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
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.
1 answer
1 answer