Sin theta of 30 degrees is1/2
It's possible
4 sin(theta) = 2 => sin(theta) = 2/4 = 0.5. Therefore theta = 30 + k*360 degrees or 150 + k*360 degrees where k is any integer.
Since theta is in the second quadrant, sin(theta) is positive. sin2(theta) = 1 - cos2(theta) = 0.803 So sin(theta) = +sqrt(0.803) = 0.896.
I shall use x instead of theta since I cannot be bothered to paste it at each step.sin(x) + 2*cos2(x) = sin(x) + 2*[1 - sin2(x)] = sin(x) + 2 - 2sin2(x) which cannot be simplified further.I shall use x instead of theta since I cannot be bothered to paste it at each step.sin(x) + 2*cos2(x) = sin(x) + 2*[1 - sin2(x)] = sin(x) + 2 - 2sin2(x) which cannot be simplified further.I shall use x instead of theta since I cannot be bothered to paste it at each step.sin(x) + 2*cos2(x) = sin(x) + 2*[1 - sin2(x)] = sin(x) + 2 - 2sin2(x) which cannot be simplified further.I shall use x instead of theta since I cannot be bothered to paste it at each step.sin(x) + 2*cos2(x) = sin(x) + 2*[1 - sin2(x)] = sin(x) + 2 - 2sin2(x) which cannot be simplified further.
It's 1/2 of sin(2 theta) .
2 sin^2 theta = 1/4 sin^2 theta = 1/8 sin theta = sqrt(1/8) theta = arcsin(sqrt(1/8))
It is 2*sin(theta)*sin(theta) because that is how multiplication is defined!
Sin theta of 30 degrees is1/2
It's possible
4 sin(theta) = 2 => sin(theta) = 2/4 = 0.5. Therefore theta = 30 + k*360 degrees or 150 + k*360 degrees where k is any integer.
because sin(2x) = 2sin(x)cos(x)
[]=theta 1. sin[]=0.5sin[] Subtract 0.5sin[] from both sides.2. 0.5sin[]=0. Divide both sides by 0.5.3. Sin[] =0.[]=0 or pi (radians)
If sin (theta) is 3/5, then sin2 (theta) is (3/5)2, or 9/25.
Since theta is in the second quadrant, sin(theta) is positive. sin2(theta) = 1 - cos2(theta) = 0.803 So sin(theta) = +sqrt(0.803) = 0.896.
2 sin (Θ) + 1 = 0sin (Θ) = -1/2Θ = 210°Θ = 330°
'csc' = 1/sin'tan' = sin/cosSo it must follow that(cos) (csc) / (tan) = (cos) (1/sin)/(sin/cos) = (cos) (1/sin) (cos/sin) = (cos/sin)2