Problem 124 - Precalculus

Question

For the following exercises, find the requested value. If \(\cos (t)=\frac{1}{7}\) and \(t\) is in the \(4^{\text { th }}\) quadrant, find \(\sin (t)\)

Step-by-Step Solution

Answer

\( \sin(t) = -\frac{4\sqrt{3}}{7} \)

Step 1: Understanding the Relationship between Sine and Cosine

We begin by recalling the Pythagorean identity for trigonometric functions: \( \sin^2(t) + \cos^2(t) = 1 \). In this exercise, we are given \( \cos(t) = \frac{1}{7} \), and we need to find \( \sin(t) \) in the fourth quadrant.

Step 2: Apply the Pythagorean Identity

Use the identity to solve for \( \sin^2(t) \):\[ \sin^2(t) = 1 - \cos^2(t) \]Substitute \( \cos(t) = \frac{1}{7} \):\[ \sin^2(t) = 1 - \left(\frac{1}{7}\right)^2 \]\[ \sin^2(t) = 1 - \frac{1}{49} \]\[ \sin^2(t) = \frac{48}{49} \]

Step 3: Solve for \( \sin(t) \) and Consider the Quadrant

To find \( \sin(t) \), take the square root of both sides:\[ \sin(t) = \pm \sqrt{\frac{48}{49}} \]Since \( t \) is in the fourth quadrant where sine is negative, we take the negative root:\[ \sin(t) = -\frac{\sqrt{48}}{7} \]Simplify \( \sqrt{48} \) to \( 4\sqrt{3} \):\[ \sin(t) = -\frac{4\sqrt{3}}{7} \]

Key Concepts

Pythagorean Identity

The Pythagorean identity is a fundamental relation in trigonometry. It states that for any angle, the square of the sine plus the square of the cosine equals one. This can be written as:

• \( \sin^2(t) + \cos^2(t) = 1 \)

This identity is incredibly useful because it allows us to find one trigonometric function if we know the other. In this exercise, we are given that \( \cos(t) = \frac{1}{7} \). By plugging this value into the Pythagorean identity, we can solve for \( \sin^2(t) \):

• \( \sin^2(t) = 1 - \cos^2(t) \)

Since \( \cos(t) = \frac{1}{7} \), when we square it, we get:

• \( \cos^2(t) = \left(\frac{1}{7}\right)^2 = \frac{1}{49} \)

Substituting into our formula gives us:

• \( \sin^2(t) = 1 - \frac{1}{49} = \frac{48}{49} \)

These steps make it possible to determine \( \sin(t) \), using just one trigonometric value and an identity.

Cosine in the Fourth Quadrant

Understanding in which quadrant an angle lies is essential for determining the sign of trigonometric functions. The fourth quadrant of the unit circle ranges from 270° to 360° (or from \( \frac{3\pi}{2} \) to \( 2\pi \) radians). Here, the cosine value is positive.So when we see a value like \( \cos(t) = \frac{1}{7} \) in the fourth quadrant, we immediately know it should remain positive. This quadrant's characteristics can be summarized as follows:

• Cosine values are positive
• Sine values are negative
• Tangent values are negative

It's crucial to recognize these aspects when solving trigonometric problems. Knowing which functions maintain positive or negative values helps direct the path of your calculations, just as knowing cosine remains positive in this scenario guides the solution.

Negative Sine Values

Sine values represent the y-coordinate of a point on the unit circle corresponding to an angle from the origin. In the fourth quadrant, these values are always negative due to the position below the x-axis. Therefore, when we determine that \( \sin(t) = \pm \sqrt{\frac{48}{49}} \) and know \( t\) is in the fourth quadrant, we must select the negative form.By choosing \( \sin(t) = -\frac{\sqrt{48}}{7} \), we confirm our understanding of the quadrant's properties. Simplifying the square root \( \sqrt{48} \) gives us \( 4\sqrt{3} \). This brings us to our final value:

• \( \sin(t) = -\frac{4\sqrt{3}}{7} \)

Always remembering which function values are negative in specific quadrants aids trigonometric equations' these calculations. This particular location for \( t \) mandates the sine's negativity, ensuring the solution adheres to trigonometric principles.