To find the exact value of sine theta, we can use the Pythagorean identity:
sin^2(theta) = 1 - cos^2(theta)
sin^2(theta) = 1 - (-5/21)^2
sin^2(theta) = 1 - 25/441
sin^2(theta) = (441-25)/441
sin^2(theta) = 416/441
Taking the square root of both sides, we get:
sin(theta) = ±√(416/441)
Since theta is in Quadrant II, sin(theta) is positive.
sin(theta) = √(416/441)
The square root of 416 can be simplified:
416 = 4 * 104 = 4 * (4 * 26) = 4 * (2 * 2 * 13) = 4 * 2^2 * 13 = 4 * 2^2 * √13
So, sin(theta) = (√(4 * 2^2 * √13))/√441
Simplifying further:
sin(theta) = (2 * 2 * √13)/21
sin(theta) = (4√13)/21
Therefore, the exact value of sin(theta) is (4√13)/21.
Given cosine, theta, equals, minus, start fraction, square root of, 21, end square root, divided by, 5, end fractioncosθ=−
5
21
and angle thetaθ is in Quadrant II, what is the exact value of sine, thetasinθ in simplest form? Simplify all radicals if needed.
1 answer