+118608 If cos(x) = 1/3 and x is an acute angle, find the value of sin(x/2)
let
\(\theta=\frac{x}{2}\qquad x=2\theta\\ \)
The question becomes.
\(If\;\; cos(2\theta) = \frac{1}{3} \text{ and }x \text{ is less between 0 and 45 degrees, find the value of } sin(\theta)\\~\\ 1-sin^2\theta-sin^2\theta=\frac{1}{3}\\ 1-2sin^2\theta=\frac{1}{3}\\ -2sin^2\theta=\frac{-2}{3}\\ sin^2\theta=\frac{1}{3}\\ sin\theta=\frac{1}{\sqrt3}\\ sin(\frac{x}{2})=\frac{1}{\sqrt3}\\ \)
Melody: I don't know much about Trig. But would this direct "solution" be valid? If not, why not? Just curious to know. Thanks.
Cos(x) = 1 /3
x = arccos(1/3)
x =70.529 - degrees
x / 2 = 70.529 / 2
x = 35. 2645 - degrees.
Sin(35.2645) =1 / sqrt(3) ???.
+118608 It is a really good way to check the answer but I doubt it would be given many marks in a test.
You got down to sin(35.2645) degrees and from that you can get the approximation but that is not an exact answer.
There is no way you can go from that and know that the answer is 1/sqrt3
(you can use it for checking by going the other way around though)
Thanks for asking this question. Answerers greatly appreciate it when their answers are thought about and learned from. 
