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)
No. Cos squared x is not the same as cos x squared. Cos squared x means cos (x) times cos (x) Cos x squared means cos (x squared)
It is cotangent(theta).
cosec(q)*cot(q)*cos(q) = 1/sin(q)*cot(q)*cos(q) = cot2(q)
zero
It is cos2x that is, "cos-squared x".
Tan^2
Cos theta squared
Zero. Anything minus itself is zero.
You can use the Pythagorean identity to solve this:(sin theta) squared + (cos theta) squared = 1.
cosine (90- theta) = sine (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
1
cos2(theta) = 1 so cos(theta) = ±1 cos(theta) = -1 => theta = pi cos(theta) = 1 => theta = 0
The question contains an expression but not an equation. An expression cannot be solved.
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 ).
Until an "equals" sign shows up somewhere in the expression, there's nothing to prove.
COS squared Theta + SIN squared Theta = 1; where Theta is the angles measurement in degrees.