Given cosine, theta, equals, minus, start fraction, square root of, 35, end square root, divided by, 6, end fractioncosθ=−
6
35
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.
The Pythagorean identity states that for any angle theta, sin^2(theta) + cos^2(theta) = 1.
Since we are given cos(theta) = -sqrt(35)/6, we can square it to find sin^2(theta).
(sin^2(theta) )+ (cos^2(theta)) = 1
sin^2(theta) + (-sqrt(35)/6)^2 = 1
sin^2(theta) + 35/36 = 1
sin^2(theta) = 1 - 35/36
sin^2(theta) = (36/36) - (35/36)
sin^2(theta) = 1/36
Taking the square root of both sides, we get:
sin(theta) = sqrt(1/36)
sin(theta) = 1/6
Therefore, the exact value of sin(theta) is 1/6.