Problem 51 - Precalculus Mathematics for Calculus

Question

Find the values of the trigonometric functions of \(\theta\) from the information given. $$\cos \theta=-\frac{2}{7}, \quad \tan \theta<0$$

Step-by-Step Solution

Answer

\( \sin \theta = \frac{3\sqrt{5}}{7}, \; \cos \theta = -\frac{2}{7}, \; \tan \theta = -\frac{3\sqrt{5}}{2} \)

Step 1: Understand the Given Information

We are given \( \cos \theta = -\frac{2}{7} \) and \( \tan \theta < 0 \). Knowing that \( \cos \theta \) is negative tells us that \( \theta \) is in either the second or third quadrant. The condition \( \tan \theta < 0 \) further narrows \( \theta \)'s location to the second quadrant since tangent is positive in the third quadrant.

Step 2: Use Pythagorean Theorem to Find \( \sin \theta \)

Use the identity \( \sin^2 \theta + \cos^2 \theta = 1 \). With \( \cos \theta = -\frac{2}{7} \), substitute this into the identity: \( \sin^2 \theta + \left(-\frac{2}{7}\right)^2 = 1 \). Simplifying gives \( \sin^2 \theta + \frac{4}{49} = 1 \). Therefore, \( \sin^2 \theta = \frac{45}{49} \). Thus, \( \sin \theta = \sqrt{\frac{45}{49}} = \frac{\sqrt{45}}{7} \). In the second quadrant, sine is positive, so \( \sin \theta = \frac{3\sqrt{5}}{7} \).

Step 3: Calculate \( \tan \theta \)

Use the identity \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). Substituting \( \sin \theta = \frac{3\sqrt{5}}{7} \) and \( \cos \theta = -\frac{2}{7} \) into this equation gives \( \tan \theta = \frac{\frac{3\sqrt{5}}{7}}{-\frac{2}{7}} = -\frac{3\sqrt{5}}{2} \).

Key Concepts

Cosine

When discussing trigonometric functions, the cosine of an angle, often written as \( \cos \theta \), is a fundamental concept. Cosine represents the ratio of the adjacent side to the hypotenuse in a right-angled triangle. In the Cartesian coordinate system, it is associated with the x-coordinate on the unit circle.

In our exercise, it is given that \( \cos \theta = -\frac{2}{7} \). A negative cosine value indicates that the angle \( \theta \) is located in either the second or third quadrant. This is because cosine values are negative in these quadrants.- Second Quadrant: Cosine is negative, sine is positive- Third Quadrant: Cosine is negative, sine is negative

Since further conditions indicate that tangent is negative, it helps zero in on the second quadrant. Remember:

• Cosine measures how far left or right along the x-axis the angle is.
• If \( \cos \theta \) is negative, angle \( \theta \) lies in the quadrants where x is negative.

Sine

Sine, denoted as \( \sin \theta \), is another primary trigonometric function. It defines the ratio of the length of the opposite side to the hypotenuse in a right-angled triangle. Sine corresponds to the y-coordinate on the unit circle.

In the exercise, we need to find \( \sin \theta \) based on the known cosine value. This can be done using the Pythagorean identity: \[ \sin^2 \theta + \cos^2 \theta = 1 \]This identity lets us solve for \( \sin^2 \theta \) when \( \cos \theta \) is known. Substituting \( \cos \theta = -\frac{2}{7} \) gives:- \( \sin^2 \theta + \left(-\frac{2}{7}\right)^2 = 1 \)- Simplifies to \( \sin^2 \theta + \frac{4}{49} = 1 \)- \( \sin^2 \theta = \frac{45}{49} \)- Therefore, \( \sin \theta = \frac{\sqrt{45}}{7} = \frac{3\sqrt{5}}{7} \)

In the second quadrant, sine is positive, so we take the positive root. It’s crucial to consider quadrants to determine the correct sign of sine.

Tangent

Tangent, shown as \( \tan \theta \), is the trigonometric ratio of sine to cosine: \[ \tan \theta = \frac{\sin \theta}{\cos \theta} \]This ratio reflects the slope of the line formed by the hypotenuse relative to the x-axis.

In our scenario, tangent is negative. Given \( \sin \theta = \frac{3\sqrt{5}}{7} \) and \( \cos \theta = -\frac{2}{7} \), we find \( \tan \theta \) as:- \[ \tan \theta = \frac{\frac{3\sqrt{5}}{7}}{-\frac{2}{7}} \]- Simplifying yields \( \tan \theta = -\frac{3\sqrt{5}}{2} \)

Since tangent is negative in the second quadrant (where sine is positive and cosine is negative), this confirms the position of \( \theta \). Remember:

• Tangent is positive when the signs of sine and cosine are the same.
• Tangent is negative when sine and cosine have opposite signs.