Given sine, theta, equals, minus, start fraction, square root of, 35, end square root, divided by, 6, end fractionsinθ=−
6
35
and angle thetaθ is in Quadrant III, what is the exact value of cosine, thetacosθ in simplest form? Simplify all radicals if needed.
Since sine is negative in Quadrant III, we can use the Pythagorean identity to find the cosine of theta.
The Pythagorean identity states that for any angle theta, sin^2(theta) + cos^2(theta) = 1.
In this case, we know that sin(theta) = -6/35.
Let's plug this into the Pythagorean identity and solve for cos(theta):
(-6/35)^2 + cos^2(theta) = 1
36/1225 + cos^2(theta) = 1
cos^2(theta) = 1 - 36/1225
cos^2(theta) = (1225 - 36)/1225
cos^2(theta) = 1189/1225
Now we need to find the square root of 1189/1225. Simplifying the radical:
cos(theta) = ±√(1189/1225)
cos(theta) = ±√(1189)/√(1225)
cos(theta) = ±√(1189)/35
So the exact value of cosine(theta) is ±√(1189)/35.
However, since theta is in Quadrant III (where cosine is negative), the simplified answer is:
cos(theta) = -√(1189)/35