It is 2*sin(theta)*sin(theta) because that is how multiplication is defined!
Your question is insufficiently precise, but I'll try to answer anyway. "Sine squared theta" usually means "the value of the sine of theta, quantity squared". "Sine theta squared" usually means "the value of the sine of the quantity theta*theta". The two are not at all the same.
If r-squared = theta then r = ±sqrt(theta)
If there is a plus in between, that would be equal to 1, as a result of the Pythagorean Theorem. Otherwise, you can convert this into other forms with some of the trigonometric identities for multiplication, but you won't really get it into a simpler form.
cos2(theta) = 1 cos2(theta) + sin2(theta) = 1 so sin2(theta) = 0 cos(2*theta) = cos2(theta) - sin2(theta) = 1 - 0 = 1
2 sin^2 theta = 1/4 sin^2 theta = 1/8 sin theta = sqrt(1/8) theta = arcsin(sqrt(1/8))
Cosine squared theta = 1 + Sine squared theta
because sin(2x) = 2sin(x)cos(x)
The derivative of (sin (theta))^.5 is (cos(theta))/(2sin(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
(/) = theta sin 2(/) = 2sin(/)cos(/)
R^2sin(theta)d(theta)d(phi)(r-hat)
4Sin(x)Cos(x) = 2(2Sin(x)Cos(x)) = 2Sin(2x) ( A Trig. identity.
Tan^2
Your question is insufficiently precise, but I'll try to answer anyway. "Sine squared theta" usually means "the value of the sine of theta, quantity squared". "Sine theta squared" usually means "the value of the sine of the quantity theta*theta". The two are not at all the same.
If r-squared = theta then r = ±sqrt(theta)
It is 1.
Until an "equals" sign shows up somewhere in the expression, there's nothing to prove.