/*! 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 29 Verify the identity. $$\frac{1... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 29

Verify the identity. $$\frac{1+\sin \theta}{\cos \theta}+\frac{\cos \theta}{1+\sin \theta}=2 \sec \theta$$

Short Answer

Expert verified
After simplifying and applying trigonometric identities in the steps, the LHS of the equation simplifies to \(1 + \sin 2\theta\) which holds for all values of \(2\theta\) making the identity valid.

Step by step solution

01

Simplify the LHS

Let's start with the left-hand side (LHS). Notice there are terms in the form of a fraction. Generally if there are fractions it is good to clear the denominators down to one fraction. Multiply the fractions by \((1+\sin\theta)\) and \(\cos\theta\) respectively which are their respective denominators in order to clear the denominators, and you get \(\cos^2\theta + \sin\theta \cos\theta + \cos\theta + \sin\theta\) after the multiplication and simplification.
02

Apply Trigonometric Identity

Notice in the first term there is \(\cos^2\theta\). A common trigonometric identity is \(\cos^2\theta = 1-\sin^2\theta\). Substituting that identity into the expression replaces \(\cos^2\theta\) and the updated expression is now \(1-\sin^2\theta + \sin\theta \cos\theta + \cos\theta + \sin\theta\). By further simplifying and combining like terms, this gives \(1 + 2\sin\theta \cos\theta\).
03

Apply another Trigonometric Identity

You can see in the simplified expression that there is a term \(2\sin\theta \cos\theta\), which can be rewritten using the double angle identity for sine, \(2\sin\theta \cos\theta = \sin 2\theta\). The LHS now becomes \(1 + \sin 2\theta\). The sine of any angle cannot exceed 1, so the maximum possible value of \(\sin 2\theta\) is 1. Hence, the equation \(1+\sin\theta \le 2\) must hold. Similarly, if we multiply both side of the initial identity by \(\cos\theta\), the right-hand side (RHS) becomes \(2\cos\theta*\sec\theta = 2\), since \(\sec\theta = \frac{1}{\cos\theta}\). The two sides of the equation are now equal, so 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

Explain what would happen if you divided each side of the equation \(\cot x \cos ^{2} x=2 \cot x\) by \(\cot x .\) Is this a correct method to use when solving equations?

Consider the equation \(2 \sin x-1=0\). Explain the similarities and differences between finding all solutions in the interval \(\left[0, \frac{\pi}{2}\right)\), finding all solutions in the interval \([0,2 \pi),\) and finding the general solution.

Find all solutions of the equation in the interval \([0,2 \pi)\). $$2 \sec ^{2} x+\tan ^{2} x-3=0$$

Use inverse functions where needed to find all solutions of the equation in the interval \([0,2 \pi)\). $$\sec ^{2} x-4 \sec x=0$$

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

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Calculus

Read Explanation

Pure Maths

Read Explanation

Geometry

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.