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Chapter 6: Problem 1

Find all solutions of the equation. $$\sin x=-\frac{\sqrt{2}}{2}$$

Short Answer

Expert verified
The solutions are \(x = \frac{5\pi}{4} + 2k\pi\) and \(x = \frac{7\pi}{4} + 2k\pi\), where \(k\) is any integer.

Step by step solution

01

Identify Reference Angle

The value \(-\frac{\sqrt{2}}{2}\) is known to correspond to the reference angle \(\theta = \frac{\pi}{4}\) radians, or \(45^\circ\), for the sine function.
02

Determine Quadrants for Negative Sine

Since \( \sin x = -\frac{\sqrt{2}}{2}\), we consider the quadrants where sine is negative. Sine is negative in Quadrants III and IV.
03

Find Principal Solutions

In Quadrant III, the angle is \(\pi + \frac{\pi}{4} = \frac{5\pi}{4}\). In Quadrant IV, the angle is \(2\pi - \frac{\pi}{4} = \frac{7\pi}{4}\). These angles correspond to the positions where the sine of the angle is the negative value of the reference angle.
04

Write General Solutions

The general solution for \(x\) is given by \(x = \frac{5\pi}{4} + 2k\pi\) and \(x = \frac{7\pi}{4} + 2k\pi\), where \(k\) is any integer, to account for the periodicity of the sine function.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Reference Angle
A reference angle is a crucial concept in solving trigonometric equations like \( \sin x = -\frac{\sqrt{2}}{2} \). It helps us determine the angle related to the trigonometric function's value without considering the sign. The reference angle is always between 0 and \( \frac{\pi}{2} \) radians or 0 and 90 degrees, making it the smallest angle formed with the x-axis.
  • The reference angle for \( \sin x = \frac{\sqrt{2}}{2} \) is \( \theta = \frac{\pi}{4} \) radians or 45掳.
  • This is because \( \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} \).

Knowing this, we can determine where in the coordinate plane the sine of the angle can be equal to \( -\frac{\sqrt{2}}{2} \), by looking at positive and negative signs of sine in different quadrants.
Sine Function
The sine function is one of the primary trigonometric functions that relate the angle in a right-angled triangle to the ratio of the length of the side opposite to the angle and the length of the hypotenuse. When understanding the sine function in terms of equations like \( \sin x = -\frac{\sqrt{2}}{2} \), it's important to remember:
  • Sine is positive in the first two quadrants and negative in the last two quadrants.
  • The periodicity of sine is \( 2\pi \), meaning its values repeat every \( 2\pi \) interval.
  • For a given arc on the unit circle, the vertical coordinate corresponds to the sine of the angle.

To solve the equation \( \sin x = -\frac{\sqrt{2}}{2} \), we note the negative sign indicates we're looking in the lower half of the unit circle (Quadrants III and IV) where sine is negative.
Quadrants
Understanding quadrants is essential for resolving trigonometric equations by determining the possible angles that satisfy a trigonometric function鈥檚 value. The coordinate plane is divided into four quadrants:
  • Quadrant I: Both sine and cosine are positive.
  • Quadrant II: Sine is positive, cosine is negative.
  • Quadrant III: Sine is negative, cosine is negative.
  • Quadrant IV: Sine is negative, cosine is positive.

For the equation \( \sin x = -\frac{\sqrt{2}}{2} \), since sine is negative, we focus on Quadrants III and IV. In these quadrants:
  • In Quadrant III, the related angle is \( \pi + \frac{\pi}{4} = \frac{5\pi}{4} \).
  • In Quadrant IV, the angle is \( 2\pi - \frac{\pi}{4} = \frac{7\pi}{4} \).

These calculations help find the principal solutions, which are then generalized by adding \( 2k\pi \), where \( k \) is an integer, to account for the cyclical nature of the sine function.

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