Problem 34 - Algebra 2 and Trigonometry

Question

In \(33-38,\) find the exact radian measure \(\theta\) of an angle with the smallest absolute value that satisfies the equation. $$ \cos \theta=\frac{\sqrt{3}}{2} $$

Step-by-Step Solution

Answer

\( \theta = \frac{\pi}{6} \)

Step 1: Understanding the cosine function

The cosine function is positive in the first and fourth quadrants. We need to determine the angle \( \theta \) such that \( \cos \theta = \frac{\sqrt{3}}{2} \).

Step 2: Identifying possible angles in radians

The angle whose cosine value is \( \frac{\sqrt{3}}{2} \) is \( \theta = \frac{\pi}{6} \) and \( \theta = -\frac{\pi}{6} \). These correspond to the angle measures in the first and fourth quadrants.

Step 3: Selecting the angle with the smallest absolute value

Between \( \frac{\pi}{6} \) and \( -\frac{\pi}{6} \), the smallest absolute value is \( \frac{\pi}{6} \) since \( \left| \frac{\pi}{6} \right| < \left| -\frac{\pi}{6} \right| \).

Step 4: Confirming solution

Verify by checking that \( \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} \), which confirms that \( \theta = \frac{\pi}{6} \) is correct.

Key Concepts

Cosine Function

The cosine function, often abbreviated as \( \cos \), is a fundamental trigonometric function. It is used across various fields, including geometry, physics, and engineering, to relate the angle of a triangle to the ratio of the length of the adjacent side over the hypotenuse in a right triangle. The cosine function is cyclical and periodic, meaning it repeats its values in a regular interval, specifically every \(2\pi\). This cyclic nature is vital for solving trigonometric equations effectively.Understanding where the cosine function is positive is crucial for solving problems like the one given in the exercise. Cosine is positive in the:

• First quadrant: This includes angles between 0 and \(\frac{\pi}{2}\).
• Fourth quadrant: Here angles range between \(\frac{3\pi}{2}\) and \(2\pi\), but as negative angles, it corresponds to \(-\frac{\pi}{2}\) to \(0\).

Recognizing the positivity of the cosine function within these quadrants allows us to solve trigonometric equations by identifying potential angle candidates that satisfy the equation. This understanding aids in pinpointing solutions that include both positive and negative angles having the same absolute value.

Radian Measure

Radian measure is a way of expressing angles that is based on the radius of a circle. Unlike degrees, which divide a circle into 360 parts, radians offer a more natural unit linked to the arc length of a circle. In fact, one complete circle is \(2\pi\) radians.The conversion between degrees and radians is given by:Degrees to radians: \( \theta_{\text{radians}} = \theta_{\text{degrees}} \times \frac{\pi}{180} \)
Radians to degrees: \( \theta_{\text{degrees}} = \theta_{\text{radians}} \times \frac{180}{\pi} \) Given that trigonometric functions are periodic with \(2\pi\), using radians simplifies the analysis of these functions and yes, makes the solutions elegant! The mentioned problem uses angles such as \(\frac{\pi}{6}\) due to this nature.
When we talk about the smallest absolute value of an angle, it indicates choosing the angle in radians which is closest to zero, thereby simplifying computation and application.

First and Fourth Quadrants

The first and fourth quadrants are parts of the Cartesian coordinate plane, especially significant in trigonometry for evaluating the sign of trigonometric functions like cosine.

• First Quadrant: In this quadrant, all trigonometric functions - sine, cosine, and tangent - are positive. Angles here range from \(0\) to \(\frac{\pi}{2}\).
• Fourth Quadrant: Here, the sine is negative, the cosine is positive, and the tangent is negative. The range of angles is usually about \(\frac{3\pi}{2}\) to \(2\pi\) or, in terms of negative angles, from \(-\frac{\pi}{2}\) to \(0\).

When you're solving equations like \( \cos \theta = \frac{\sqrt{3}}{2} \), both the first and fourth quadrants are relevant because cosine values in both these quadrants are positive. The angle \(\frac{\pi}{6}\) in the first quadrant and \(-\frac{\pi}{6}\) in the fourth quadrant both satisfy the equation. However, choosing \(\frac{\pi}{6}\) gives the angle with the smallest absolute value which is often preferred when solving such problems.