Problem 20 - Precalculus with Calculus Previews

Question

Find the exact value of the given trigonometric expression. Do not use a calculator. $$ \tan \left(\sin ^{-1}\left(-\frac{1}{6}\right)\right) $$

Step-by-Step Solution

Answer

\( \tan(\sin^{-1}(-\frac{1}{6})) = -\frac{\sqrt{35}}{35} \).

Step 1: Understand the Problem

We need to find the exact value of the expression \( \tan \left(\sin^{-1}\left(-\frac{1}{6}\right)\right) \). First, we understand that \( \sin^{-1}\left(-\frac{1}{6}\right) \) represents an angle \( \theta \) such that \( \sin(\theta) = -\frac{1}{6} \). Our task is to find \( \tan(\theta) \).

Step 2: Draw a Right Triangle

To solve for \( \tan(\theta) \), we will draw a right triangle. Assume \( \theta \) is in the right triangle with \( \sin(\theta) = -\frac{1}{6} \). This means the opposite side is \(-1\) and the hypotenuse is \(6\) (considering absolute values to find the cosine first). The negative sign indicates the angle is below the x-axis (in a coordinate plane).

Step 3: Find the Missing Side Using Pythagorean Theorem

The Pythagorean Theorem allows us to find the adjacent side: \( x^2 + (-1)^2 = 6^2 \Rightarrow x^2 + 1 = 36 \Rightarrow x^2 = 35 \Rightarrow x = \sqrt{35} \). Thus, the adjacent side of the triangle is \( \sqrt{35} \).

Step 4: Calculate \( \tan(\theta) \)

Now, using the definition of tangent: \( \tan(\theta) = \frac{\text{opposite side}}{\text{adjacent side}} = \frac{-1}{\sqrt{35}} \). To express this without a radical in the denominator, multiply the numerator and the denominator by \( \sqrt{35} \): \( \frac{-1 \cdot \sqrt{35}}{\sqrt{35} \cdot \sqrt{35}} = \frac{-\sqrt{35}}{35} \).

Step 5: Confirm the Range and Sign of \( \tan(\theta) \)

Since \( \sin^{-1}(-\frac{1}{6}) \) represents an angle in the range of \([-\frac{\pi}{2}, \frac{\pi}{2}]\), it indicates a fourth quadrant angle (where sine is negative and cosine is positive). Therefore, \( \tan(\theta) \) is indeed negative, confirming our solution is appropriately signed.

Key Concepts

Trigonometric Identities

Trigonometric identities are fundamental relationships between trigonometric functions. They help simplify and calculate various expressions. Common identities include those for sine, cosine, and tangent. A crucial identity involves the tangent function:

• \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \)

This identity is instrumental when dealing with inverse trigonometric functions. They are especially useful in situations where given angles must be evaluated without a calculator, as they provide exact, simplified forms of trigonometric expressions.After finding the sine of an angle, we can use identities to uncover other trigonometric values, such as cosine or tangent. In our problem, identifying \( \tan(\theta) \) through its identity was key in attaining the final solution.

Pythagorean Theorem

The Pythagorean Theorem is a fundamental concept in mathematics used to relate the sides of a right triangle. For a triangle with legs of lengths \(a\) and \(b\), and hypotenuse \(c\), the theorem states that:

• \( a^2 + b^2 = c^2 \)

In the given problem, we apply this theorem to find the missing side of a right triangle. When solving \( \tan(\sin^{-1}(-\frac{1}{6})) \), we identify the hypotenuse as \(6\) and the opposite side as \(-1\). Thus:

• \( x^2 + (-1)^2 = 6^2 \)

Solving for \(x\), we find:

• \( x^2 = 35 \Rightarrow x = \sqrt{35} \)

This process utilizes the Pythagorean Theorem to conclude that the adjacent side is \( \sqrt{35} \), providing the necessary components to calculate the tangent of the angle \( \theta \).

Right Triangle Trigonometry

Right triangle trigonometry is a critical approach when dealing with angles and their respective trigonometric functions. In right triangles, each angle and side is interrelated through functions like sine, cosine, and tangent, which form the foundation for identifying unknown elements of the triangle.The right triangle helps us visualize and solve for the tangent of \( \theta \) once \( \sin(\theta) = -\frac{1}{6} \) is known. By drawing the triangle and placing \( \theta \) within it, we see:

• Opposite side = \(-1\)
• Hypotenuse = \(6\)
• Adjacent side = \(\sqrt{35}\)

From this geometric picture, \( \tan(\theta) \) is defined as the ratio of the opposite side to the adjacent side:

• \( \tan(\theta) = \frac{-1}{\sqrt{35}} \)

By rationalizing, we simplify it to \( \frac{-\sqrt{35}}{35} \). This step ensures that the tangent is correctly expressed. This approach elucidates how trigonometric problems are solved through a firm grasp of right triangle trigonometry.