The derivative of (sin (theta))^.5 is (cos(theta))/(2sin(theta))
For such simplifications, it is usually convenient to convert any trigonometric function that is not sine or cosine, into sine or cosine. In this case, you have: sin theta / sec theta = sin theta / (1/cos theta) = sin theta cos theta.
sin (theta) = [13* sin (32o)]/8 = 13*0.529919264/8 = 0.861118804 [theta] = sin-1 (0.861118804) [theta] = 59.44o
The identity for tan(theta) is sin(theta)/cos(theta).
It is not! So the question is irrelevant.
Argument Hopes this help!!
It's 1/2 of sin(2 theta) .
The derivative of (sin (theta))^.5 is (cos(theta))/(2sin(theta))
It is 2*sin(theta)*sin(theta) because that is how multiplication is defined!
For such simplifications, it is usually convenient to convert any trigonometric function that is not sine or cosine, into sine or cosine. In this case, you have: sin theta / sec theta = sin theta / (1/cos theta) = sin theta cos theta.
sin (theta) = [13* sin (32o)]/8 = 13*0.529919264/8 = 0.861118804 [theta] = sin-1 (0.861118804) [theta] = 59.44o
(Sin theta + cos theta)^n= sin n theta + cos n theta
The identity for tan(theta) is sin(theta)/cos(theta).
2 sin^2 theta = 1/4 sin^2 theta = 1/8 sin theta = sqrt(1/8) theta = arcsin(sqrt(1/8))
Remember that tan = sin/cos. So your expression is sin/cos times cos. That's sin(theta).
Sin theta of 30 degrees is1/2
(/) = theta sin 2(/) = 2sin(/)cos(/)