(/) = theta sin 2(/) = 2sin(/)cos(/)
- cos theta
Tan^2
You can use the Pythagorean identity to solve this:(sin theta) squared + (cos theta) squared = 1.
sin/cos
Let 'theta' = A [as 'A' is easier to type] sec A - 1/(sec A) = 1/(cos A) - cos A = (1 - cos^2 A)/(cos A) = (sin^2 A)/(cos A) = (tan A)*(sin A) Then you can swap back the 'A' with theta
It's 1/2 of sin(2 theta) .
Cos theta squared
4*cos2(theta) = 1 cos2(theta) = 1/4 cos(theta) = sqrt(1/4) = ±1/2 Now cos(theta) = 1/2 => theta = 60 + 360k or theta = 300 + 360k while Now cos(theta) = -1/2 => theta = 120 + 360k or theta = 240 + 360k where k is an integer.
(/) = theta sin 2(/) = 2sin(/)cos(/)
- cos theta
'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
cos2(theta) = 1 so cos(theta) = ±1 cos(theta) = -1 => theta = pi cos(theta) = 1 => theta = 0
because sin(2x) = 2sin(x)cos(x)
Remember that tan = sin/cos. So your expression is sin/cos times cos. That's sin(theta).
Tan^2
cos2(theta) = 1 cos2(theta) + sin2(theta) = 1 so sin2(theta) = 0 cos(2*theta) = cos2(theta) - sin2(theta) = 1 - 0 = 1