Cos theta squared
if alpha is the solid angle subtended by a cone and theta is the 2d projection angle of this solid angle...Then----(both alpha and theta are in radians then) cos(theta) = 1 - (alpha/2*pi)
Angular velocity is equal to the change in theta / change in time theta equals the arc length/ radius
sin integral is -cos This is so because the derivative of cos x = -sin x
If the angle is measured in degrees, then cos(33) = 0.8387, approx.
7
Remember that tan = sin/cos. So your expression is sin/cos times cos. That's sin(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
Tan^2
cos(t) - cos(t)*sin2(t) = cos(t)*[1 - sin2(t)] But [1 - sin2(t)] = cos2(t) So, the expression = cos(t)*cos2(t) = cos3(t)
To integrate ( \cos^2 \theta \sin \theta ), you can use a substitution method. Let ( u = \cos \theta ), then ( du = -\sin \theta , d\theta ). The integral becomes ( -\int u^2 , du ), which evaluates to ( -\frac{u^3}{3} + C ). Substituting back, the final result is ( -\frac{\cos^3 \theta}{3} + C ).
cos2(theta) = 1 so cos(theta) = ±1 cos(theta) = -1 => theta = pi cos(theta) = 1 => theta = 0
A - WORKWork = F.s cos (theta)
-Sin^(2)(Theta) + Cos^(2)Theta => Cos^(2)Theta - Sin^(2)Theta Factor (Cos(Theta) - Sin(Theta))( Cos(Theta) + Sin(Theta)) #Is the Pythagorean factors . Or -Sin^(2)Theta = -(1 - Cos^(2)Theta) = Cos(2)Theta - 1 Substitute Cos^(2)Thetqa - 1 + Cos^(2) Theta = 2Cos^(2)Theta - 1
(Sin theta + cos theta)^n= sin n theta + cos n theta
The identity for tan(theta) is sin(theta)/cos(theta).
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 is cotangent(theta).