Problem 15 - Precalculus with Calculus Previews

Question

Find the exact value of the given trigonometric expression. Do not use a calculator. $$ \sin \left(\cos ^{-1} \frac{3}{5}\right) $$

Step-by-Step Solution

Answer

The exact value is \( \frac{4}{5} \).

Step 1: Understand Cosine Inverse

We are given the expression \( \sin \left( \cos^{-1} \frac{3}{5} \right) \). Here, \( \cos^{-1} \frac{3}{5} \) represents an angle \( \theta \) such that \( \cos(\theta) = \frac{3}{5} \). We need to find \( \sin(\theta) \).

Step 2: Use the Pythagorean Identity

The Pythagorean identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \) helps us find \( \sin(\theta) \). Since \( \cos(\theta) = \frac{3}{5} \), we can substitute this value into the equation, giving us \( \sin^2(\theta) + \left(\frac{3}{5}\right)^2 = 1 \).

Step 3: Simplify the Equation

Calculating \( \left(\frac{3}{5}\right)^2 \), we get \( \frac{9}{25} \). Substitute this value back into the equation: \( \sin^2(\theta) + \frac{9}{25} = 1 \). To find \( \sin^2(\theta) \), subtract \( \frac{9}{25} \) from 1.

Step 4: Solve for \( \sin^2(\theta) \)

1 can be expressed as \( \frac{25}{25} \), so \( \sin^2(\theta) = \frac{25}{25} - \frac{9}{25} = \frac{16}{25} \).

Step 5: Determine \( \sin(\theta) \)

Taking the square root of both sides to solve for \( \sin(\theta) \), we get \( \sin(\theta) = \sqrt{\frac{16}{25}} = \frac{4}{5} \). Note that since \( \cos^{-1} \) yields a principal value of \( \theta \) between 0 and \( \pi \), \( \sin(\theta) \) should be positive.

Key Concepts

Inverse Trigonometry

Inverse trigonometry is where you find angles from known trigonometric values. Consider the expression \( \sin \left( \cos^{-1} \frac{3}{5} \right) \). Here, \( \cos^{-1} \frac{3}{5} \) indicates an angle, \( \theta \), where \( \cos(\theta) = \frac{3}{5} \). The domain of \( \cos^{-1} x \) is from -1 to 1, and the range is from 0 to \( \pi \). This means \( \theta \) is between 0 and \( \pi \), where the cosine and sine are positive in the first quadrant. In this expression, we use the inverse cosine function to find the exact angle value, which we then use to find the sine of that angle. Hence, inverse trig functions help find unknown angles when a trigonometric value is known.

Pythagorean Identity

The Pythagorean identity is a fundamental formula in trigonometry written as \( \sin^2(\theta) + \cos^2(\theta) = 1 \). This identity helps in finding one trig function knowing the other. Given \( \cos(\theta) = \frac{3}{5} \), you can find \( \sin(\theta) \) using this identity. By substituting \( \cos(\theta) = \frac{3}{5} \) into the identity, we have \( \sin^2(\theta) + \left(\frac{3}{5}\right)^2 = 1 \). Calculate \( \left(\frac{3}{5}\right)^2 \) to get \( \frac{9}{25} \). Then isolate \( \sin^2(\theta) \) by subtracting the square of cosine from 1, giving you \( \sin^2(\theta) = \frac{16}{25} \). This results from rearranging known values, allowing us to solve trigonometric problems with interdependence of sine and cosine.

Exact Trigonometric Values

Exact trigonometric values refer to precise calculations, not approximations, of trig functions without using a calculator. In our exercise, you need to determine \( \sin(\theta) \) from \( \sin^2(\theta) = \frac{16}{25} \). To retrieve \( \sin(\theta) \), take the positive square root, yielding \( \sin(\theta) = \frac{4}{5} \). The decision to choose the positive root stems from \( \cos^{-1} \)'s range of 0 to \( \pi \), where sine is naturally positive. Practicing this skill enhances efficiency in solving trig problems by enabling one to recall specific angle and ratio relationships quickly. Knowing these exact values helps streamline solutions, as seen here in bypassing the need for calculators by leveraging foundational trigonometric identities and relationships.