I am currently working on a problem relating to trigonometric identities and I seem to be stuck. The question wants me to find sin(x/2) when cosx = -(12/13) in the interval (pi,3pi/2). I know that the eventual answer will be 5/sqrt26, however, I am not sure how to get there. Any help would be appreciated :).

Yes, I get that, but which formula do I rearrange to find sin(x/2)? Is it the one for sin2x? I am not too familiar with the formulas as I have been forced to learn trig identities on short notice.

you look up you half-angle formulas!

sin(x/2) = √((1-cosx)/2)
= √(1+12/13)/2)
and just simplify that.

To find sin(x/2) when cos(x) = -(12/13) in the given interval, you can use the half-angle identity for sine.

The half-angle identity for sine is: sin(x/2) = ±√[(1 - cosx) / 2]

However, in your case, since x is in the interval (pi, 3pi/2), which means it lies in the third quadrant, we can infer that sin(x/2) will be negative.

So, we can modify the half-angle identity for sine to: sin(x/2) = - √[(1 - cosx) / 2]

Now, let's substitute the given value of cosx = -(12/13) into the equation:

sin(x/2) = - √[(1 - cosx) / 2]
= - √[(1 - (-12/13)) / 2]
= - √[(1 + 12/13) / 2]
= - √[(25/13) / 2]
= - √(25/26)
= - (√25 / √26)
= - (5 / √26)
= - (5 / √(26) * (√(26) / √(26))
= - (5√26 / 26)

Thus, sin(x/2) = - (5√26 / 26).