0
i am not sure about how to solve the values for a theta graph but i know it looks like a spiral and you can plot these points if you draw a square and halve it and in each of the centres of the square is another point of theta.
then you need the gradient of the line so if you have f(x)=θ
then what is f(x)' which would be the gradient by the way f(x)' is another way of saying differentiated or dy/dθ
What else can I help you with?
How do you solve r squared equals theta?
If r-squared = theta then r = ±sqrt(theta)
Can you solve sin 2 theta equals 0?
If sin2(theta) = 0, then theta is N pi, N being any integer
How do you solve sin2 theta if sin theta equals 3 over 5 and theta is in the first quadrant?
If sin (theta) is 3/5, then sin2 (theta) is (3/5)2, or 9/25.
How do you solve theta if cos squared theta equals 1 and 0 is less than or equal to theta which is less than 2pi?
cos2(theta) = 1 so cos(theta) = ±1 cos(theta) = -1 => theta = pi cos(theta) = 1 => theta = 0
How would you solve and show work for cos2 theta if cos squared theta equals 1 and theta is in the 4th quadrant?
cos2(theta) = 1 cos2(theta) + sin2(theta) = 1 so sin2(theta) = 0 cos(2*theta) = cos2(theta) - sin2(theta) = 1 - 0 = 1
How do you solve cos theta subtract cos squared theta divide 1 plus cos theta?
The question contains an expression but not an equation. An expression cannot be solved.
How do you solve 2 sin squared theta equals one fourth?
2 sin^2 theta = 1/4 sin^2 theta = 1/8 sin theta = sqrt(1/8) theta = arcsin(sqrt(1/8))
How do you solve 1- 2cos squared theta without a calculator?
-1
How do you solve 4 cosine squared theta equals 1?
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.
How would you solve d theta divided by 4.28?
d theta divided by 4.28 is not an equation or inequality: it is an expression. An expression cannot be solved.
How do you solve for x and y in an angle?
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)
How do you solve sin2 theta equals 0.75?
To solve the equation (\sin^2 \theta = 0.75), first take the square root of both sides to get (\sin \theta = \pm \sqrt{0.75} = \pm \frac{\sqrt{3}}{2}). Then, find the angles (\theta) for which (\sin \theta = \frac{\sqrt{3}}{2}) and (\sin \theta = -\frac{\sqrt{3}}{2}). The solutions are (\theta = \frac{\pi}{3} + 2k\pi) and (\theta = \frac{2\pi}{3} + 2k\pi) for the positive case, and (\theta = \frac{7\pi}{6} + 2k\pi) and (\theta = \frac{4\pi}{3} + 2k\pi) for the negative case, where (k) is any integer.
