/*! 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 42 Verify the identity. \(\tan \f... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 0: Problem 42

Verify the identity. \(\tan \frac{t}{2}=\frac{1-\cos t}{\sin t}\)

Short Answer

Expert verified
To verify the trigonometric identity \(\tan\frac{t}{2} = \frac{1-\cos t}{\sin t}\), we first use the double-angle formulas to rewrite the left side of the equation as \(\frac{\frac{1-\cos t}{2}}{\frac{1+\cos t}{2}} = \frac{1-\cos t}{1+\cos t}\). Then, we use the Pythagorean identity to simplify the expression, resulting in the fraction \(\frac{(1-\cos t)^2}{\sin^2t}\). Finally, we simplify the expression to obtain the right side of the equation, proving the identity: \(\tan\frac{t}{2} = \frac{1-\cos t}{\sin t}\).

Step by step solution

01

Start with the left side of the equation

We will begin by working with the left side of the equation, which is \(\tan\frac{t}{2}\).
02

Use the double-angle formula for sine and cosine

We know that \(\sin\frac{t}{2}=\frac{1-\cos t}{2}\) and \(\cos\frac{t}{2}=\frac{1+\cos t}{2}\), so we will subsitute these expressions into the left side of the equation: \[ \tan\frac{t}{2} = \frac{\sin\frac{t}{2}}{\cos\frac{t}{2}} \] \[ \frac{\frac{1-\cos t}{2}}{\frac{1+\cos t}{2}} = \frac{1-\cos t}{1+\cos t} \]
03

Use the Pythagorean identity

To continue simplifying the expression, we need the relation between sine and cosine: \(\sin^2x + \cos^2x = 1\). We can use this identity to convert the expression into a fraction where the numerator is sine and the denominator is cosine: \[ \frac{1-\cos t}{1+\cos t} = \frac{1-\cos t}{1+\cos t} \times \frac{1-\cos t}{1-\cos t} = \frac{(1-\cos t)^2}{1-\cos^2 t} \] By applying the Pythagorean identity (\(\sin^2t + \cos^2t = 1\)) on the denominator, we can rewrite it as: \[ \frac{(1-\cos t)^2}{\sin^2 t} \]
04

Simplify and obtain the right side of the equation

Now we can simplify the expression by dividing the top and bottom of the fraction by \((1-\cos t)\), which gives: \[ \frac{1-\cos t}{\sin t} \] This matches the right side of the identity, and we have successfully verified the trigonometric identity: \[ \tan\frac{t}{2} = \frac{1-\cos t}{\sin t} \]

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

Plot the graph of the function \(f\) in an appropriate viewing window. (Note: The answer is not unique.) $$ f(x)=\frac{x^{3}}{x^{3}+1} $$

Find the exact value of the given expression. $$ \cos ^{-1} \frac{1}{2} $$

Find the inverse of \(f .\) Then use a graphing utility to plot the graphs of \(f\) and \(f^{-1}\) using the same viewing window. $$ f(x)=\sqrt[3]{x-1} $$

Plot the graph of the function \(f\) in an appropriate viewing window. (Note: The answer is not unique.) $$ f(x)=x^{2} \sin \frac{1}{x} $$

Show that \(f\) and \(g\) are inverses of each other by verifying that \(f[g(x)]=x\) and \(g[f(x)]=x\). $$ f(x)=\frac{1}{3} x^{3} ; \quad g(x)=\sqrt[3]{3 x} $$
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Decision Maths

Read Explanation

Statistics

Read Explanation

Logic and Functions

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.