Problem 18 - Precalculus with Calculus Previews

Question

Find the exact value of the given trigonometric expression. Do not use a calculator. $$ \sin \left(\arctan \frac{1}{4}\right) $$

Step-by-Step Solution

Answer

\( \sin \left( \arctan \frac{1}{4} \right) = \frac{\sqrt{17}}{17} \).

Step 1: Understand the Problem

We are asked to find the sine of the angle whose tangent is \( \frac{1}{4} \). This involves using trigonometric identities to express sine in terms of tangent.

Step 2: Use the Definition of Arctan

The expression \( \arctan \frac{1}{4} \) represents an angle \( \theta \) such that \( \tan \theta = \frac{1}{4} \). This implies that the opposite side is 1 and the adjacent side is 4 in a right triangle.

Step 3: Use the Pythagorean Theorem to Find the Hypotenuse

To find the hypotenuse \( h \) of the right triangle, use the Pythagorean theorem: \[ h = \sqrt{1^2 + 4^2} = \sqrt{1 + 16} = \sqrt{17}. \]

Step 4: Calculate the Sine of the Angle

The sine of the angle \( \theta \) is the ratio of the opposite side to the hypotenuse:\[ \sin \theta = \frac{1}{\sqrt{17}}. \]

Step 5: Rationalize the Denominator

To express \( \sin \theta \) with a rationalized denominator, multiply the numerator and the denominator by \( \sqrt{17} \): \[ \sin \theta = \frac{1}{\sqrt{17}} \cdot \frac{\sqrt{17}}{\sqrt{17}} = \frac{\sqrt{17}}{17}. \]

Key Concepts

Arctan Function

The arctan function, also known as the inverse tangent function, is an essential concept in trigonometry. It is used to determine the angle whose tangent is a given number. Specifically, if you have \( \arctan x \), it means you are finding the angle \( \theta \) such that \( \tan \theta = x \). In the context of trigonometric identities, this function plays a crucial role as it connects angles with their trigonometric ratios.

When dealing with arctan, it is important to remember:

• The output is an angle, often denoted in radians or degrees.
• The range of the arctan function is usually from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\), meaning it can produce angles from -90° to 90°.
• Arctan is particularly useful when working with right triangles, as it allows finding angles using known side ratios.

In our exercise, \( \arctan \frac{1}{4} \) signifies an angle \( \theta \) where the tangent of \( \theta \) equals \( \frac{1}{4} \). This gives us the foundation to dive deeper into solving the problem using other trigonometric identities and theorems.

Pythagorean Theorem

The Pythagorean theorem is an age-old mathematical principle that relates the sides of a right-angled triangle. It states that in such a triangle, the square of the hypotenuse \( c \) is equal to the sum of the squares of the other two sides, often labeled \( a \) and \( b \). The formula is expressed as \( c^2 = a^2 + b^2 \). This theorem is not only foundational in geometry but is also continuously useful in solving trigonometry problems.

In the given exercise, we used the Pythagorean theorem to find the hypotenuse of the triangle, where the opposite side is 1 and the adjacent side is 4. By applying the theorem:

• Calculate \( h^2 = 1^2 + 4^2 \).
• This results in \( h^2 = 1 + 16 = 17 \).
• The hypotenuse \( h \) is the square root of 17, \( h = \sqrt{17} \).

Finding this hypotenuse is critical, as it allows us to calculate other trigonometric functions for the angle \( \theta \). For instance, to find the sine of \( \theta \), we need to know the ratio of the opposite side to the hypotenuse.

Rationalizing Denominators

Rationalizing denominators is a technique used in mathematics to eliminate irrational numbers from the denominator of a fraction, making it easier to interpret and often simpler to use in further calculations. This process involves multiplying the numerator and denominator by a suitable value to create a rational denominator.

Consider our solution from the exercise: initially, the sine of \( \theta \) is given by \( \sin \theta = \frac{1}{\sqrt{17}} \). Here, \( \sqrt{17} \) is an irrational number in the denominator. To rationalize it, we multiply both the numerator and the denominator by \( \sqrt{17} \):

• Multiply to get \( \sin \theta = \frac{1 \cdot \sqrt{17}}{\sqrt{17} \cdot \sqrt{17}} \).
• This simplifies to \( \frac{\sqrt{17}}{17} \), ridding the fraction of the square root in the denominator.

This method provides a cleaner, more conventional expression of the trigonometric function and ensures that calculations involving these expressions are straightforward and standard-compliant.