Question

How do you use an appropriate Half-Angle Formula to find the exact value of the expression `sin ((13pi)/12)`?

Answer 1

Use `sin(1/2((13pi)/6)) = -sqrt((1 - cos((13pi)/6))/2)` to get `-(sqrt(2-sqrt3))/2`. (Which can be written in other forms.)

Explanation:

Because we want a sine, the appropriate Half-Angle Formula is:

`sin(x/2) - +- sqrt((1-cosx)/2)`

We note that `(13pi)/12 > pi = (12pi)/12` but `< (3pi)/2 = (18pi)/12`
So it is in the third quadrant and has negative sine.

To find `x`, we ask oursolves "Ok, so `(13pi)/12` is half of what?"

We could solve `1/2 x = (13 pi)/12`, but it is pretty clear that `1/2 * (13pi)/6 = (13pi)/12`, so `x = (13pi)/6`
(Just double the number we're interested in.)

Finally, we note that `cos((13pi)/6) = sqrt3/2`

So we get:
`sin(1/2((13pi)/6)) = -sqrt((1 - cos((13pi)/6))/2)`

`= -sqrt((1 - sqrt3/2)/2)`

`= -sqrt((2 - sqrt3)/4)`

`= -sqrt(2 - sqrt3)/2`