Question

How can I calculate the exact value of cos(1/2alpha) if alpha=arcsin1/3.

Answer 1

alpha=arcsin1/3 ---> sin α = 1/3 , then y = 1, r = 3
since sin α = y/r, construct a right-angled triangle using those values
x^2 + y^2 = r^2
x^2 + 1 = 9
x = ± √8 = ± 2√2
and we have cos α = ± 2√2/3

from cos 2A = 2cos α - 1
cos α = 2cos^2 (α/2) - 1
2√2/3 = 2cos^2 (α/2) - 1
2√2/3 + 1 = 2cos^2 (α/2)
2cos^2 (α/2) = (2√2 + 3)/3
cos^2 (α/2) = (2√2 + 3)/6
cos (α/2) = ± √[ (2√2 + 3)/6 ]

Answer 2

half-way down my solution,
from cos 2A = 2cos α - 1, should have been from cos 2A = 2cos^2 A - 1
(does not affect rest of solution), checked answer with calculator

Answer 3

That looks kind of messy, but note that
2(3 + 2√2) = 6+4√2 = (2+√2)^2
So, (2√2 + 3)/6 = (2+√2)^2/12
That makes
cos(α/2) = (2+√2)/(2√3) = 1/√3 + 1/√6

Answer 4

And a purist would still not be satisfied with 1/√3 + 1/√6 , insisting that
we rationalize the denominators
1/√3 = √3/3
1/√6 = √6/6

then 1/√3 + 1/√6 = (2√3 + √6)/6

Answer 5

rationalizing denominators is for Algebra I
I find it annoying and of no real use.