Question

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

Answer 1

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`.