Problem 71 - Algebra and Trigonometry

Question

Find the values of the trigonometric functions of \(t\) from the given information. \(\sin t=-\frac{1}{4}, \quad \sec t<0\)

Step-by-Step Solution

Answer

\( \cos t = -\frac{\sqrt{15}}{4} \), \( \tan t = \frac{1}{\sqrt{15}} \), \( \csc t = -4 \), \( \sec t = -\frac{4}{\sqrt{15}} \), \( \cot t = \sqrt{15} \).

Step 1: Analyze the Given Information

We know that \( \sin t = -\frac{1}{4} \) and \( \sec t < 0 \). The value of \( \sin t \) indicates that the sine of angle \( t \) is negative, and as \( \sec t = \frac{1}{\cos t} \), the cosine of angle \( t \) must also be negative. Thus, \( t \) lies in the third quadrant, where both sine and cosine are negative.

Step 2: Calculate Cosine Using Pythagorean Identity

In the third quadrant, we have \( \sin^2 t + \cos^2 t = 1 \). Substitute the known value of \( \sin t \): \[ \left(-\frac{1}{4}\right)^2 + \cos^2 t = 1 \] Simplifying gives \( \frac{1}{16} + \cos^2 t = 1 \), or \( \cos^2 t = 1 - \frac{1}{16} = \frac{15}{16} \). Therefore, \( \cos t = -\sqrt{\frac{15}{16}} = -\frac{\sqrt{15}}{4} \) as cosine is negative in the third quadrant.

Step 3: Determine Other Trigonometric Functions

Now we use \( \cos t \) and \( \sin t \) to find other trigonometric functions.ewline1. \( \tan t = \frac{\sin t}{\cos t} = \frac{-\frac{1}{4}}{-\frac{\sqrt{15}}{4}} = \frac{1}{\sqrt{15}} \).ewline2. \( \csc t = \frac{1}{\sin t} = \frac{1}{-\frac{1}{4}} = -4 \)ewline3. \( \sec t = \frac{1}{\cos t} = \frac{1}{-\frac{\sqrt{15}}{4}} = -\frac{4}{\sqrt{15}} \)ewline4. \( \cot t = \frac{1}{\tan t} = \sqrt{15} \).

Key Concepts

Sine Function

The sine function, denoted as \( \sin(\theta) \), expresses the ratio between the opposite side and the hypotenuse in a right triangle. It reflects how steep a line is, ranging between -1 and 1.
In the given exercise, \( \sin t = -\frac{1}{4} \), indicating that the sine value is negative. This is critical because it helps determine in which quadrant angle \( t \) lies. The sine function is negative in the third and fourth quadrants, but as we have additional information from \( \sec t < 0 \), we narrow \( t \) down to the third quadrant.

• In any right triangle, \( \sin(t) = \frac{opposite}{hypotenuse} \).
• In the unit circle, \( \sin(t) \) corresponds to the y-coordinate of the point where the angle intercepts the circle.

Understanding the behavior of the sine function across different quadrants is crucial for correctly solving trigonometric equations and identities.

Cosine Function

The cosine function, written as \( \cos(\theta) \), measures the ratio of the adjacent side to the hypotenuse in a right triangle. It ranges from -1 to 1, similar to the sine function.
The cosine of our angle, \( t \), is calculated using the given information on sine, and involves applying the Pythagorean Identity. With \( \sec t < 0 \) indicating a negative cosine, we calculate it as \( \cos t = -\frac{\sqrt{15}}{4} \).

• For a right triangle, \( \cos(t) = \frac{adjacent}{hypotenuse} \).
• In the unit circle, \( \cos(t) \) aligns with the x-coordinate of the circle point.

Finding these trigonometric values in heavier tasks, like determining all trigonometric function values from a list, strengthens angle comprehension.

Pythagorean Identity

The Pythagorean Identity states that for any angle \( t \), \( \sin^2(t) + \cos^2(t) = 1 \). This is a fundamental identity amongst trigonometric functions and is derived from the Pythagorean Theorem in a unit circle context.
It's particularly useful when one trigonometric value is known, and you need to find another, as it was used in the exercise to find \( \cos t \) from \( \sin t \). By substituting \( \sin t = -\frac{1}{4} \) into the identity and solving for \( \cos^2 t \), we find \( \cos t = -\frac{\sqrt{15}}{4} \).

• The Pythagorean Identity: \( \sin^2(t) + \cos^2(t) = 1 \).
• Helps verify correctness and interdependency of calculated trigonometric values.

This identity forms the basis for more complex trigonometric explorations, allowing students to progress confidently.