Problem 67 - Algebra and Trigonometry

Question

Find the values of the trigonometric functions of \(t\) from the given information. sec \(t=3,\) terminal point of \(t\) is in Quadrant IV

Step-by-Step Solution

Answer

\(\sin(t) = -\frac{2\sqrt{2}}{3}, \cos(t) = \frac{1}{3}, \tan(t) = -2\sqrt{2}, \csc(t) = -\frac{3\sqrt{2}}{4}, \sec(t) = 3, \cot(t) = -\frac{\sqrt{2}}{4}.\)

Step 1: Understand Secant and the Reciprocal Relationship

The secant function sec \(t\) is the reciprocal of the cosine function, \[ sec(t) = \frac{1}{\cos(t)}. \]Given sec \(t = 3\), we find the cosine by taking the reciprocal:\[ \cos(t) = \frac{1}{3}. \]

Step 2: Identify the Sine Function Value

In Quadrant IV, the cosine value is positive, but the sine value is negative. We use the Pythagorean identity:\[ \sin^2(t) + \cos^2(t) = 1. \]Substitute \(\cos(t) = \frac{1}{3}\):\[ \sin^2(t) + \left(\frac{1}{3}\right)^2 = 1, \]\[ \sin^2(t) + \frac{1}{9} = 1, \]\[ \sin^2(t) = 1 - \frac{1}{9} = \frac{8}{9}. \]\(\sin(t)\) must be negative:\(\sin(t) = -\sqrt{\frac{8}{9}} = -\frac{\sqrt{8}}{3} = -\frac{2\sqrt{2}}{3}.\)

Step 3: Determine the Tan Function

Use the identities and values found to calculate \(\tan(t)\):\[ \tan(t) = \frac{\sin(t)}{\cos(t)}. \]Substitute \(\sin(t) = -\frac{2\sqrt{2}}{3}\) and \(\cos(t) = \frac{1}{3}\):\[ \tan(t) = \frac{-\frac{2\sqrt{2}}{3}}{\frac{1}{3}} = -2\sqrt{2}. \]

Step 4: Calculate the Cotangent, Cosecant, and Values

For \(\csc(t)\), which is the reciprocal of \(\sin(t)\): \[ \csc(t) = \frac{1}{\sin(t)} = -\frac{3}{2\sqrt{2}} = -\frac{3\sqrt{2}}{4}. \]For \(\cot(t)\): \[ \cot(t) = \frac{1}{\tan(t)} = -\frac{1}{2\sqrt{2}} = -\frac{\sqrt{2}}{4}. \]

Key Concepts

Secant Function

The secant function, denoted as \( \sec(t) \), is one of the six fundamental trigonometric functions. It is the reciprocal of the cosine function. This means if you know secant, you can easily find cosine by taking the reciprocal. To put it mathematically, \( \sec(t) = \frac{1}{\cos(t)} \).
For example, in the given problem, we're told that \( \sec(t) = 3 \). So, to find \( \cos(t) \), we flip it to get \( \cos(t) = \frac{1}{3} \).
This relationship highlights how closely connected these trigonometric functions are, allowing us to switch perspectives and solve for different unknowns within trigonometric problems. Understanding the secant and its relationship with cosine is crucial as it frequently appears in various mathematical contexts.

Reciprocal Identities

Reciprocal identities are a handy set of tools in trigonometry that show how basic functions can be expressed as reciprocals of one another. They help in quickly finding values for different functions when given one.
These identities include:

• \( \sec(t) = \frac{1}{\cos(t)} \)
• \( \csc(t) = \frac{1}{\sin(t)} \)
• \( \cot(t) = \frac{1}{\tan(t)} \)

Using this concept in our example, given \( \sec(t) = 3 \), we found \( \cos(t) = \frac{1}{3} \).
Similarly, knowing \( \sin(t) = -\frac{2\sqrt{2}}{3} \) allowed us to find \( \csc(t) = -\frac{3\sqrt{2}}{4} \).
These identities simplify the solving process by linking different trigonometric functions through simple reciprocal relationships.

Pythagorean Identity

The Pythagorean identity is a key formula in trigonometry. It relates the squares of the sine and cosine functions to one, encapsulated in the equation \( \sin^2(t) + \cos^2(t) = 1 \).
This identity is very useful for calculating unknown trigonometric functions when another is known. In the problem, we know \( \cos(t) = \frac{1}{3} \). Plugging this into the identity, we have: \[\sin^2(t) + \left(\frac{1}{3}\right)^2 = 1\]Which simplifies to: \[\sin^2(t) = 1 - \frac{1}{9} = \frac{8}{9}\]Taking the square root gives \( \sin(t) = -\frac{2\sqrt{2}}{3} \), considering we're in Quadrant IV where sine is negative.
The Pythagorean identity not only links sine and cosine but also helps to solve various trigonometric scenarios efficiently.

Quadrants in Trigonometry

Trigonometry divides the coordinate plane into four quadrants, which influence the sign of trigonometric functions.
This division is fundamental as it determines whether outputs for sine, cosine, and tangent are positive or negative based on the angle's terminal side position:

• 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 the exercise, the terminal point of \( t \) is in Quadrant IV, dictating that \( \cos(t) \) is positive and \( \sin(t) \) is negative.
Understanding the quadrant setting is vital for correctly assessing the signs of trigonometric function values, which ensures accurate calculations and results.