Question

Given that $\theta$ is in quadrant 2 and $\tan(\theta) = -7/2$, give an exact answer for the following:

A) $\sin(\theta) = -\sqrt{53}/2, \cos(\theta) = -1/2$
B) $\sin(\theta) = \sqrt{53}/2, \cos(\theta) = -1/2$
C) $\sin(\theta) = -\sqrt{53}/2, \cos(\theta) = 1/2$
D) $\sin(\theta) = \sqrt{53}/2, \cos(\theta) = 1/2$

Answer 1

The correct option is B) sin(θ) = √53/2, cos(θ) = -1/2, found by applying the properties of trigonometric functions in quadrant 2 and using the Pythagorean identity to solve for sine and cosine with a given tangent value.

Step-by-step explanation:

The question involves determining the sine and cosine of an angle θ in quadrant 2 with a given tangent value of -7/2. To find the sine and cosine, we first consider the properties of trigonometric functions in different quadrants and use the Pythagorean identity:

• In quadrant 2, sine is positive and cosine is negative.
• The Pythagorean identity is sin^2(θ) + cos^2(θ) = 1.

We know that tan(θ) = sin(θ)/cos(θ), so given tan(θ) = -7/2, we can assume sin(θ) = 7k and cos(θ) = -2k for some positive k. This gives us (7k)^2 + (-2k)^2 = 1, which simplifies to 53k^2 = 1, and k = 1/√53.

We then get sin(θ) = 7/√53 = √53/53 &=radic;53/2 and cos(θ) = -2/√53 = -√53/53 = -1/2 since we know that sine is positive and cosine is negative in quadrant 2. Thus, the correct option is B) sin(θ) = √53/2, cos(θ) = -1/2.