Question 753010
The identity you refer to is 
(1) sin^2(theta) + cos^2(theta) = 1, independent of the quadrant in which theta lies. Using the value of sin(theta) we get
(2) (60/61)^2 + cos^2(theta) = 1 or
(3) cos^2(theta) = 1 - (60/61)^2 or
(4) cos^2(theta) = (61^2 - 60^2)/(61)^2.
The numerator of the right side of (4) is the difference of two perfect squares, therefore can be factored into the product of the sum and difference or
(5) cos^2(theta) = [(61-60)*(61+60)]/(61)^2 or
(6) cos^2(theta) = (1*121)/(61)^2 or
(7) cos^2(theta) = (11)^2/(61)^2 or taking the square root of both sides we get
(8) cos(theta) = +/- 11/61.
Since theta is in the second quadrant we select the negative value of the cosine and we get
(9) cos(theta) = -11/61
Answer: the cos(theta) = -11/61
Look Mom, "no calculator".
By the way we can check our answer using (1).
Is ((60/61)^2 + (-11/61)^2 = 1)?
Is (3600/3721 + 121/3721 = 1)?
Is (3721/3721 = 1)?
Is (1 = 1)? Yes