ove that the following identity is true. sin t 1+ cos t -1- cos t sin t We begin on the left side of the equation by multiplying the numerator and denominator by the conjugate of the denominator. We can then use a Pythagorean Ident reduce. 1- cos t sin t 1+ cos t sin t 1+ cos t %3D sin t(1 - cos t) 1. sin t(1 - cos t) sin 1- cos t sin t
ove that the following identity is true. sin t 1+ cos t -1- cos t sin t We begin on the left side of the equation by multiplying the numerator and denominator by the conjugate of the denominator. We can then use a Pythagorean Ident reduce. 1- cos t sin t 1+ cos t sin t 1+ cos t %3D sin t(1 - cos t) 1. sin t(1 - cos t) sin 1- cos t sin t
Trigonometry (11th Edition)
11th Edition
ISBN:9780134217437
Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Chapter1: Trigonometric Functions
Section: Chapter Questions
Problem 1RE:
1. Give the measures of the complement and the supplement of an angle measuring 35°.
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Question
![**Prove that the following identity is true.**
\[
\frac{\sin t}{1 + \cos t} = \frac{1 - \cos t}{\sin t}
\]
We begin on the left side of the equation by multiplying the numerator and denominator by the conjugate of the denominator. We can then use a Pythagorean identity on the denominator, and reduce.
\[
\frac{\sin t}{1 + \cos t} = \frac{\sin t}{1 + \cos t} \times \frac{1 - \cos t}{1 - \cos t} = \frac{\sin t (1 - \cos t)}{(1 + \cos t)(1 - \cos t)}
\]
The denominator becomes:
\[
(1 + \cos t)(1 - \cos t) = 1 - \cos^2 t
\]
\[
= \sin^2 t
\]
Therefore,
\[
\frac{\sin t (1 - \cos t)}{\sin^2 t} = \frac{1 - \cos t}{\sin t}
\]
Thus proving the identity:
\[
\frac{\sin t}{1 + \cos t} = \frac{1 - \cos t}{\sin t}
\]
**Explanation of the Process:**
1. Begin with the left side: \(\frac{\sin t}{1 + \cos t}\).
2. Multiply by the conjugate: \(\frac{1 - \cos t}{1 - \cos t}\) to clear the denominator.
3. Simplify the denominator using the Pythagorean identity: \(1 - \cos^2 t = \sin^2 t\).
4. Simplify the resulting expression to match the right side of the equation.
This confirms the identity is true.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F792a3d2b-2c57-4584-9301-f01e5203845f%2F7140908e-4d73-49cc-99e5-ff45ffe2c8da%2Fs0p5g9_processed.png&w=3840&q=75)
Transcribed Image Text:**Prove that the following identity is true.**
\[
\frac{\sin t}{1 + \cos t} = \frac{1 - \cos t}{\sin t}
\]
We begin on the left side of the equation by multiplying the numerator and denominator by the conjugate of the denominator. We can then use a Pythagorean identity on the denominator, and reduce.
\[
\frac{\sin t}{1 + \cos t} = \frac{\sin t}{1 + \cos t} \times \frac{1 - \cos t}{1 - \cos t} = \frac{\sin t (1 - \cos t)}{(1 + \cos t)(1 - \cos t)}
\]
The denominator becomes:
\[
(1 + \cos t)(1 - \cos t) = 1 - \cos^2 t
\]
\[
= \sin^2 t
\]
Therefore,
\[
\frac{\sin t (1 - \cos t)}{\sin^2 t} = \frac{1 - \cos t}{\sin t}
\]
Thus proving the identity:
\[
\frac{\sin t}{1 + \cos t} = \frac{1 - \cos t}{\sin t}
\]
**Explanation of the Process:**
1. Begin with the left side: \(\frac{\sin t}{1 + \cos t}\).
2. Multiply by the conjugate: \(\frac{1 - \cos t}{1 - \cos t}\) to clear the denominator.
3. Simplify the denominator using the Pythagorean identity: \(1 - \cos^2 t = \sin^2 t\).
4. Simplify the resulting expression to match the right side of the equation.
This confirms the identity is true.
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