Problem 4 - Precalculus with Calculus Previews

Question

Given that \(\cos t=\frac{3}{4}\) and that \(P(t)\) is a point in the fourth quadrant, find \(\sin t\)

Step-by-Step Solution

Answer

\( \sin t = -\frac{\sqrt{7}}{4} \)

Step 1: Understand the Problem

We are given that \( \cos t = \frac{3}{4} \). We need to find \( \sin t \) for a point \( P(t) \) in the fourth quadrant. Recall that in the fourth quadrant, the sine is negative.

Step 2: Use the Pythagorean Identity

Recall the Pythagorean identity: \( \sin^2 t + \cos^2 t = 1 \). We can use this identity to find \( \sin t \).

Step 3: Substitute the Given Cosine Value

Substitute \( \cos t = \frac{3}{4} \) into the Pythagorean identity: \( \sin^2 t + \left(\frac{3}{4}\right)^2 = 1 \).

Step 4: Simplify the Equation

Calculate \( \left(\frac{3}{4}\right)^2 = \frac{9}{16} \). Thus, the equation becomes \( \sin^2 t + \frac{9}{16} = 1 \).

Step 5: Solve for \( \sin^2 t \)

Isolate \( \sin^2 t \) in the equation: \( \sin^2 t = 1 - \frac{9}{16} = \frac{16}{16} - \frac{9}{16} = \frac{7}{16} \).

Step 6: Find \( \sin t \)

Since we solved \( \sin^2 t = \frac{7}{16} \), take the square root to find \( \sin t \). Since \( P(t) \) is in the fourth quadrant, \( \sin t \) is negative. Therefore, \( \sin t = -\sqrt{\frac{7}{16}} = -\frac{\sqrt{7}}{4} \).

Key Concepts

Pythagorean Identity

The Pythagorean Identity is a cornerstone of trigonometry. It establishes a fundamental relationship between sine and cosine. This identity is expressed as \[\sin^2 t + \cos^2 t = 1\]This equation comes directly from observing a right triangle inscribed in a circle. Here, the hypotenuse is 1, meaning this identity also relates to the unit circle.
When using this identity, you can find one trigonometric function if you know the other. In this problem, using the known value of \( \cos t = \frac{3}{4} \), we substitute into the identity, allowing us to solve for \( \sin t \). This demonstrates the utility of the identity in confirming the relationship between the two functions. It's crucial always to consider the quadrant to apply the correct sign for the trigonometric function. In the fourth quadrant, sine is negative as noted in the problem.

Quadrants in Trigonometry

The concept of quadrants is essential in understanding the signs of trigonometric functions as they relate to angles. The coordinate plane is divided into four quadrants:

• First Quadrant: Both sine and cosine are positive.
• Second Quadrant: Sine is positive, cosine is negative.
• Third Quadrant: Both sine and cosine are negative.
• Fourth Quadrant: Cosine is positive, sine is negative.

This system helps in determining the signs of trigonometric functions. In the exercise, the point \( P(t) \) is in the fourth quadrant. Here, while cosine is positive, sine is negative. This directly affects our solution for \( \sin t \). Understanding which quadrant an angle is in will guide not only the signs of the trigonometric functions but also the approach to solving equations involving them.

Circular Functions

Circular functions refer to the sine and cosine functions as they relate to the unit circle. The unit circle is a circle with a radius of one, centered at the origin of a coordinate plane.
Sine and cosine are defined based on the coordinates of points on the unit circle. For an angle \( t \):

• The x-coordinate corresponds to \( \cos t \).
• The y-coordinate corresponds to \( \sin t \).

These definitions help in visualizing how trigonometric functions behave as the angle changes. The exercise uses the value of \( \cos t \) (given as \( \frac{3}{4} \)) and finds \( \sin t \) by leveraging the symmetry and identity properties of circular functions. Through this visualization, one can comprehend how these functions are interconnected across different segments of the circle, which reflects various quadrants, affecting their sign and value dynamics.