Any random individual might have some prayer of answering that question if he knew
what angle theta is. Without that information, the best anyone can do is guess, and
if he's smart, he won't even try to do that much.
Sin theta of 30 degrees is1/2
(/) = theta sin 2(/) = 2sin(/)cos(/)
You can use the Pythagorean identity to solve this:(sin theta) squared + (cos theta) squared = 1.
Since sin(theta) = 1/cosec(theta) the first two terms simply camcel out and you are left with 1 divided by tan(theta), which is cot(theta).
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.
You put the acceleration on the x-axis, and sin theta on the y-axis
inclined planes can be used in the investigation of acceleration. specificaly using m*g*sin(theta)=a (well i think that was the equation) acceleration is equal to mass*gravity*sin(theta) where sin(theta) is equal to opposite(o) over hypotenuse(h) or theta = (1/sin) * o/h
One relationship is: cos(x) = sin(90° - x) if you use degrees. Or in radians: cos(x) = sin(pi/2 - x) Another relationship is the pythagorean identity.
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).