Find the value for sin(theta), cos(2 theta)=3/4 and 270 degrees<0<360 degrees

sinθ = √((1-cos2θ)/2) = √(1/8)

since 2θ is in QIV, θ is in QII

To find the value of sin(theta) given cos(2 theta) = 3/4 and 270 degrees < theta < 360 degrees, we can follow these steps:

1. Recall the double angle identity for cosine: cos(2 theta) = 2 cos^2(theta) - 1
2. Substitute the given value of cos(2 theta) = 3/4 into the equation: 3/4 = 2 cos^2(theta) - 1
3. Move the constant term to the other side of the equation: 2 cos^2(theta) = 3/4 + 1
4. Simplify the right side of the equation: 2 cos^2(theta) = 7/4
5. Divide both sides of the equation by 2 to isolate cos^2(theta): cos^2(theta) = 7/8
6. Take the square root of both sides of the equation to solve for cos(theta): cos(theta) = ±√(7/8)

Since we have the restriction 270 degrees < theta < 360 degrees, we know that cosine is negative in this interval. Therefore, cos(theta) = -√(7/8).

Now, to find sin(theta), we can use the trigonometric identity sin^2(theta) + cos^2(theta) = 1. Rearranging the equation, we get:

sin^2(theta) = 1 - cos^2(theta)
sin^2(theta) = 1 - (-√(7/8))^2
sin^2(theta) = 1 - 7/8
sin^2(theta) = 8/8 - 7/8
sin^2(theta) = 1/8

Since we are looking for the value of sin(theta) in the given interval, where sine is positive, we take the positive square root:

sin(theta) = √(1/8)
sin(theta) = √(1)/√(8)
sin(theta) = 1/√(8)
To rationalize the denominator, we multiply both the numerator and denominator by √(8):

sin(theta) = (1/√(8)) * (√(8)/√(8))
sin(theta) = √(8)/8

Therefore, the value of sin(theta) in the given interval is √(8)/8.