Problem 29 - Calculus of a Single Variable

Question

Write the expression in algebraic form. (Hint: Sketch a right triangle, as demonstrated in Example 3.) \(\tan \left(\operatorname{arcsec} \frac{x}{3}\right)\)

Step-by-Step Solution

Answer

\(\tan (\operatorname{arcsec} \frac{x}{3}) = \frac{{\sqrt{x^2 - 9}}}{3}\)

Step 1: Interpret the Inverse Secant function

To interpret the inverse secant function, it will be helpful to sketch a right triangle. To do this, consider a right triangle, in which one of the angles is \(\theta\) satisfying \( \operatorname{arcsec} \frac{x}{3} = \theta \). This implies that \(\sec{\theta}=\frac{x}{3} \). By definition of secant function as the reciprocal of cosine function, this implies that \(\cos{\theta} = \frac{3}{x} \). Now, from the Pythagorean theorem applied to right triangle, the third side, representing the opposite of angle \(\theta \), can be obtained as \(\sqrt{x^2 - 9} \). Note that using the Pythagorean theorem depends on assigning the hypotenuse a value of \(x\) and the adjacent side (base) a value of \(3\) because secant \(= \frac{hypotenuse}{adjacent} \).

Step 2: Compute the Tangent of arcsec

Now, using the so-obtained side lengths from the right triangle sketch, the tangent of the angle \(\theta \) is computed. The tangent of an angle in a right triangle is given by the ratio of the length of the side opposite the angle to the length of the side adjacent to the angle. Therefore, \(\tan \theta = \frac{{\sqrt{x^2 - 9}}}{3} \). Since \(\theta\) was defined to be equal to \(\operatorname{arcsec} \frac{x}{3}\), this means that \(\tan (\operatorname{arcsec} \frac{x}{3}) = \frac{{\sqrt{x^2 - 9}}}{3} \).

Step 3: Writing the final answer

The algebraic form of the given trigonometric expression is thus \(\tan (\operatorname{arcsec} \frac{x}{3}) = \frac{{\sqrt{x^2 - 9}}}{3}\)

Key Concepts

Trigonometric Identities

Understanding trigonometric identities is crucial in solving problems involving trigonometric functions. Trigonometric identities are equations that relate different trigonometric functions of an angle. They are essential tools for simplifying expressions, solving equations, and transforming formulas.
Here are some fundamental trigonometric identities:

• Pythagorean Identities:
  • \( ext{sin}^2 \theta + ext{cos}^2 \theta = 1 \)
  • \( 1 + ext{tan}^2 \theta = ext{sec}^2 \theta \)
  • \( ext{cot}^2 \theta + 1 = ext{csc}^2 \theta \)
• Reciprocal Identities:
  • \( ext{sec} \theta = \frac{1}{\text{cos} \theta} \)
  • \( ext{csc} \theta = \frac{1}{\text{sin} \theta} \)
  • \( ext{cot} \theta = \frac{1}{\text{tan} \theta} \)

These identities help in determining unknown sides or angles in right triangles and are instrumental in the algebraic manipulation of trigonometric expressions. In the given exercise, the reciprocal identity of secant, \( \text{sec} \theta = \frac{x}{3} \), is used to find the cosine value, \( \text{cos} \theta = \frac{3}{x} \), and thereby solve the problem step-by-step.

Pythagorean Theorem

The Pythagorean theorem is a vital mathematical principle used to relate the sides of a right triangle. It states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. The formula can be written as:

• \( a^2 + b^2 = c^2 \)

where:

• \( a \) and \( b \) are the lengths of the legs of the triangle
• \( c \) is the length of the hypotenuse

In the step-by-step solution provided for the original exercise, this theorem is applied to find the missing side of the right triangle. With the hypotenuse as \( x \) and one leg as \( 3 \), the other leg is calculated as \( \sqrt{x^2 - 9} \). This calculation is crucial for determining the tangent of the angle, which is the ratio of the opposite side to the adjacent side. Understanding how the Pythagorean theorem is applied helps to deduce the algebraic form \( \frac{\sqrt{x^2 - 9}}{3} \).

Right Triangles

Right triangles are a type of triangle where one angle measures exactly 90 degrees. These triangles are fundamental in trigonometry as they help in defining basic trigonometric ratios: sine, cosine, and tangent. Understanding these ratios within a right triangle context is key to solving trigonometric problems.
Trigonometric ratios are simply the relationships between the sides of a right triangle relative to one of its acute angles:

• Sine: \( \text{sin} \theta = \frac{\text{opposite}}{\text{hypotenuse}} \)
• Cosine: \( \text{cos} \theta = \frac{\text{adjacent}}{\text{hypotenuse}} \)
• Tangent: \( \text{tan} \theta = \frac{\text{opposite}}{\text{adjacent}} \)

In the original problem, a right triangle is sketched to aid in visualizing the scenario where \( \text{arcsec} \) function is involved. The exercise utilizes the ratios derived from the triangle to find the tangent of angle \( \theta \). By understanding the structure of right triangles and their properties, students can easily navigate through the solution and comprehend how each side and angle relate to one another in the context of inverse trigonometric functions.