Given sine, theta, equals, start fraction, square root of, 35, end square root, divided by, 6, end fractionsinθ=
6
35
and angle thetaθ is in Quadrant II, what is the exact value of cosine, thetacosθ in simplest form? Simplify all radicals if needed.
In Quadrant II, sine and cosine are both positive.
We know that sine(theta) = 6/35.
To find cosine(theta), we can use the Pythagorean Identity:
sin^2(theta) + cos^2(theta) = 1.
Plugging in the value for sine(theta), we get:
(6/35)^2 + cos^2(theta) = 1.
36/1225 + cos^2(theta) = 1.
cos^2(theta) = 1 - 36/1225.
cos^2(theta) = 1225/1225 - 36/1225.
cos^2(theta) = (1225 - 36) / 1225.
cos^2(theta) = 1189 / 1225.
Taking the square root of both sides, we find:
cos(theta) = sqrt(1189) / sqrt(1225).
cos(theta) = sqrt(1189) / 35.
Therefore, the exact value of cosine(theta) is sqrt(1189) / 35, in simplest form.