Cos(360 - X) =
Trig. Identity
Cos(360)Cos(x) + Sin(360)Sin(x) =>
1CosX + 0Sinx =>
CosX + o =>
CosX
Tan^2
-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
You can use the Pythagorean identity to solve this:(sin theta) squared + (cos theta) squared = 1.
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 ).
sin/cos
Zero. Anything minus itself is zero.
cosine (90- theta) = sine (theta)
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)
Cos theta squared
cos(theta) = 0.7902 arcos(0.7902) = theta = 38 degrees you find complimentary angles
108.435 degrees 288.435 degrees (decimal is rounded)
cos2(theta) = 1 so cos(theta) = ±1 cos(theta) = -1 => theta = pi cos(theta) = 1 => theta = 0
Remember that tan = sin/cos. So your expression is sin/cos times cos. That's sin(theta).
You cannot prove it because it is not true! cos(0) = 1 cos(2*pi) = 1 cos(4*pi) = 1 ... cos(2*k*pi) = 1 for all integers k or, if you still work in degrees, cos(0) = 1 cos(360) = 1 cos(720) = 1 ... cos(k*360) = 1 for all integers k
(Sin theta + cos theta)^n= sin n theta + cos n theta
The identity for tan(theta) is sin(theta)/cos(theta).