[]=theta 1. sin[]=0.5sin[] Subtract 0.5sin[] from both sides.
2. 0.5sin[]=0. Divide both sides by 0.5.
3. Sin[] =0.
[]=0 or pi (radians)
sin(0)=0 and sin(very large number) is approximately equal to that same very large number.
It is not! So the question is irrelevant.
The equation cannot be proved because of the scattered parts.
Any value for which sin(theta) = 0, i.e. theta = N*180, N being an Integer.
The sine and cosine of complementary angles are related through the identity (\sin(90^\circ - \theta) = \cos(\theta)) and (\cos(90^\circ - \theta) = \sin(\theta)). This means that the sine of an angle is equal to the cosine of its complementary angle, and vice versa. Therefore, for any angle (\theta), the values of sine and cosine are essentially swapped when considering complementary angles.
sin(0)=0 and sin(very large number) is approximately equal to that same very large number.
It is 2*sin(theta)*sin(theta) because that is how multiplication is defined!
It is not! So the question is irrelevant.
because sin(2x) = 2sin(x)cos(x)
answer is 2.34 degrees answer is 2.34 degrees
The equation cannot be proved because of the scattered parts.
The half angle formula is: sin theta/2 = ± sqrt (1 - cos theta/2)
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
Any value for which sin(theta) = 0, i.e. theta = N*180, N being an Integer.
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
-0.5736