for Sin(-7pi/12) I got sqrt(2)-sqrt(3)/2 I must be doing something wrong and I must have not understood what I read in the text. This area of the text seems more difficult than the rest of the material we have covered. Please explain the half angle formula so even I can understand it.
Keith Stevens

2. Heloo, Keith!

The formulas are:

. . $\sin^2\!\theta \:=\:\frac{1 - \cos2\theta}{2}\quad\Rightarrow\quad \sin\theta \:=\:\pm\sqrt{\frac{1-\cos2\theta}{2}}$

. . $\cos^2\!\theta \:=\:\frac{1 + \cos2\theta}{2}\quad\Rightarrow\quad\cos\theta \:=\:\pm\sqrt{\frac{1+\cos2\theta}{2}}$

and we must choose the sign carefully.

Use the half angle formulas:

$(1)\;\;\sin\left(\text{-}\frac{7\pi}{12}\right)\quad(2)\;\sin\!\left(\frac {7\pi}{12}\right)\quad(3)\;\cos\!\left(\frac{13\pi }{12}\right)\quad(4)\;\cos\!\left(\frac{3\pi}{8}\r ight)$

$(1)\;\text{-}\frac{7\pi}{12}$ is in Quadrant 3, where sine is negative.
$\sin\left(\text{-}\frac{7\pi}{12}\right)\:=\:-\sqrt{\frac{1 - \cos\left(\text{-}\frac{7\pi}{6}\right)}{2}} \:=\:$ $-\sqrt{\frac{1 - \left(\text{-}\frac{\sqrt{3}}{2}\right)}{2}} \:=\:-\sqrt{\frac{2 + \sqrt{3}}{4}}\:=\:-\frac{\sqrt{2 + \sqrt{3}}}{2}$

$(2)\;\;\frac{7\pi}{12}$ is in Quadrant 2, where sine is positive.
$\sin\left(\frac{7\pi}{12}\right)\:=\:\sqrt{\frac{1 - \cos\left(\frac{7\pi}{6}\right)}{2}} \:=\:$ $\sqrt{\frac{1 - \left(\text{-}\frac{\sqrt{3}}{2}\right)}{2}} \:=\:\sqrt{\frac{2 + \sqrt{3}}{4}}\:=\:\frac{\sqrt{2 + \sqrt{3}}}{2}$

$(3)\;\;\frac{13\pi}{12}$ is in Quadrant 3, where cosine is negative.
$\cos\left(\frac{13\pi}{12}\right)\:=\:-\sqrt{\frac{1 + \cos\left(\frac{13\pi}{6}\right)}{2}} \:=\:$ $-\sqrt{\frac{1 + \left(\frac{\sqrt{3}}{2}\right)}{2}} \:=\:-\sqrt{\frac{2 + \sqrt{3}}{4}}\:=\:-\frac{\sqrt{2 + \sqrt{3}}}{2}$

$(4)\;\;\frac{3\pi}{8}$ is in Quadrant 1, where cosine is positive.
$\cos\left(\frac{3\pi}{8}\right)\:=\:\sqrt{\frac{1 + \cos\left(\frac{3\pi}{4}\right)}{2}} \:=\:$ $\sqrt{\frac{1 + \left(\text{-}\frac{\sqrt{2}}{2}\right)}{2}} \:=\:\sqrt{\frac{2 - \sqrt{2}}{4}}\:=\:\frac{\sqrt{2 - \sqrt{2}}}{2}$