/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 63 Evaluate the sine, cosine, and t... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 63

Evaluate the sine, cosine, and tangent of the angle without using a calculator. $$-\frac{\pi}{6}$$

Short Answer

Expert verified
The sine of \(-\frac{\pi}{6}\) is \(-\frac{1}{2}\), the cosine is \(\frac{\sqrt{3}}{2}\), and the tangent is \(-\frac{1}{\sqrt{3}}\).

Step by step solution

01

Evaluate sine function

The sine of the angle can be found by noticing that it is the negative of an angle with sin known: \(\sin(-\theta)= -\sin(\theta)\). Hence, \(\sin\left(-\frac{\pi}{6}\right) = -\sin\left(\frac{\pi}{6}\right)\), where \(\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\). Therefore, \(\sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2}\)
02

Evaluate cosine function

The cosine of the angle can be found by knowing the fact that it is equal to cosine of the positive of that angle: \(\cos(-\theta)= \cos(\theta)\). Hence, \(\cos\left(-\frac{\pi}{6}\right) = \cos\left(\frac{\pi}{6}\right)\), where \(\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\). Therefore, \(\cos\left(-\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\)
03

Evaluate tangent function

The tangent of the angle can be found by noticing that it is the negative of an angle with tan known: \(\tan(-\theta)= -\tan(\theta)\). Therefore, \(\tan\left(-\frac{\pi}{6}\right) = -\tan\left(\frac{\pi}{6}\right)\), where \(\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}\). Hence, \(\tan\left(-\frac{\pi}{6}\right) = -\frac{1}{\sqrt{3}}\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Use a graphing utility to graph the function. $$f(x)=\pi-\sin ^{-1}\left(\frac{2}{3}\right)$$

Is a degree or a radian the greater unit of measure? Explain.

The numbers of hours \(H\) of daylight in Denver, Colorado, on the 15 th of each month are: \(1(9.67), 2(10.72), \quad 3(11.92), \quad 4(13.25)\) \(5(14.37), \quad 6(14.97), \quad 7(14.72), \quad 8(13.77), \quad 9(12.48)\) \(10(11.18), \quad 11(10.00), \quad 12(9.38) . \quad\) The month is represented by \(t,\) with \(t=1\) corresponding to January. A model for the data is \(H(t)=12.13+2.77 \sin \left(\frac{\pi t}{6}-1.60\right)\) (a) Use a graphing utility to graph the data points and the model in the same viewing window. (b) What is the period of the model? Is it what you expected? Explain. (c) What is the amplitude of the model? What does it represent in the context of the problem? Explain.

Define the inverse secant function by restricting the domain of the secant function to the intervals \([0, \pi / 2)\) and \((\pi / 2, \pi],\) and sketch the graph of the inverse trigonometric function.

Finding Arc Length Find the length of the are on a circle of radius \(r\) intercepted by a central angle \(\boldsymbol{\theta}\). $$r=15 \text { inches, } \theta=120^{\circ}$$
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Logic and Functions

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation

Geometry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.