Problem 8 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-22: Find the exact value of the expression whenever it is defined. (a) \(\sin \left(\sin ^{-1} \frac{2}{3}\right)\) (b) \(\cos \left[\cos ^{-1}\left(-\frac{1}{5}\right)\right]\) (c) \(\tan \left[\tan ^{-1}(-9)\right]\)

Step-by-Step Solution

Answer

(a) \(\frac{2}{3}\), (b) \(-\frac{1}{5}\), (c) \(-9\).

Step 1: Recognize the Inverse Function Property

Start by understanding the property of inverse trigonometric functions. When you apply a trigonometric function to its own inverse, the operations cancel each other out, returning the original value, as long as it is within the defined range for inverse functions.

Step 2: Solve Part (a) for Sine

Given expression is \( \sin(\sin^{-1}(\frac{2}{3})) \). Since \( \sin^{-1} \) maps values in the interval \([-1, 1]\) to \([-\frac{\pi}{2}, \frac{\pi}{2}]\), applying \( \sin \) to \( \sin^{-1} \) of a value just returns that value. Thus, \( \sin(\sin^{-1}(\frac{2}{3})) = \frac{2}{3} \).

Step 3: Solve Part (b) for Cosine

For the expression \( \cos(\cos^{-1}(-\frac{1}{5})) \), observe that \( \cos^{-1} \) maps \([-1, 1]\) to \([0, \pi]\). Applying \( \cos \) to \( \cos^{-1} \) of a value returns that value, provided it is within the valid range. Therefore, \( \cos(\cos^{-1}(-\frac{1}{5})) = -\frac{1}{5} \).

Step 4: Solve Part (c) for Tangent

Given the expression \( \tan(\tan^{-1}(-9)) \). \( \tan^{-1} \) function maps all real numbers to \( (-\frac{\pi}{2}, \frac{\pi}{2}) \). Applying the \( \tan \) function to its inverse will return the argument, so \( \tan(\tan^{-1}(-9)) = -9 \).

Key Concepts

Sine Function

The sine function is one of the fundamental trigonometric functions. It is defined as the ratio of the opposite side to the hypotenuse in a right triangle. The inverse sine function, often written as \( \sin^{-1} \) or arcsin, reverses this process. This function takes a value from the range \([-1, 1]\) and returns an angle, typically measured in radians, from \([-\frac{\pi}{2}, \frac{\pi}{2}]\).

• When you see an expression such as \( \sin(\sin^{-1}(x)) \), it implies that the sine of an angle that gives \( x \) returns \( x \) itself, assuming \( x \) is within the range of \([-1, 1]\).
• For example, \( \sin(\sin^{-1}(\frac{2}{3})) \) equals \( \frac{2}{3} \), as \( \frac{2}{3} \) is within this range.

This reflects the property that applying the sine function to its inverse effectively cancels them out and restores the original value.

Cosine Function

The cosine function is another pillar of trigonometry, representing the ratio of the adjacent side to the hypotenuse of a right triangle. The inverse cosine function, denoted as \( \cos^{-1} \) or arccos, does the inverse operation by assigning the ratio back to an angle.

• The inverse cosine function accepts inputs from the range \([-1, 1]\) and returns angles in radians within \([0, \pi]\).
• Therefore, if you have a function like \( \cos(\cos^{-1}(x)) \), it will simply return \( x \) if \( x \) is valid within the range.
• In the case of \( \cos(\cos^{-1}(-\frac{1}{5})) \), since \(-\frac{1}{5}\) is indeed within \([-1, 1]\), the cosine of the angle whose cosine is \(-\frac{1}{5}\) is naturally \(-\frac{1}{5}\).

This showcases that the cosine applied to its inverse effectively negates their operations, bringing us back to the initial rational value.

Tangent Function

The tangent function is particularly unique in that it is defined as the ratio of the sine to the cosine of an angle. This function can take any real number and its range is from \(-\infty\) to \(\infty\). The inverse tangent function, \( \tan^{-1} \) or arctan, will take real numbers and provide an angle ranging from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\).

• When you encounter the expression \( \tan(\tan^{-1}(y)) \), it simplifies to \( y \) if \( y \) is a real number, as the tan and its inverse cancel each other out.
• For example, \( \tan(\tan^{-1}(-9)) = -9 \), as \(-9\) is within the real numbers domain that the inverse tangent function accepts.

This illustrates how for tangent functions, the application of its inverse immediately brings you back to the starting value, making it straightforward in operations.