find the exact value of sinu/2, given that cosu=-4/5 and pi/2less than u less than pi
use the formula
cos 2A = 1 - 2sin^2 A
cos u = 1 - 2sin^2 (u/2) , π/2 < u < π , so u is in quad II
-4/5 = 1 - 2sin^2 (u/2)
2 sin^2 (u/2) = 9/5
sin^2 (u/2) = 9/10
sin u/2 = √(9/10)
= 3/√10 or 3√10/10
To find the exact value of sin(u/2), we need to use the half-angle identity for sine. The half-angle identity for sine is defined as:
sin(u/2) = ± √[(1 - cos(u))/2]
First, let's find the value of cos(u). We are given that cos(u) = -4/5.
Next, we can use the given value of cos(u) to find sin(u) using the Pythagorean identity for sine and cosine. The Pythagorean identity is defined as:
sin^2(u) + cos^2(u) = 1
Plugging in the value of cos(u) = -4/5, we have:
sin^2(u) + (-4/5)^2 = 1
sin^2(u) + 16/25 = 1
sin^2(u) = 1 - 16/25
sin^2(u) = 9/25
Since u is in the second quadrant where the sine is positive, we take the positive square root:
sin(u) = √(9/25) = 3/5
Now, substituting the value of sin(u) = 3/5 into the half-angle identity for sine, we have:
sin(u/2) = ± √[(1 - cos(u))/2]
sin(u/2) = ± √[(1 - (-4/5))/2]
sin(u/2) = ± √[(1 + 4/5)/2]
sin(u/2) = ± √[(9/5)/2]
sin(u/2) = ± √(9/10)
sin(u/2) = ± (3/√10)
Therefore, the exact value of sin(u/2) is ± (3/√10).