Problem 23 - Algebra 2 and Trigonometry

Question

In \(15-26,\) find each exact value in radians, expressing each answer in terms of \(\pi\) \(y=\arctan (-\sqrt{3})\)

Step-by-Step Solution

Answer

The exact value of \( y = \arctan(-\sqrt{3}) \) is \(-\frac{\pi}{3}\).

Step 1: Understand the Arc Tangent Function

The function \( y = \arctan(x) \) gives us the angle \( y \) in radians for which \( \tan(y) = x \). The range of \( \arctan \) is \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \). For this problem, we need to find the angle \( y \) so that \( \tan(y) = -\sqrt{3} \).

Step 2: Recall Reference Angles

For \( \tan(y) = -\sqrt{3} \), recall the reference angle where \( \tan(y) = \sqrt{3} \). This angle is \( \frac{\pi}{3} \) radians, since \( \tan(\frac{\pi}{3}) = \sqrt{3} \).

Step 3: Determine the Correct Quadrant

Since \( \tan(y) = -\sqrt{3} \) is negative and the range of \( \arctan \) is \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \), the angle \( y \) must be in the fourth quadrant. Angles in the fourth quadrant can be represented by negative angles. The equivalent negative angle of \( \frac{\pi}{3} \) is \(-\frac{\pi}{3}\).

Step 4: Verify the Answer

Verify by checking \( \tan(-\frac{\pi}{3}) \). Calculating \( \tan(-\frac{\pi}{3}) = -\tan(\frac{\pi}{3}) = -\sqrt{3} \). This confirms that our value of \( y = -\frac{\pi}{3} \) is correct.

Key Concepts

Understanding Reference Angles

Reference angles are a fundamental concept when working with trigonometric functions. Simply put, a reference angle is the smallest angle that a given angle makes with the x-axis. It is always positive and less than or equal to 90 degrees (or \(\frac{\pi}{2}\) radians).
A reference angle helps us find the corresponding trigonometric function values for angles in different quadrants. It's like a helper, letting us translate an angle into a familiar form that we already know how to work with.
Think of it this way:

• For \(\tan(\theta) = \sqrt{3}\), the reference angle \(\theta\) is \(\frac{\pi}{3}\). This is because we know that at \(\frac{\pi}{3}\), the value of tangent is \(\sqrt{3}\).
• The reference angle is always taken as a positive acute angle.
• When we have a negative value, the reference angle helps us find equivalent angles in other quadrants where the function value might be negative.

Basics of Trigonometric Functions

Trigonometric functions are mathematical relationships between the angles and sides of a triangle. They are essential for understanding various geometrical properties and phenomena that involve rotations and periodic motions.
We mainly focus on three primary functions:

• Sine \(\sin(\theta)\): Represents the y-coordinate of a point on the unit circle.
• Cosine \(\cos(\theta)\): Represents the x-coordinate of a point on the unit circle.
• Tangent \(\tan(\theta)\): Represents the ratio of \(\sin(\theta)\) to \(\cos(\theta)\).

These functions are periodic, meaning they repeat their values in regular intervals.
For tangent, it repeats every \(\pi\) radians, and it can be positive or negative depending on the angle and its quadrant.
The arcsin, arccos, and arctan functions generate angles from these ratios, helping us move from known side ratios back to angles themselves.

Exploring Quadrants in Trigonometry

The concept of quadrants is essential when dealing with trigonometric functions and determining the signs of their values. The Cartesian plane is divided into four quadrants, each indicating whether the x and y coordinates are positive or negative.

• First Quadrant: Both x and y are positive, thus sine, cosine, and tangent are all positive.
• Second Quadrant: x is negative, y is positive. Sine is positive, cosine and tangent are negative.
• Third Quadrant: Both x and y are negative, meaning sine and cosine are negative, but tangent is positive.
• Fourth Quadrant: x is positive, y is negative. Cosine is positive, while sine and tangent are negative.

Understanding which quadrant an angle falls into helps us determine the sign of the trigonometric functions.
In our scenario with \(\tan(y) = -\sqrt{3}\), knowing that tangent is negative points us towards the second or fourth quadrant.
However, since the range of \(\arctan\) is from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\), referring to angles between \(-90°\) and \(90°\), we focus on the fourth quadrant where both negative angles and negative tangents are valid. So, \(y = -\frac{\pi}{3}\) neatly fits here.