The equation cannot be proved because of the scattered parts.
Until an "equals" sign shows up somewhere in the expression, there's nothing to prove.
Yes. (Theta in radians, and then approximately, not exactly.)
It is not! So the question is irrelevant.
The expression (\cos^2(90^\circ - \theta)) can be simplified using the co-function identity, which states that (\cos(90^\circ - \theta) = \sin(\theta)). Therefore, (\cos^2(90^\circ - \theta) = \sin^2(\theta)). This means that (\cos^2(90^\circ - \theta)) is equal to the square of the sine of (\theta).
There is none because it is not true.
Until an "equals" sign shows up somewhere in the expression, there's nothing to prove.
Yes. (Theta in radians, and then approximately, not exactly.)
Cosine squared theta = 1 + Sine squared theta
cos2(theta) = 1 so cos(theta) = ±1 cos(theta) = -1 => theta = pi cos(theta) = 1 => theta = 0
It is 2*sin(theta)*sin(theta) because that is how multiplication is defined!
-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
4Sin(theta) = 2 Sin(Theta) = 2/4 = 1/2 - 0.5 Theta = Sin^(-1) [0.5] Theta = 30 degrees.
answer is 2.34 degrees answer is 2.34 degrees
It is not! So the question is irrelevant.
assuming that you mean what is theta if sin 4 theta = 0 then then theta=0, 0.25pi, 0.5pi, 0.75pi... if not then without additional information the best answer you can get is sin4theta=sin4theta
tan (phi)= (V* sin (theta) + Ia*Xs)/(V*cos (theta) +Ia*ra) theta is power factor angle torque angle= phi-theta
There is none because it is not true.