Question

How do you evaluate sin(13π/12)?

Answer 1

13pi/12 lies between pi and 2pi, which means sin(13pi/12) < 0

Recall the double angle identity,

sin^2(x) = (1 - cos(2x))/2

If we let x = 13pi/12, then

sin(13pi/12) = - sqrt[(1 - cos(13pi/6))/2]

where we took the negative square root because we expect a negative value.

Now, because cosine has a period of 2pi, we have

cos(13pi/6) = cos(2pi + pi/6) = cos(pi/6) = sqrt[3]/2

Then

sin(13pi/12) = - sqrt[(1 - sqrt[3]/2)/2]

sin(13pi/12) = - sqrt[2 - sqrt[3]]/2