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How do you finish the equation sin 2 theta?


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Answered 2010-07-24 01:46:07

(/) = theta

sin 2(/) = 2sin(/)cos(/)

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It's 1/2 of sin(2 theta) .


2 sin^2 theta = 1/4 sin^2 theta = 1/8 sin theta = sqrt(1/8) theta = arcsin(sqrt(1/8))


It is 2*sin(theta)*sin(theta) because that is how multiplication is defined!


Sin theta of 30 degrees is1/2


2 sin (Θ) + 1 = 0sin (Θ) = -1/2Θ = 210°Θ = 330°


4 sin(theta) = 2 => sin(theta) = 2/4 = 0.5. Therefore theta = 30 + k*360 degrees or 150 + k*360 degrees where k is any integer.




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)


'csc' = 1/sin'tan' = sin/cosSo it must follow that(cos) (csc) / (tan) = (cos) (1/sin)/(sin/cos) = (cos) (1/sin) (cos/sin) = (cos/sin)2


It is a mathematical expression.


2 sin(x) + 1 = 0 2 sin(x) = -1 sin(x) = -1/2 x = 210° and 330°


1 cot(theta)=cos(theta)/sin(theta) cos(45 degrees)=sqrt(2)/2 AND sin(45 degrees)=sqrt(2)/2 cot(45 deg)=cos(45 deg)/sin(deg)=(sqrt(2)/2)/(sqrt(2)/2)=1



Let 'theta' = A [as 'A' is easier to type] sec A - 1/(sec A) = 1/(cos A) - cos A = (1 - cos^2 A)/(cos A) = (sin^2 A)/(cos A) = (tan A)*(sin A) Then you can swap back the 'A' with theta


The half angle formula is: sin theta/2 = ± sqrt (1 - cos theta/2)


One way would be as follows: Let b represent the length of the base, l the length of each of the two sides, and theta the angle between the base and the two sides of length l. Now drop a perpendicular line from each vertex at the top of the trapezoid to the base. This yields two right triangles and a rectangle in the middle. The height of each right triangle (as well as the height of the rectangle) equals l*sin(theta) [because sin(theta)=opposite/hypotenuse] and the length of the base of each right triangle is l*cos(theta). The base of the rectangle is b minus the lengths of the two right triangles. Area of the trapezoid=2*area of each right triangle+area of the rectangle=2*(1/2)*(l*sin(theta)*l*cos(theta))+(b-2*l*cos(theta))(l*sin(theta))=)*(l*sin(theta)*l*cos(theta))+(b-2*l*cos(theta))(l*sin(theta))=b*l*sin(theta)-l2*sin(theta)*cos(theta)


If sin2(theta) = 0, then theta is N pi, N being any integer


The only real solution is theta = 0For theta 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 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 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 For theta > 0, sin theta increases slower than 3*theta and so the sum is always negative.


It is a simple trigonometric equation. However, without information on whether the angles are measures in degrees or radians, and with no domain for theta, the equation cannot be solved.


L=1/2m(r'^2+r^2*Sin[theta]^2*phi'^2+r^2*theta'^2)-mgr Sin[theta]. Where theta is the polar angle from the z axis, and phi is the azimuthal. I assumed a uniform grav field down.


Use the equation square root of (gravity times distance)/(2 sin theta*cos theta) when the height difference between the initial and final is negligible, meaning the same. If different heights, use the same without the 2 on the bottom. Use the equation square root of (gravity times distance)/(2 sin theta*cos theta) when the height difference between the initial and final is negligible, meaning the same. If different heights, use the same without the 2 on the bottom.


96 degrees Let theta represent the measure of the angle we are trying to find and theta' represent the measure of its supplement. From the problem, we know: theta=theta'+12 Because supplementary angles sum to 180 degrees, we also know: theta+theta'=180 Substituting the value from theta in the first equation into the second, we get: (theta'+12)+theta'=180 2*theta'+12=180 2*theta'=180-12=168 theta'=168/2=84 Substituting this value for theta' back into the first equation, we get: theta+84=180 theta=180-84=96


I shall use x instead of theta since I cannot be bothered to paste it at each step.sin(x) + 2*cos2(x) = sin(x) + 2*[1 - sin2(x)] = sin(x) + 2 - 2sin2(x) which cannot be simplified further.I shall use x instead of theta since I cannot be bothered to paste it at each step.sin(x) + 2*cos2(x) = sin(x) + 2*[1 - sin2(x)] = sin(x) + 2 - 2sin2(x) which cannot be simplified further.I shall use x instead of theta since I cannot be bothered to paste it at each step.sin(x) + 2*cos2(x) = sin(x) + 2*[1 - sin2(x)] = sin(x) + 2 - 2sin2(x) which cannot be simplified further.I shall use x instead of theta since I cannot be bothered to paste it at each step.sin(x) + 2*cos2(x) = sin(x) + 2*[1 - sin2(x)] = sin(x) + 2 - 2sin2(x) which cannot be simplified further.



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