Problem 65 - Precalculus Mathematics for Calculus

Question

Find the values of the trigonometric functions of \(t\) from the given information. \(\sin t=\frac{3}{5}, \quad\) terminal point of \(t\) is in Quadrant II

Step-by-Step Solution

Answer

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

Step 1: Identify Known Information

We know that \( \sin t = \frac{3}{5} \) and the terminal point of \( t \) is in the second quadrant.

Step 2: Determine \( \cos t \) Using the Pythagorean Identity

In any right triangle, \( \sin^2 t + \cos^2 t = 1 \). We have \( \sin t = \frac{3}{5} \) so \( \sin^2 t = \left(\frac{3}{5}\right)^2 = \frac{9}{25} \). Thus, \( \cos^2 t = 1 - \sin^2 t = 1 - \frac{9}{25} = \frac{16}{25} \). Since we are in Quadrant II, where cosine is negative, \( \cos t = -\frac{4}{5} \).

Step 3: Find \( \tan t \) Using \( \tan t = \frac{\sin t}{\cos t} \)

We have \( \sin t = \frac{3}{5} \) and \( \cos t = -\frac{4}{5} \). Thus, \( \tan t = \frac{\sin t}{\cos t} = \frac{\frac{3}{5}}{-\frac{4}{5}} = -\frac{3}{4} \).

Step 4: Determine \( \csc t \) as the Reciprocal of \( \sin t \)

Since \( \csc t = \frac{1}{\sin t} \), we compute \( \csc t = \frac{1}{\frac{3}{5}} = \frac{5}{3} \).

Step 5: Calculate \( \sec t \) as the Reciprocal of \( \cos t \)

Since \( \sec t = \frac{1}{\cos t} \), we compute \( \sec t = \frac{1}{-\frac{4}{5}} = -\frac{5}{4} \).

Step 6: Find \( \cot t \) Using \( \cot t = \frac{1}{\tan t} \)

Since \( \tan t = -\frac{3}{4} \), \( \cot t = \frac{1}{-\frac{3}{4}} = -\frac{4}{3} \).

Key Concepts

Pythagorean identity

The Pythagorean identity is a fundamental concept in trigonometry. It is expressed as \( \sin^2 t + \cos^2 t = 1 \) for any angle \( t \). This identity is derived from the Pythagorean theorem, applying it to the unit circle, where the radius is 1. Given this equation:

• \( \sin^2 t \) represents the square of the sine of the angle.
• \( \cos^2 t \) represents the square of the cosine of the angle.

In our exercise, we use this identity to find the value of \( \cos t \) knowing \( \sin t = \frac{3}{5} \). We start by squaring the sine value to get \( \frac{9}{25} \). Subtracting this value from 1 gives \( \cos^2 t = \frac{16}{25} \), highlighting the interdependent relationship between sine and cosine.

quadrants and signs

Understanding the quadrants of the Cartesian plane is essential for determining the signs of trigonometric functions. The plane is divided into four quadrants:

• Quadrant I: All trigonometric functions are positive.
• Quadrant II: Sine is positive, cosine and tangent are negative.
• Quadrant III: Tangent is positive, sine and cosine are negative.
• Quadrant IV: Cosine is positive, sine and tangent are negative.

In this exercise, the angle \( t \) is located in Quadrant II. Therefore, \( \sin t \) is positive, and \( \cos t \) is negative, which explains why we find \( \cos t = -\frac{4}{5} \). These quadrant rules help in predicting the behavior of trigonometric functions across the unit circle.

reciprocal trigonometric functions

Reciprocal trigonometric functions provide greater flexibility and are used widely across various problems. They are defined as the inverse of the primary trigonometric functions:

• Cosecant \( \csc t \) is the reciprocal of sine, \( \csc t = \frac{1}{\sin t} \).
• Secant \( \sec t \) is the reciprocal of cosine, \( \sec t = \frac{1}{\cos t} \).
• Cotangent \( \cot t \) is the reciprocal of tangent, \( \cot t = \frac{1}{\tan t} \).

In this exercise, we used these relationships to find \( \csc t = \frac{5}{3} \), \( \sec t = -\frac{5}{4} \), and \( \cot t = -\frac{4}{3} \). Knowing one trigonometric function allows easy calculation of its reciprocal.

right triangle

The concept of a right triangle is central to understanding trigonometric functions. In a right triangle, one angle is always \(90^\circ\), and the properties of these triangles closely relate to the definitions of sine, cosine, and tangent:

• Sine represents the ratio of the opposite side to the hypotenuse.
• Cosine represents the ratio of the adjacent side to the hypotenuse.
• Tangent represents the ratio of the opposite side to the adjacent side.

In this exercise, we can visualize a right triangle setup that aids in understanding how the given \( \sin t = \frac{3}{5} \) translates to an opposite side of 3 units and a hypotenuse of 5 units. By using the Pythagorean identity, we were able to find the value for the adjacent side, leading us to find \( \cos t \) and other necessary trigonometric functions.

trigonometric identities

Trigonometric identities are equations involving trigonometric functions that are true for every value of the occurring variables. They are essential for simplifying and solving trigonometric equations:

• Pythagorean identities are perhaps the most known, such as \( \sin^2 t + \cos^2 t = 1 \).
• Reciprocal identities help switch between functions, like \( \csc t = \frac{1}{\sin t} \).
• Quotient identities include \( \tan t = \frac{\sin t}{\cos t} \).

In our problem, these identities played a crucial role. We used the Pythagorean identity to discover \( \cos t \). The quotient identity helped in determining \( \tan t = -\frac{3}{4} \). Understanding these identities allows students to navigate the complexities of trigonometric relationships with ease.