If cscx=2 and x greater 90 degrees and x less than 180 degrees, then sin x/2 =
To find the value of sin(x/2) given that csc(x) = 2 and x > 90° and x < 180°, we can use the reciprocal relationship between csc(x) and sin(x).
Recall that csc(x) is the reciprocal of sin(x), which means csc(x) = 1/sin(x).
Given that csc(x) = 2, we can write:
2 = 1/sin(x)
To solve for sin(x), we can take the reciprocal of both sides of the equation:
1/2 = sin(x)
Now, we need to find sin(x/2). We can use the half-angle formula for sin(x/2):
sin(x/2) = ±sqrt[(1 - cos(x))/2]
To determine the sign (±), we need to consider the given conditions: x > 90° and x < 180°.
Since x is in the second quadrant (90° < x < 180°), sin(x/2) will be positive.
Therefore, we can simplify the half-angle formula for sin(x/2) as follows:
sin(x/2) = sqrt[(1 - cos(x))/2]
Now, we need to find the value of cos(x) to substitute it into the formula. To do that, we can use the Pythagorean Identity:
sin^2(x) + cos^2(x) = 1
Since we know sin(x) = 1/2 from the given information, we can substitute this value into the equation:
(1/2)^2 + cos^2(x) = 1
1/4 + cos^2(x) = 1
cos^2(x) = 3/4
Taking the square root of both sides, we get:
cos(x) = ±sqrt(3)/2
Since x is in the second quadrant (90° < x < 180°), cos(x) will be negative. Therefore, we have:
cos(x) = -sqrt(3)/2
Now, we can substitute this value of cos(x) into the formula for sin(x/2):
sin(x/2) = sqrt[(1 - cos(x))/2]
sin(x/2) = sqrt[(1 - (-sqrt(3)/2))/2]
sin(x/2) = sqrt[(1 + sqrt(3)/2)/2]
sin(x/2) = sqrt[(2 + sqrt(3))/4]
sin(x/2) = sqrt(2 + sqrt(3))/2
So, sin(x/2) = sqrt(2 + sqrt(3))/2.
So sinx = 1/2 and x must be in 2nd quadrant.
then by Pythagoras cos x = -√3/2
we know cos 2A = 1 - 2sin^2A
then cosx = 1 - 2sin^2(x/2)
-√3/2 = 1 - 2sin^2(x/2)
2sin^2(x/2) = 1 + √3/2
sin^2(x/2) = (2 + √3)/4
sin(x/2) = (√(2+√3))/2