It is not! So the question is irrelevant.
The sine of a negative angle is equal to the negative sine of the corresponding positive angle. This is expressed mathematically as (\sin(-\theta) = -\sin(\theta)). This property reflects the odd function nature of the sine function, which means that its graph is symmetric about the origin. Therefore, if you know the sine of a positive angle, you can easily determine the sine of its negative counterpart.
[]=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.
The equation cannot be proved because of the scattered parts.
The only real solution is theta = 0For theta < 0 square root of 3 theta is not defined.For theta > 0, sin theta increases slower than 3*theta and so the sum is always negative.The only real solution is theta = 0For theta < 0 square root of 3 theta is not defined.For theta > 0, sin theta increases slower than 3*theta and so the sum is always negative.The only real solution is theta = 0For theta < 0 square root of 3 theta is not defined.For theta > 0, sin theta increases slower than 3*theta and so the sum is always negative.The only real solution is theta = 0For theta < 0 square root of 3 theta is not defined.For theta > 0, sin theta increases slower than 3*theta and so the sum is always negative.
The sine of a negative angle is equal to the negative sine of the corresponding positive angle. This is expressed mathematically as (\sin(-\theta) = -\sin(\theta)). This property reflects the odd function nature of the sine function, which means that its graph is symmetric about the origin. Therefore, if you know the sine of a positive angle, you can easily determine the sine of its negative counterpart.
you have to do the arcsin which is sin-1 on your calculator. i have not met anyone in my life who can do sin or arcsin in their head. not even my college teachers. your theta is equal to 20degrees
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
because sin(2x) = 2sin(x)cos(x)
Since theta is in the second quadrant, sin(theta) is positive. sin2(theta) = 1 - cos2(theta) = 0.803 So sin(theta) = +sqrt(0.803) = 0.896.
[]=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)
answer is 2.34 degrees answer is 2.34 degrees
sin(0)=0 and sin(very large number) is approximately equal to that same very large number.
The equation cannot be proved because of the scattered parts.
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
4Sin(theta) = 2 Sin(Theta) = 2/4 = 1/2 - 0.5 Theta = Sin^(-1) [0.5] Theta = 30 degrees.