Math and Arithmetic
Algebra
Trigonometry

# How do you finish the equation sin 2 theta?

###### Wiki User

(/) = theta

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

๐
0
๐คจ
0
๐ฎ
0
๐
0

## Related Questions

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!

2 sin (&Icirc;&tilde;) + 1 = 0sin (&Icirc;&tilde;) = -1/2&Icirc;&tilde; = 210&Acirc;&deg;&Icirc;&tilde; = 330&Acirc;&deg;

4 sin(theta) = 2 =&gt; 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

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 = &plusmn; 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.

###### Math and ArithmeticAlgebraTrigonometryGeometryPhysics

Copyright ยฉ 2021 Multiply Media, LLC. All Rights Reserved. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply.