Problem 58 - Contemporary Precalculus

Question

In Exercises \(55-60\), find the values of all six trigonometric functions at \(t\) if the given conditions are true. $$\sin t=-2 / 3 \quad \text { and } \quad \sec t>0$$

Step-by-Step Solution

Answer

Question: Given \(\sin t = -\frac{2}{3}\) and \(\sec t > 0\), find the values of all six trigonometric functions at t. Answer: The trigonometric functions at t are: 1. \(\sin t = -\frac{2}{3}\) 2. \(\cos t = \frac{\sqrt{5}}{3}\) 3. \(\tan t = -\frac{2\sqrt{5}}{5}\) 4. \(\cot t = -\frac{\sqrt{5}}{2}\) 5. \(\csc t = -\frac{3}{2}\) 6. \(\sec t = \frac{3\sqrt{5}}{5}\)

Step 1: Determine the quadrant and use the Pythagorean identity to find cosine

Since \(\sin t = -2/3\) and \(\sec t > 0\), we can find in which quadrant the angle t lies. Since the sine is negative and the secant (reciprocal of cosine) is positive, the angle t must be in the fourth quadrant, where cosine is positive and sine is negative. Now we can use the Pythagorean identity: \(\sin^2 t + \cos^2 t = 1\), to find the value of cosine: $$\cos^2 t = 1 - \sin^2 t = 1 - (-\frac{2}{3})^2 = 1 - \frac{4}{9} = \frac{5}{9}$$ Since t is in the fourth quadrant and cosine is positive there, we have: $$\cos t = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}$$

Step 2: Find the values of tangent, cotangent, cosecant, and secant using relationships

Now that we have the values of sine and cosine, we can find the values of the other four trigonometric functions by using their relationships with sine and cosine: 1. Tangent: \(\tan t = \frac{\sin t}{\cos t} = \frac{-2/3}{\sqrt{5}/3} = -\frac{2}{\sqrt{5}} = -\frac{2\sqrt{5}}{5}\) (by rationalizing the denominator) 2. Cotangent: \(\cot t = \frac{1}{\tan t} = -\frac{\sqrt{5}}{2}\) 3. Cosecant: \(\csc t = \frac{1}{\sin t} = -\frac{3}{2}\) 4. Secant: \(\sec t = \frac{1}{\cos t} = \frac{3}{\sqrt{5}} = \frac{3\sqrt{5}}{5}\) (by rationalizing the denominator) So we have found the values of all six trigonometric functions at t: 1. \(\sin t = -\frac{2}{3}\) 2. \(\cos t = \frac{\sqrt{5}}{3}\) 3. \(\tan t = -\frac{2\sqrt{5}}{5}\) 4. \(\cot t = -\frac{\sqrt{5}}{2}\) 5. \(\csc t = -\frac{3}{2}\) 6. \(\sec t = \frac{3\sqrt{5}}{5}\)

Key Concepts

Pythagorean Identity

The Pythagorean identity is one of the fundamental relationships in trigonometry. It relates the squares of the sine and cosine of an angle to the number one. This identity is written as \( \sin^2 t + \cos^2 t = 1 \).
This equation can help us find one trigonometric function if we know the other. It is useful in many calculations where we only have a partial set of information.
For example, in the given exercise, knowing that \( \sin t = -\frac{2}{3} \), we can use this identity to find \( \cos t \).
We rearrange the formula to solve for \( \cos^2 t \):

• \( \cos^2 t = 1 - \sin^2 t \)

Substituting the known value:

• \( \cos^2 t = 1 - \left(-\frac{2}{3}\right)^2 = \frac{5}{9} \)

Taking the square root gives us the possible values for \( \cos t \). However, we need to apply knowledge from other trigonometric identities and quadrants to make the correct choice.

Trigonometric Identities

Trigonometric identities are equations that hold true for any angle \( t \), providing relationships between the six trigonometric functions. Two key identities often used are:

• The Pythagorean identities, such as \( \sin^2 t + \cos^2 t = 1 \).
• The reciprocal identities: \( \csc t = \frac{1}{\sin t} \), \( \sec t = \frac{1}{\cos t} \), \( \cot t = \frac{1}{\tan t} \).

These identities are the building blocks used for solving complex trigonometric problems.
If you know one function, you can figure out others using these relationships.
For instance, once you have \( \sin t \) and \( \cos t \), you can find the tangent using \( \tan t = \frac{\sin t}{\cos t} \).
Similarly, you can find the other three functions (cosecant, secant, and cotangent) through their reciprocal identities. This process involves simple division and, at times, rationalization, especially when roots are involved.

Trigonometric Quadrants

Understanding trigonometric quadrants is crucial for correctly applying trigonometric functions. The coordinate plane is divided into four quadrants. Each quadrant has specific signs for trigonometric functions.

• Quadrant I: all 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 our exercise, since \( \sin t = -\frac{2}{3} \) and \( \sec t > 0 \), the angle \( t \) must be in the fourth quadrant.
Knowing this, we can determine which functions are positive and negative without having to remember all the rules separately.
In essence, understanding which quadrant your angle lies in will automatically lead you to realize the signs of all six trigonometric functions at that angle. This skill is vital for accurately solving trigonometric problems.