Problem 68 - Algebra and Trigonometry

Question

Find the values of the trigonometric functions of \(t\) from the given information. \(\sec t=2, \quad \sin t<0\)

Step-by-Step Solution

Answer

The values are \( \cos t = \frac{1}{2} \), \( \sin t = -\frac{\sqrt{3}}{2} \), \( \tan t = -\sqrt{3} \), \( \cot t = -\frac{\sqrt{3}}{3} \), and \( \csc t = -\frac{2\sqrt{3}}{3} \).

Step 1: Analyze Given Information

We are given that \( \sec t = 2 \) and \( \sin t < 0 \). The secant function is the reciprocal of the cosine function, so \( \sec t = \frac{1}{\cos t} \). Therefore, \( \cos t = \frac{1}{\sec t} = \frac{1}{2} \).

Step 2: Determine the Quadrant for Angle \( t \)

Since \( \cos t = \frac{1}{2} \) is positive and \( \sin t < 0 \), we must determine the quadrant in which \( t \) is located. \( \cos t \) is positive in the first and fourth quadrants, but \( \sin t < 0 \) only in the fourth quadrant. Therefore, \( t \) lies in the fourth quadrant.

Step 3: Calculate \( \sin t \) Using Pythagorean Identity

Using the Pythagorean identity \( \sin^2 t + \cos^2 t = 1 \), we can find \( \sin t \). We have \( \cos t = \frac{1}{2} \), so \( \cos^2 t = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \). Thus, \( \sin^2 t = 1 - \frac{1}{4} = \frac{3}{4} \) and \( \sin t = -\sqrt{\frac{3}{4}} = -\frac{\sqrt{3}}{2} \) since \( \sin t < 0 \) in the fourth quadrant.

Step 4: Calculate the Remaining Trigonometric Functions

We now calculate \( \tan t \), \( \cot t \), \( \csc t \), and \( \cos t \) using the values found:- \( \tan t = \frac{\sin t}{\cos t} = \frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}} = -\sqrt{3} \).- \( \cot t = \frac{1}{\tan t} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3} \) (after rationalizing).- \( \csc t = \frac{1}{\sin t} = -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3} \) (after rationalizing).- From Step 1, \( \cos t = \frac{1}{2} \).

Key Concepts

Secant

In trigonometry, the secant function, denoted as \( \sec \theta \), is the reciprocal of the cosine function. Simply put, it is given by \( \sec \theta = \frac{1}{\cos \theta} \). The secant function is undefined wherever the cosine is zero since division by zero is not possible.

• It can be larger than 1 or smaller than -1, depending on the angle.
• Usually, secant is considered when dealing with angles in a right triangle or unit circle.

For example, if the secant of an angle \( t \) is 2, like in our given problem, it means that \( \cos t = \frac{1}{2} \). This reciprocal relationship is foundational in solving trigonometric problems.

Quadrants

The concept of quadrants is an essential part of understanding angles within a unit circle. The unit circle is divided into four quadrants, each representing a specific range of angle measures and signs for sine and cosine.

• Quadrant I: Both sine and cosine are positive.
• Quadrant II: Sine is positive, and cosine is negative.
• Quadrant III: Both sine and cosine are negative.
• Quadrant IV: Sine is negative, and cosine is positive.

Since we know \( \cos t = \frac{1}{2} \) is positive and \( \sin t < 0 \), the angle \( t \) must be in the fourth quadrant, where the cosine is positive, and sine is negative.

Pythagorean Identity

The Pythagorean identity is a cornerstone in trigonometry. It is expressed as \( \sin^2 \theta + \cos^2 \theta = 1 \). This identity allows us to relate sine and cosine in any right triangle.
By plugging in known values, such as \( \cos t = \frac{1}{2} \) from our problem, we find \( \cos^2 t = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \).

• We can then substitute to find \( \sin^2 t = 1 - \frac{1}{4} = \frac{3}{4} \).
• This gives two potential answers for \( \sin t \): \( \sqrt{\frac{3}{4}} \) or \( -\sqrt{\frac{3}{4}} \).

Considering the angle lies in the fourth quadrant, where sine is negative, we conclude \( \sin t = -\frac{\sqrt{3}}{2} \).

Rationalizing

Rationalizing involves rewriting a fraction so that there are no square roots in the denominator. This process is used to simplify expressions.When calculating trigonometric functions such as \( \csc t \) and \( \cot t \), if our expression is like \( -\frac{2}{\sqrt{3}} \), we multiply numerator and denominator by \( \sqrt{3} \) to eliminate the square root in the denominator.

• This transforms \( -\frac{2}{\sqrt{3}} \) into \( -\frac{2\sqrt{3}}{3} \).
• The same method applies to \( \cot t \) converting \( -\frac{1}{\sqrt{3}} \) to \( -\frac{\sqrt{3}}{3} \).

Rationalizing makes it easier to understand mathematical expressions and ensures consistency when reporting results.