Half-Angle and Exact Value

[PowerXtremeBaby]

Okay, promise this is the last time I'll ask about trig! My test is tomorrow, and I've realized that there was a few things I missed.

Half-Angle:
1) sin(-105degrees) identity= sin(x/2)=+/-sqrt (1-cos(x)/2)
I've gotten as far as
sin(-210/2)=+/-sqrt (1-cos(-210x)/2)
sin(-105degrees)=+/- sqrt ((1-rt(3)/2)/2)
=+/-sqrt ((2/2-rt(3)/2)/2)
=+/-sqrt ((2-rt(3)/2)/2)
=+/-sqrt (2-rt(3)/2) (1/2)
=+/-sqrt (2-rt(3)/4)
Is that right? if so, is it negative or positive?

2)Cos(-5pi/12) identity= cos(x/2)=+/-sqrt (1-cos(x)/2)


Exact Value:
I don't know how to do these at all. These are 4 problems that look like they are done differently to me. Sorry, I don't know where to start.
1) cos(11pi/12)

2) cos(2pi/5)cos(3pi/5)-sin(2pi/5)sin(3pi/5)

3) sin (-17pi/12)+sin(pi/12)

4) cos(13pi/24)sin(31pi/24)


Thank you in advance!

 

[Aladdin]

#1)Half angle:
\(\displaystyle sin\frac{theta}{2}=\pm\sqrt\frac{1-cos(theta)}{2}\)

\(\displaystyle sin(-105)=-sin(105)=-sin(210/2)=\pm\sqrt\frac{1-cos(210)}{2}\)

Exact Value:
#2)
cos(A + B) = cos A cos B - sin A sin B
cos (11?/12) = cos (2?/3 + ?/4)

cos (2?/3 + ?/4) = cos 2?/3 cos ?/4 - sin 2?/3 sin ?/4 =

\(\displaystyle -\frac{1}{2}\frac{\sqrt2}{2}-\frac{\sqrt3}{2}\frac{\sqrt2}{2}\) = \(\displaystyle \frac{-\sqrt2-\sqrt6}{4}\)