4Sin(x)Cos(x) =
2(2Sin(x)Cos(x)) =
2Sin(2x) ( A Trig. identity.
4Sin(theta) = 2 Sin(Theta) = 2/4 = 1/2 - 0.5 Theta = Sin^(-1) [0.5] Theta = 30 degrees.
-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 ).
Cos(360 - X) = Trig. Identity Cos(360)Cos(x) + Sin(360)Sin(x) => 1CosX + 0Sinx => CosX + o => CosX
Cos theta squared
4Sin(theta) = 2 Sin(Theta) = 2/4 = 1/2 - 0.5 Theta = Sin^(-1) [0.5] Theta = 30 degrees.
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).
-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).
It is cotangent(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
Zero. Anything minus itself is zero.
The question contains an expression but not an equation. An expression cannot be solved.
You can use the Pythagorean identity to solve this:(sin theta) squared + (cos theta) squared = 1.