If theta is a rotation angle that terminates in quadrant ll and sin of theta is 5/13, find cos of theta and tan of theta. Use the first Pythagorean identity to solve for cos of theta and use quotient identity to solve for tan of theta.
To find the value of cosine (cos) of theta and tangent (tan) of theta, we can use the given information that sine (sin) of theta is 5/13 and the fact that theta terminates in quadrant II.
First, let's use the first Pythagorean identity:
sin^2(theta) + cos^2(theta) = 1
Plugging in the value of sin(theta) as 5/13, we get:
(5/13)^2 + cos^2(theta) = 1
25/169 + cos^2(theta) = 1
Next, rearrange the equation to solve for cos^2(theta):
cos^2(theta) = 1 - 25/169
cos^2(theta) = (169 - 25)/169
cos^2(theta) = 144/169
Taking the square root of both sides, we find:
cos(theta) = Âħsqrt(144/169)
Since theta terminates in quadrant II, cos(theta) is negative. Therefore:
cos(theta) = -sqrt(144/169) = -12/13
Now, let's use the quotient identity for tangent (tan):
tan(theta) = sin(theta) / cos(theta)
Plugging in the values we have:
tan(theta) = (5/13) / (-12/13)
Dividing fractions, we get:
tan(theta) = (5/13) * (-13/12)
tan(theta) = -5/12
Therefore, the value of cos(theta) is -12/13 and the value of tan(theta) is -5/12.