Find the exact value of cos(x/2) if sin x = 3/4 and x is in quadrant II.

==> I got �ã(4-�ã7)/2. Is this correct?

I can't read your symbols, but

if sinx = 3/4 in II then
cos x = -√7/4

cosx = 2cos^2 (x/2) - 1
-√7/4 + 1 = 2cos^2 (x/2)
(4-√7)/8 = cos^2 (x/2)

cos x/2 = √[(4-√7)/8]

Oops, yeah those were supposed to be radicals :) thanks so much!!!!

To find the exact value of cos(x/2), we can use the double-angle formula for cos:

cos(2θ) = 1 - 2sin²(θ)

Let's solve this step by step:

1. We know that sin(x) = 3/4. To find cos(x), we can use the Pythagorean identity:

cos²(x) = 1 - sin²(x)

cos²(x) = 1 - (3/4)²
cos²(x) = 1 - 9/16
cos²(x) = 16/16 - 9/16
cos²(x) = 7/16

Taking the square root of both sides:
cos(x) = ±√(7/16)

Since x is in quadrant II, where cos(x) is negative, we take the negative sign:
cos(x) = -√(7/16)

2. Now, we can use the half-angle formula for cos:

cos(x/2) = ±√((1 + cos(x))/2)

cos(x/2) = ±√((1 - √(7/16))/2)

Since x is in quadrant II, where sine is positive, we choose the positive square root:
cos(x/2) = √((1 - √(7/16))/2)

Now, let's simplify further:

cos(x/2) = √((16/16 - √7/4)/2)
cos(x/2) = √((16 - √7*4)/32)
cos(x/2) = √((16 - 2√7)/32)
cos(x/2) = √((4 - √7)/8)

So, the correct value of cos(x/2) if sin x = 3/4 and x is in quadrant II is √((4 - √7)/8).