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.
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.