<s>[INST] If cosB=3/4, how do I find cos 2B and cos(B/2)? I got $\cos 2B$;
$\cos 2B=2 \cos^2 B - 1
             =9/8 -1
              =1/8$
but when I tried cos1/2B, I got: COS1/2B=cos^2*1/4B-1, then I solved it, and didn't get the answer which was in the book, why?
 [/INST] $\cos 2B = 2 \cos^2 B - 1 
= \frac9{8} - 1 = \frac{1}{8}$
Second answer -
$\cos B = 2 \cos^2(\frac B2) - 1$
$\frac34 = 2 \cos^2(\frac B2) - 1$
$\frac 74 =  2 \cos^2(\frac B2)$
$\frac 78 =  \cos^2(\frac B2)$
$\sqrt{\frac78} =  \cos (\frac B2)$
$\frac12 \sqrt{\frac72} = \cos (\frac B2)$
</s>