cos2(theta) = 1
so cos(theta) = ±1
cos(theta) = -1 => theta = pi
cos(theta) = 1 => theta = 0
If r-squared = theta then r = ±sqrt(theta)
To determine what negative sine squared plus cosine squared is equal to, start with the primary trigonometric identity, which is based on the pythagorean theorem...sin2(theta) + cos2(theta) = 1... and then solve for the question...cos2(theta) = 1 - sin2(theta)2 cos2(theta) = 1 - sin2(theta) + cos2(theta)2 cos2(theta) - 1 = - sin2(theta) + cos2(theta)
Cosine squared theta = 1 + Sine squared theta
You can use the Pythagorean identity to solve this:(sin theta) squared + (cos theta) squared = 1.
It also equals 13 12.
Until an "equals" sign shows up somewhere in the expression, there's nothing to prove.
2 sin^2 theta = 1/4 sin^2 theta = 1/8 sin theta = sqrt(1/8) theta = arcsin(sqrt(1/8))
cos2(theta) = 1 cos2(theta) + sin2(theta) = 1 so sin2(theta) = 0 cos(2*theta) = cos2(theta) - sin2(theta) = 1 - 0 = 1
It is 2*sin(theta)*sin(theta) because that is how multiplication is defined!
Yes. (Theta in radians, and then approximately, not exactly.)
-1
Since there is no equation, there is nothing that can be solved.