Problem 43 - Algebra and Trigonometry

Question

43–50 Find the values of the trigonometric functions of \(\theta\) from the information given. \(\sin \theta=\frac{3}{5}, \quad \theta\) in quadrant II

Step-by-Step Solution

Answer

\( \cos \theta = -\frac{4}{5}, \tan \theta = -\frac{3}{4}, \csc \theta = \frac{5}{3}, \sec \theta = -\frac{5}{4}, \cot \theta = -\frac{4}{3} \).

Step 1: Understand the given information

We know that \( \sin \theta = \frac{3}{5} \) and \( \theta \) is in the second quadrant. In the second quadrant, sine is positive, and cosine and tangent are negative.

Step 2: Use the Pythagorean identity to find cosine

The Pythagorean identity is \( \sin^2 \theta + \cos^2 \theta = 1 \). Substitute \( \sin \theta \) into this identity: \( \left(\frac{3}{5}\right)^2 + \cos^2 \theta = 1 \). Calculate \( \left(\frac{3}{5}\right)^2 = \frac{9}{25} \). Thus, \( \frac{9}{25} + \cos^2 \theta = 1 \). Solve for \( \cos^2 \theta \): \( \cos^2 \theta = 1 - \frac{9}{25} = \frac{16}{25} \). Therefore, \( \cos \theta = -\frac{4}{5} \) (negative in the second quadrant).

Step 3: Find tangent using sine and cosine

Tangent can be found using the formula \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). Substitute \( \sin \theta = \frac{3}{5} \) and \( \cos \theta = -\frac{4}{5} \): \( \tan \theta = \frac{\frac{3}{5}}{-\frac{4}{5}} = -\frac{3}{4} \).

Step 4: Determine other trigonometric functions

Now we find the remaining functions. \( \csc \theta = \frac{1}{\sin \theta} = \frac{5}{3} \). \( \sec \theta = \frac{1}{\cos \theta} = -\frac{5}{4} \). \( \cot \theta = \frac{1}{\tan \theta} = -\frac{4}{3} \).

Key Concepts

Pythagorean Identity

The Pythagorean identity is a fundamental relationship in trigonometry. This identity connects the sine and cosine of an angle. It is given by the equation:

• \( \sin^2 \theta + \cos^2 \theta = 1 \)

This equation shows that the square of the sine of an angle added to the square of the cosine of the same angle will always equal one. This is true for any angle \( \theta \).

In the given exercise, we start with \( \sin \theta = \frac{3}{5} \). To use the Pythagorean identity, you substitute this value into the identity:

• \( \left(\frac{3}{5}\right)^2 + \cos^2 \theta = 1 \)
• Calculate \( \left(\frac{3}{5}\right)^2 = \frac{9}{25} \)
• Then solve: \( \cos^2 \theta = 1 - \frac{9}{25} = \frac{16}{25} \)

Taking the square root gives \( \cos \theta = \pm \frac{4}{5} \). But because \( \theta \) is in quadrant II where cosine is negative, \( \cos \theta = -\frac{4}{5} \). This illustrates how to determine cosine from sine using the Pythagorean identity.

Tangent Function

Tangent is one of the basic trigonometric functions, and it can be defined in terms of sine and cosine. The formula for tangent is:

• \( \tan \theta = \frac{\sin \theta}{\cos \theta} \)

This formula shows how tangent is derived from sine and cosine.

From the problem, we have \( \sin \theta = \frac{3}{5} \) and \( \cos \theta = -\frac{4}{5} \) for \( \theta \) in quadrant II. Substitute these into the tangent formula:

• \( \tan \theta = \frac{\frac{3}{5}}{-\frac{4}{5}} = -\frac{3}{4} \)

In quadrant II, the tangent of an angle is negative because sine is positive and cosine is negative. Understanding this relationship helps compute tangent accurately.

Sine and Cosine Relationships

Sine and cosine are closely related. These functions shift in value depending on which quadrant of the unit circle the angle falls.

Here are some fundamental aspects:

• In quadrant I, both \( \sin \theta \) and \( \cos \theta \) are positive.
• In quadrant II, \( \sin \theta \) is positive, while \( \cos \theta \) is negative.
• In quadrant III, both are negative.
• In quadrant IV, \( \cos \theta \) is positive, while \( \sin \theta \) is negative.

This concept is important because it helps determine the signs of sine and cosine in different quadrants. Knowing whether a trigonometric function is positive or negative based on the angle's quadrant allows you to accurately find other trigonometric values.

In the exercise, \( \theta \) is in quadrant II so \( \sin \theta = \frac{3}{5} \) is positive, but \( \cos \theta = - \frac{4}{5} \) is negative. This relationship helps ensure accurate calculations and results.