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Chapter 7: Problem 7

Find the exact value of the trigonometric function at the given real number. (a) \(\sin \frac{7 \pi}{6}\) \(\quad\) (b) \(\sin \left(-\frac{\pi}{6}\right)\) \(\quad\) (c) \(\sin \frac{11 \pi}{6}\)

Short Answer

Expert verified
(a) \( -\frac{1}{2} \), (b) \( -\frac{1}{2} \), (c) \( -\frac{1}{2} \).

Step by step solution

01

Identify the Quadrant

For each angle, we will first identify in which quadrant the angle lies on the unit circle.(a) The angle \( \frac{7\pi}{6} \) is in the third quadrant.(b) The angle \( -\frac{\pi}{6} \) is in the fourth quadrant.(c) The angle \( \frac{11\pi}{6} \) is in the fourth quadrant.
02

Determine Reference Angle

Next, find the reference angle for each given angle by determining the equivalent angle in the first quadrant.(a) For \( \frac{7\pi}{6} \), the reference angle is \( \pi - \frac{7\pi}{6} = \frac{\pi}{6} \).(b) For \( -\frac{\pi}{6} \), the absolute value is already a reference angle: \( \frac{\pi}{6} \).(c) For \( \frac{11\pi}{6} \), the reference angle is found as \( 2\pi - \frac{11\pi}{6} = \frac{1\pi}{6} \).
03

Find Sine of Reference Angle

Find the sine value of the reference angle \( \frac{\pi}{6} \) which is known:\( \sin \frac{\pi}{6} = \frac{1}{2} \).
04

Apply Sign According to Quadrant

Apply the correct sign to the sine value based on the quadrant of the original angles:(a) Since \( \frac{7\pi}{6} \) is in the third quadrant where sine is negative, \( \sin \frac{7\pi}{6} = -\frac{1}{2} \).(b) Since \( -\frac{\pi}{6} \) is in the fourth quadrant where sine is negative, \( \sin \left(-\frac{\pi}{6}\right) = -\frac{1}{2} \).(c) Since \( \frac{11\pi}{6} \) is in the fourth quadrant where sine is negative, \( \sin \frac{11\pi}{6} = -\frac{1}{2} \).

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Key Concepts

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

Unit Circle
The unit circle is a fundamental concept in trigonometry, offering a simple way to reach deeper understanding of trigonometric functions like sine, cosine, and tangent. It is a circle with a radius of 1 centered at the origin of the coordinate plane.
  • Although the circumference of the unit circle is not particularly large, only a value of \(2\pi\), it is crucial for calculating the values of trigonometric functions at various angles.
  • The unit circle is divided into quadrants, each representing a 90-degree section of the circle.
  • In trigonometry, angles are usually measured in radians, where the whole circle is \(2\pi\) radians.
Using the unit circle, you can easily find reference angles and quickly deduce the sign and value of trigonometric functions in any quadrant. It greatly simplifies calculations needed for solving problems like finding \(\sin \frac{7\pi}{6}\).
Reference Angle
A reference angle is a helpful way of understanding the original angle's position on the unit circle. It is the acute angle (between 0 and \(\frac{\pi}{2}\) radians) that the terminal side of the given angle makes with the x-axis.
  • Reference angles are always taken as positive and allow us to use known values from the first quadrant for calculations.
  • For example, in our exercise, the reference angle for \(\frac{7\pi}{6}\) is calculated as \(\frac{\pi}{6}\), which occurs in the first quadrant and can be easily solved using the unit circle.
  • If an angle is given in the negative, like \(-\frac{\pi}{6}\), its reference angle would just be \(\frac{\pi}{6}\), being the same positive distance from the x-axis.
Finding these reference angles is crucial, as it influences both the value and sign of trigonometric functions by considering the quadrant.
Quadrants in Trigonometry
Understanding the quadrants in trigonometry is vital to accurately determine the values of trigonometric functions like sine and cosine. The unit circle is divided into four quadrants:
  • Quadrant I (0 to \(\frac{\pi}{2}\)): All trigonometric function values are positive here.
  • Quadrant II (\(\frac{\pi}{2}\) to \(\pi\)): Sine is positive, but cosine and tangent are negative.
  • Quadrant III (\(\pi\) to \(\frac{3\pi}{2}\)): Only tangent is positive, while sine and cosine are negative.
  • Quadrant IV (\(\frac{3\pi}{2}\) to \(2\pi\)): Cosine is positive but sine and tangent are negative.
In our problem, the values \(\frac{7\pi}{6}\), \(-\frac{\pi}{6}\), and \(\frac{11\pi}{6}\) fall into different quadrants, impacting the sine values to result in negatives for all given angles. Knowing which quadrant an angle lies in helps to determine the sign of the function directly as part of solving trigonometric problems.

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