/*! 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 48 Verify the identity. $$\cos \l... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 48

Verify the identity. $$\cos \left(\sin ^{-1} x\right)=\sqrt{1-x^{2}}$$

Short Answer

Expert verified
From the given exercise we have successfully verified that \(\cos \left(\sin ^{-1} x\right)=\sqrt{1-x^{2}}\). This has been accomplished by translating the problem to a geometric interpretation, using Pythagorean theorem and trigonometric definitions.

Step by step solution

01

Translate to geometric interpretation and build a right triangle

Start by setting \(\sin^{-1}(x) = \theta\). This representation means that \(x\) is the sine of \(\theta\). From the right triangle interpretation of sine, it's known that sine equals opposite over hypotenuse. So, one can consider a right triangle where the side opposite to angle \(\theta\) is \(x\) and the hypotenuse is \(1\).
02

Find the length of the adjacent side using Pythagorean theorem

After setting up the right triangle, find the length of the other side using the Pythagorean theorem. Since the lengths of the other two sides are \(x\) and \(1\), the length of the remaining side is \(\sqrt{1 - x^2}\).
03

Find the cosine of 胃

In a right triangle, the cosine of an angle is found by dividing the length of the adjacent side by the length of the hypotenuse. From the mentioned triangle, \(\cos(\theta) = \frac{\sqrt{1-x^2}}{1}\).
04

Substituting back

Now, we substitute back \(\theta\) with \(\sin^{-1}(x)\). Hence, \(\cos(\sin^{-1}(x)) = \sqrt{1-x^{2}}\) and the identity is verified.

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

Determine whether the statement is true or false. Justify your answer. If you correctly solve a trigonometric equation to the statement \(\sin x=3.4\), then you can finish solving the equation by using an inverse function.

Use a graphing utility to approximate the solutions (to three decimal places) of the equation in the given interval. $$4 \cos ^{2} x-2 \sin x+1=0, \quad\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$$

A Ferris wheel is built such that the height \(h\) (in feet) above ground of a seat on the wheel at time \(t\) (in minutes) can be modeled by \(h(t)=53+50 \sin \left(\frac{\pi}{16} t-\frac{\pi}{2}\right)\) The wheel makes one revolution every 32 seconds. The ride begins when \(t=0\). (a) During the first 32 seconds of the ride, when will a person on the Ferris wheel be 53 feet above ground? (b) When will a person be at the top of the Ferris wheel for the first time during the ride? If the ride lasts 160 seconds, how many times will a person be at the top of the ride, and at what times?

Find all solutions of the equation in the interval \([0,2 \pi)\). $$2 \cos ^{2} x+\cos x-1=0$$

Find the exact value of each expression. (a) \(\sin \left(\frac{3 \pi}{4}+\frac{5 \pi}{6}\right)\) (b) \(\sin \frac{3 \pi}{4}+\sin \frac{5 \pi}{6}\)
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Mechanics Maths

Read Explanation

Calculus

Read Explanation

Geometry

Read Explanation

Discrete Mathematics

Read Explanation

Applied Mathematics

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.