Find two values of θ, 0 ≤ θ \u003c 2𝜋, that satisfy each equation. sin θ = √2/2

Ch. 1 - Angles and the Trigonometric Functions

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Blitzer 3rd Edition

Ch. 1 - Angles and the Trigonometric Functions

문제 99

1장, 문제 99

Find two values of θ, 0 ≤ θ < 2𝜋, that satisfy each equation.
sin θ = √2/2

검증된 단계별 안내

Recall the range of the sine function: \(\sin \theta\) can only take values between \(-1\) and \(1\). Since \(\frac{\sqrt{2}}{2}\) is approximately \(0.707\), it is within this range, so solutions exist.

Recognize that \(\sin \theta = \frac{\sqrt{2}}{2}\) corresponds to a well-known angle in the unit circle. Identify the reference angle \(\alpha\) such that \(\sin \alpha = \frac{\sqrt{2}}{2}\).

From the unit circle, the reference angle \(\alpha\) is \(\frac{\pi}{4}\) because \(\sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}\).

Since sine is positive in the first and second quadrants, find the two angles \(\theta\) in \([0, 2\pi)\) where \(\sin \theta = \frac{\sqrt{2}}{2}\). These are \(\theta = \alpha\) and \(\theta = \pi - \alpha\).

Write the two solutions explicitly as \(\theta = \frac{\pi}{4}\) and \(\theta = \pi - \frac{\pi}{4}\), which simplifies to \(\theta = \frac{3\pi}{4}\).

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Unit Circle and Angle Measurement

The unit circle is a circle with radius 1 centered at the origin, used to define trigonometric functions for all angles. Angles are measured in radians from 0 to 2π for one full rotation, which helps identify the sine values corresponding to specific angles.

Sine Function Values and Their Range

The sine function outputs values between -1 and 1. Knowing that sin θ = √2/2 corresponds to specific standard angles (π/4 and 3π/4) within the interval 0 ≤ θ < 2π is essential for finding solutions.

Finding Multiple Solutions in One Period

Since sine is positive in the first and second quadrants, there are two angles between 0 and 2π where sin θ equals √2/2. Understanding the symmetry of sine values in these quadrants allows identification of both solutions.

Find two values of θ, 0 ≤ θ < 2𝜋, that satisfy each equation. sin θ = √2/2