How do you find sin (x/2), given #cos x = -5/8#, with #pi/2 < x <pi#?

1 Answer
May 18, 2015

Recall the formula for a cosine of a sum of two angles:
#cos(alpha+beta)=cos(alpha)*cos(beta)-sin(alpha)*sin(beta)#

Use it for #alpha=beta=x/2#:
#cos(x/2+x/2)=cos^2(x/2)-sin^2(x/2)#

Substitute #x/2+x/2=x#
Use the identity #sin^2(gamma)+cos^2(gamma)=1# for #gamma=x/2#
The result is:
#cos(x)=1-2sin^2(x/2)#

Now we can use the value of #cos(x)=-5/8#.
#1-2sin^2(x/2)=-5/8#
#sin^2(x/2)=(1+5/8)/2=13/16#
#sin(x/2)=+-sqrt(13)/4#

If #pi/2 < x < pi#, #pi/4 < x/2 < pi/2#. For these angles function sine is positive.
Therefore, #sin(x/2)=sqrt(13)/4#.