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.
Answer
Attempt 1 out of 2
cosine, theta, equalscosθ=
In Quadrant II, the cosine function is negative. So we have:
cosθ = -√(1 - sin^2θ)
= -√(1 - (6/35)^2)
= -√(1 - 36/1225)
= -√(1225/1225 - 36/1225)
= -√(1189/1225)
= -√1189/√1225
= -√1189/35
Therefore, the exact value of cosine(theta) is -√1189/35.