/*! 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 27 Show that $$ \cos \left(\sin... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 27

Show that $$ \cos \left(\sin ^{-1} t\right)=\sqrt{1-t^{2}} $$ whenever \(-1 \leq t \leq 1\)

Short Answer

Expert verified
To show that \(\cos(\sin^{-1}t) = \sqrt{1-t^2}\), we first consider a right-angled triangle with angle \(\theta\) opposite to the side of length \(t\) and the hypotenuse length 1. Then, we find the length of the adjacent side as \(\sqrt{1-t^2}\) using the Pythagorean theorem. By calculating the cosine of the angle, we have \(\cos\theta = \frac{\sqrt{1-t^2}}{1}\). Since \(\sin\theta = t\), we have \(\theta = \sin^{-1}t\), and thus \(\cos(\sin^{-1}t) = \sqrt{1-t^2}\), which holds for all \(t\) in the range \(-1 \leq t \leq 1\).

Step by step solution

01

Consider a right-angled triangle

Let us consider a right-angled triangle with angle \(\theta\) opposite to the side of length \(t\) and the hypotenuse length 1. By definition, we have \(\sin\theta = t\).
02

Find the length of the adjacent side

From the Pythagorean theorem, we have the relationship between the sides of a right-angled triangle: \((\text{opposite side})^2 + (\text{adjacent side})^2 = (\text{hypotenuse})^2\). Plug in the values we know: \(t^2 + (\text{adjacent side})^2 = 1^2\). Therefore, we have the equation \((\text{adjacent side})^2 = 1 - t^2\). Taking the square root of both sides, we obtain the length of the adjacent side as \(\sqrt{1-t^2}\).
03

Calculate the cosine of the angle

Now we have all the side lengths of the right-angled triangle: 1. Opposite side: \(t\) 2. Adjacent side: \(\sqrt{1-t^2}\) 3. Hypotenuse: 1 Recall that the cosine of an angle in a right-angled triangle is given by the ratio of the adjacent side length to the hypotenuse length. Here, we have \(\cos\theta = \frac{\sqrt{1-t^2}}{1}\).
04

Express the result using the inverse sine function

Since \(\sin\theta = t\), we have \(\theta = \sin^{-1}t\). Therefore, we can write the cosine as \(\cos(\sin^{-1}t) = \sqrt{1-t^2}\). This equality holds for all \(t\) in the range \(-1 \leq t \leq 1\), as required.

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

Show that $$ \sin (\pi-\theta)=\sin \theta $$ for every angle \(\theta\).

Suppose \(u\) and \(v\) are in the interval \(\left(\frac{\pi}{2}, \pi\right),\) with \(\tan u=-2\) and \(\tan v=-3\) Find exact expressions for the indicated quantities. $$ \cos (v-6 \pi) $$

Use the right triangle below for Exercises \(1-76\) This triangle is not drawn to scale corresponding to the data in the exercises. Suppose \(\sin v=\frac{1}{3} .\) Evaluate \(\tan u\)

Suppose a 19 -foot ladder is leaning against a wall, making a \(71^{\circ}\) angle with the ground (as measured from a perpendicular line from the base of the ladder to the wall). How high up the wall is the end of the ladder?

Suppose \(u\) and \(v\) are in the interval \(\left(\frac{\pi}{2}, \pi\right),\) with \(\tan u=-2\) and \(\tan v=-3\) Find exact expressions for the indicated quantities. $$ \cos v $$
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Decision Maths

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.