Problem 21 - Precalculus with Calculus Previews

Question

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

Step-by-Step Solution

Answer

The exact value is \( \frac{5}{3} \).

Step 1: Understand the Inverse Sine

The expression \( \sin^{-1} \frac{3}{5} \) represents an angle whose sine is \( \frac{3}{5} \). Let's denote this angle as \( \theta \). Thus, we have: \( \sin(\theta) = \frac{3}{5} \).

Step 2: Use the Pythagorean Identity

To find \( \csc(\theta) \), we need \( \sin(\theta) \) and \( \cos(\theta) \). We already have \( \sin(\theta) = \frac{3}{5} \). Use the identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \) to find \( \cos(\theta) \).

Step 3: Solve for \( \cos(\theta) \)

Plug \( \sin(\theta) = \frac{3}{5} \) into the identity: \[ \left(\frac{3}{5}\right)^2 + \cos^2(\theta) = 1 \] which simplifies to \[ \frac{9}{25} + \cos^2(\theta) = 1 \]. Solve for \( \cos^2(\theta) \): \[ \cos^2(\theta) = 1 - \frac{9}{25} = \frac{16}{25} \]. Thus, \( \cos(\theta) = \pm \frac{4}{5} \).

Step 4: Determine the Quadrant

Since \( \theta = \sin^{-1} \frac{3}{5} \) results in a reference angle in the first quadrant (the range of \( \sin^{-1} \) is \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\)), \( \cos(\theta) \) must be positive. Therefore, \( \cos(\theta) = \frac{4}{5} \).

Step 5: Compute \( \csc(\theta) \)

\( \csc(\theta) \) is the reciprocal of \( \sin(\theta) \). So we find \( \csc(\theta) \) by: \[ \csc(\theta) = \frac{1}{\sin(\theta)} = \frac{1}{\frac{3}{5}} = \frac{5}{3} \].

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions help us determine angles from known ratios of sides in a right triangle. They "undo" what the basic trigonometric functions do. For instance, the inverse sine function, written as \( \sin^{-1} \), takes a sine value and gives you an angle. This is handy when you know the sine of an angle, but not the angle itself. For example, if \( \sin(\theta) = \frac{3}{5} \), then \( \theta \) is \( \sin^{-1}\left(\frac{3}{5}\right) \). This identifies the angle whose sine value is \( \frac{3}{5} \). The range of \( \sin^{-1} \) is particularly important: it yields an angle in the interval \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\), ensuring that the result is always an angle between these bounds.
In the context of the problem, this means our angle is located in the first quadrant.

Pythagorean Identity

The Pythagorean identity is a fundamental relation between the sine and cosine of an angle, expressed as \( \sin^2(\theta) + \cos^2(\theta) = 1 \). This identity is derived from the Pythagorean theorem, which states that in any right-angled triangle, the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse. In terms of trigonometry:

• \( \sin(\theta) \) or opposite side over hypotenuse
• \( \cos(\theta) \) or adjacent side over hypotenuse

If you know one trigonometric function, you can use this identity to find the other.
For instance, in the exercise where \( \sin(\theta) = \frac{3}{5} \), you can find \( \cos(\theta) \) by rearranging the Pythagorean identity: \[ \cos^2(\theta) = 1 - \sin^2(\theta) \]\[ \cos^2(\theta) = 1 - \left(\frac{3}{5}\right)^2 \] Simplifying gives \( \cos(\theta) = \frac{4}{5} \), ensuring the angle is positive in the first quadrant.

Cosecant Function

The cosecant function, denoted as \( \csc \), is the reciprocal of the sine function. If you know \( \sin(\theta) \), you automatically know \( \csc(\theta) \) since it is just \( \frac{1}{\sin(\theta)} \). In simpler terms, if \( \sin(\theta) \) gives you the ratio of the opposite side to the hypotenuse, then \( \csc(\theta) \) gives you the hypotenuse to opposite ratio.

• If \( \sin(\theta) = \frac{3}{5} \), then \( \csc(\theta) = \frac{5}{3} \).

This reciprocal relationship is very useful in solving trigonometric expressions, as demonstrated in the exercise. Here, the goal was to find \( \csc \left( \sin^{-1} \frac{3}{5} \right) \), which simplifies directly to \( \frac{5}{3} \) by taking the reciprocal of \( \sin(\theta) = \frac{3}{5} \). This simplicity showcases the practical utility of the cosecant function in trigonometry.