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.
It is 1.
If sine theta is 0.28, then theta is 16.26 degrees. Cosine 2 theta, then, is 0.8432
It also equals 13 12.
Until an "equals" sign shows up somewhere in the expression, there's nothing to prove.
-1
Cosine squared theta = 1 + Sine squared theta
If r-squared = theta then r = ±sqrt(theta)
It is 1.
If sine theta is 0.28, then theta is 16.26 degrees. Cosine 2 theta, then, is 0.8432
-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
You can use the Pythagorean identity to solve this:(sin theta) squared + (cos theta) squared = 1.
It also equals 13 12.
cos2(theta) = 1 so cos(theta) = ±1 cos(theta) = -1 => theta = pi cos(theta) = 1 => theta = 0
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
If X and Y are sides of a right triangle, R is the hypoteneuse, and theta is the angle at the X-R vertex, then sin(theta) is Y / R and cosine(theta) is X / R. It follows, then, that X is R cosine(theta) and Y is R sin(theta)