Given sine, theta, equals, minus, start fraction, square root of, 13, end square root, divided by, 7, end fractionsinθ=−
7
13
and angle thetaθ is in Quadrant IV, 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=
Since we know that sine theta is negative in Quadrant IV, we can determine that cosine theta is positive.
To find the exact value of cosine theta, we can use the Pythagorean identity: sin^2(theta) + cos^2(theta) = 1.
Since sin theta = -7/13, we can substitute this value into the equation and solve for cos theta:
(-7/13)^2 + cos^2(theta) = 1
49/169 + cos^2(theta) = 1
cos^2(theta) = 1 - 49/169
cos^2(theta) = 120/169
Taking the square root of both sides, we get:
cos(theta) = sqrt(120/169)
To simplify the radical, we can factor 120 and 169:
120 = 2^3 * 3 * 5
169 = 13^2
sqrt(120/169) = (sqrt(2^3 * 3 * 5))/13
= (2 * sqrt(30))/13
Therefore, the exact value of cosine theta in simplest form is (2 * sqrt(30))/13.