If sin⁡ x=3√/2, and 90°<x<180°, what is cos⁡(x/2)?

Question

If sin⁡ x=3√/2, and 90°<x<180°, what is cos⁡(x/2)?

Answer 1

I'm guessing you meant to write sin(x) = √(3)/2 (i.e. one half of the square root of 3)

The relation between sin(x) and cos(x) is always: sin²x + cos²x = 1

Since you know sin(x), rearrange the identity above as: cos²x = 1 - sin²x

which gives: cos²x = 1 - (√(3)/2)²

= 1 - 3/4

= 1/4

therefore: cos(x) = ±1/2

Answer 2

Is this a typo? 3√/2 ??? The sine of any angle is always ≤ 1. Please clarify.