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)
Cosine squared theta = 1 + Sine squared theta
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!
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)
Yes. (Theta in radians, and then approximately, not exactly.)
-1
The equation that satisfies the condition "what divided by cosine squared theta equals one" is simply the expression itself. If we let ( x ) be the quantity, then the equation can be expressed as ( \frac{x}{\cos^2 \theta} = 1 ). Solving for ( x ) gives ( x = \cos^2 \theta ). Thus, ( \cos^2 \theta ) divided by ( \cos^2 \theta ) equals one.
4*cos2(theta) = 1 cos2(theta) = 1/4 cos(theta) = sqrt(1/4) = ±1/2 Now cos(theta) = 1/2 => theta = 60 + 360k or theta = 300 + 360k while Now cos(theta) = -1/2 => theta = 120 + 360k or theta = 240 + 360k where k is an integer.