Final answer:
To determine the sine (sin x) and cosine (cos x) when cot x = -3/2 and sec x is negative, first use the Pythagorean identity for cotangent to find sin x and then use the Pythagorean identity for sine to solve for cos x. Ultimately, sin x = √(4/13) and cos x = -3/√13.
Step-by-step explanation:
The student's question is about finding the sine and cosine of an angle where the cotangent of x is given as -3/2 and the secant of x is less than zero. This suggests that the angle x lies in either the second or third quadrant where the cosine of x is negative (since sec x = 1/cos x, and sec x < 0 indicates cos x < 0). Using the Pythagorean identity for cotangent, which is cot2(x) + 1 = csc2(x), we can find the sine of x. With cot(x) = -3/2, this becomes (3/2)2 + 1 = csc2(x), or 9/4 + 1 = 13/4 = csc2(x). Taking the square root gives us csc(x) = ± √(13/4), but since csc x = 1/sin x, the sine will have the opposite sign of the secant (which is negative), thus sin x must be positive. Therefore, sin x = √(4/13). To find the cosine of x, we can use the relation sin2(x) + cos2(x) = 1. We already determined that sin x = √(4/13), so plugging that in we get (√(4/13))2 + cos2(x) = 1, which simplifies to 4/13 + cos2(x) = 1. Solving for cos2(x) gives us cos2(x) = 9/13, and since the cosine is negative in the second and third quadrants, cos x = -√(9/13) = -3/√13.