Problem 21 - Algebra and Trigonometry

Question

Find all angles \(\theta\) between \(0^{\circ}\) and \(180^{\circ}\) satisfying the given equation. $$ \sin \theta=\frac{1}{2} $$

Step-by-Step Solution

Answer

The angles are \(30^{\circ}\) and \(150^{\circ}\).

Step 1: Understand the Sine Function

The sine function, \( \sin \theta \), oscillates between \(-1\) and \(1\). It returns a positive value in the first and second quadrants, which is where we will focus since the problem asks for solutions between \(0^{\circ}\) and \(180^{\circ}\).

Step 2: Recognize Known Sine Ratios

Recall that \( \sin 30^{\circ} = \frac{1}{2} \). Since the sine function is periodic, other angles may also yield the same sine ratio within the specified range.

Step 3: Determine Second Quadrant Solution

In the second quadrant, the sine of an angle \( \theta \) is also equal to right above \(180^{\circ} - \theta\) from the first quadrant. We find that \( \sin(180^{\circ} - 30^{\circ}) = \sin 150^{\circ} = \frac{1}{2} \).

Step 4: List All Possible Solutions

From our steps, we determined that the solutions where \( \sin \theta = \frac{1}{2} \) are \( \theta = 30^{\circ} \) and \( \theta = 150^{\circ} \) within the interval from \(0^{\circ}\) to \(180^{\circ}\).

Key Concepts

Sine Function

The sine function is a fundamental concept in trigonometry. It connects an angle within a right triangle to the ratio of the length of the side opposite the angle to the length of the hypotenuse. If a triangle has an angle \(\theta\), the sine of \(\theta\) can be expressed as:

• \(\sin \theta = \frac{\text{opposite side}}{\text{hypotenuse}}\)

The sine function is periodic with a period of \(360^{\circ}\). This means it repeats its values every \(360^{\circ}\) of rotation. It oscillates between -1 and 1, making it useful in describing wave-like phenomena.
Within the range of \(0^{\circ}\) to \(180^{\circ}\), the sine value is always positive. This is because the y-coordinate of any point on the unit circle associated with the sine is above the x-axis in these quadrants.
When solving equations like \(\sin \theta = \frac{1}{2}\), understanding the values of sine at common angles, and the function's properties, is essential to finding all possible angles \(\theta\) within a given range.

Quadrants in Trigonometry

In trigonometry, the unit circle allows us to categorize angles into four quadrants:

• First Quadrant: \(0^{\circ}\) to \(90^{\circ}\)
• Second Quadrant: \(90^{\circ}\) to \(180^{\circ}\)
• Third Quadrant: \(180^{\circ}\) to \(270^{\circ}\)
• Fourth Quadrant: \(270^{\circ}\) to \(360^{\circ}\)

Each quadrant affects the sign of trigonometric functions. Specifically, for the sine function, values are positive in both the first and second quadrants. This is crucial when determining the values of \(\theta\) that satisfy \(\sin \theta = \frac{1}{2}\).
To find angles with a positive sine, we focused on the first and second quadrants. In our exercise, \(\sin 30^{\circ}\) provides a value of \(\frac{1}{2}\) in the first quadrant. In the second quadrant, \(\sin (180^{\circ} - 30^{\circ}) = \sin 150^{\circ} = \frac{1}{2}\).
Understanding which quadrants correspond to positive or negative values of sine helps to simplify solving trigonometric equations.

Special Angles

Trigonometry often involves solving problems using special angles, which are angles with well-known sine, cosine, and tangent values. These include:

• \(30^{\circ}\)
• \(45^{\circ}\)
• \(60^{\circ}\)

Each of these angles has specific, easily recognizable trigonometric values, with \(\sin 30^{\circ}\) equal to \(\frac{1}{2}\). Understanding these angles is crucial to quickly solving trigonometric equations.
When solving \(\sin \theta = \frac{1}{2}\), recognizing that \(\theta = 30^{\circ}\) is a special angle helps us find the first solution instantly. Further, by employing the property that \(\sin(180^{\circ} - \theta) = \sin \theta\), we determine that \(150^{\circ}\) is the second solution within our specified range.
Memorizing and understanding these special angles simplify many trigonometric problems significantly, saving time and reducing mistakes.