Module 11, Trigonometric Identities Assignment In mathematics, an identity is an equation that is always true. When you learned how to factor expressions and equations, you worked with several algebraic identities, including (a + b)² = a² + 2ab + b2 and (a - b)³ = (a - b)(a² + ab + b2). These identities helped you to rewrite and simplify more complex expressions or equations. This, in turn, allowed you to more easily solve problems. In this module you were introduced to trigonometric identities, including reciprocal identities, quotient identities, Pythagorean identities, and others. Like algebraic identities, you can use trigonometric identities to help you rewrite and simplify more complex expressions or equations in order to more easily solve problems. You can also use these identities to verify whether trigonometric equations represent identities. There are several ways to show an equation may be an identity, but to prove it is an identity, you must verify it. what you know about trigonometric identities to explore the questions below. 1. Amna needs to decide if the equation = 1- cos 0 is an identity. Choose 3 possible values for 0, substitute them into this equation, and simplify. What do your results indicate about sin² 0 1+cos e the equation sin² 0 1+cos@ 1 cos 0? Show all your work. -

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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### Module 11. Trigonometric Identities Assignment

In mathematics, an identity is an equation that is always true. When you learned how to factor expressions and equations, you worked with several algebraic identities, including:

\[ (a + b)^2 = a^2 + 2ab + b^2 \]
\[ (a - b)^3 = (a - b)(a^2 + ab + b^2) \]

These identities helped you to rewrite and simplify more complex expressions or equations. This, in turn, allowed you to more easily solve problems.

In this module, you were introduced to trigonometric identities, including reciprocal identities, quotient identities, Pythagorean identities, and others. Like algebraic identities, you can use trigonometric identities to help you rewrite and simplify more complex expressions or equations in order to more easily solve problems. You can also use these identities to verify whether trigonometric equations represent identities. There are several ways to show an equation may be an identity, but to prove it is an identity, you must verify it.

Use what you know about trigonometric identities to explore the questions below.

1. Amna needs to decide if the equation 
\[
\frac{\sin^2 \theta}{1+\cos \theta} = 1 - \cos \theta
\]
is an identity. Choose 3 possible values for \( \theta \), substitute them into this equation, and simplify. What do your results indicate about the equation 
\[
\frac{\sin^2 \theta}{1+\cos \theta} = 1 - \cos \theta
\]
? Show all your work.

- When \(\theta = 0\)
\[
0/2 = 1 - 1
\]
- When …

(Continue with steps to be filled in depending on the chosen values for \( \theta \))

---

This section appears on a website as part of a module aimed at teaching and verifying trigonometric identities. The context provides guidance on using algebraic techniques to explore and confirm trigonometric expressions. By following along, students will reinforce their understanding of fundamental mathematical principles.
Transcribed Image Text:### Module 11. Trigonometric Identities Assignment In mathematics, an identity is an equation that is always true. When you learned how to factor expressions and equations, you worked with several algebraic identities, including: \[ (a + b)^2 = a^2 + 2ab + b^2 \] \[ (a - b)^3 = (a - b)(a^2 + ab + b^2) \] These identities helped you to rewrite and simplify more complex expressions or equations. This, in turn, allowed you to more easily solve problems. In this module, you were introduced to trigonometric identities, including reciprocal identities, quotient identities, Pythagorean identities, and others. Like algebraic identities, you can use trigonometric identities to help you rewrite and simplify more complex expressions or equations in order to more easily solve problems. You can also use these identities to verify whether trigonometric equations represent identities. There are several ways to show an equation may be an identity, but to prove it is an identity, you must verify it. Use what you know about trigonometric identities to explore the questions below. 1. Amna needs to decide if the equation \[ \frac{\sin^2 \theta}{1+\cos \theta} = 1 - \cos \theta \] is an identity. Choose 3 possible values for \( \theta \), substitute them into this equation, and simplify. What do your results indicate about the equation \[ \frac{\sin^2 \theta}{1+\cos \theta} = 1 - \cos \theta \] ? Show all your work. - When \(\theta = 0\) \[ 0/2 = 1 - 1 \] - When … (Continue with steps to be filled in depending on the chosen values for \( \theta \)) --- This section appears on a website as part of a module aimed at teaching and verifying trigonometric identities. The context provides guidance on using algebraic techniques to explore and confirm trigonometric expressions. By following along, students will reinforce their understanding of fundamental mathematical principles.
### Trigonometric Identity Verification

#### Problem Statement
4. If possible, algebraically verify 
\[ \frac{\sin^2 \theta}{1 + \cos \theta} = 1 - \cos \theta \]
in a different way. Show all your steps. If not possible, explain why.

#### Solution

To verify the identity \[ \frac{\sin^2 \theta}{1 + \cos \theta} = 1 - \cos \theta \], we can take the following steps:

1. **Left-Hand Side (LHS) Simplification:**

   Let's start with the left-hand side (LHS) of the equation:
   \[ \frac{\sin^2 \theta}{1 + \cos \theta} \]

   Using the Pythagorean identity: \(\sin^2 \theta = 1 - \cos^2 \theta\), we substitute:
   \[ \frac{1 - \cos^2 \theta}{1 + \cos \theta} \]

2. **Factoring the Numerator:**
   
   Next, we factor \(1 - \cos^2 \theta\) as a difference of squares:
   \[ 1 - \cos^2 \theta = (1 - \cos \theta)(1 + \cos \theta) \]

   So the LHS becomes:
   \[ \frac{(1 - \cos \theta)(1 + \cos \theta)}{1 + \cos \theta} \]

3. **Canceling the Common Factor:**

   Notice that \((1 + \cos \theta)\) in the numerator and the denominator are the same, so we can cancel this common factor:
   \[ \frac{(1 - \cos \theta)(1 + \cos \theta)}{1 + \cos \theta} = 1 - \cos \theta \]

4. **Conclusion:**

   Now we have the simplified form of the LHS:
   \[ 1 - \cos \theta \]

   Which is exactly the right-hand side (RHS) of the original equation:
   \[ 1 - \cos \theta \]

   Therefore, we have verified the identity \( \frac{\sin^2 \theta}{1 + \cos \theta} = 1 - \cos \theta \).

By showing all the steps involved, we have algebraically demonstrated that
Transcribed Image Text:### Trigonometric Identity Verification #### Problem Statement 4. If possible, algebraically verify \[ \frac{\sin^2 \theta}{1 + \cos \theta} = 1 - \cos \theta \] in a different way. Show all your steps. If not possible, explain why. #### Solution To verify the identity \[ \frac{\sin^2 \theta}{1 + \cos \theta} = 1 - \cos \theta \], we can take the following steps: 1. **Left-Hand Side (LHS) Simplification:** Let's start with the left-hand side (LHS) of the equation: \[ \frac{\sin^2 \theta}{1 + \cos \theta} \] Using the Pythagorean identity: \(\sin^2 \theta = 1 - \cos^2 \theta\), we substitute: \[ \frac{1 - \cos^2 \theta}{1 + \cos \theta} \] 2. **Factoring the Numerator:** Next, we factor \(1 - \cos^2 \theta\) as a difference of squares: \[ 1 - \cos^2 \theta = (1 - \cos \theta)(1 + \cos \theta) \] So the LHS becomes: \[ \frac{(1 - \cos \theta)(1 + \cos \theta)}{1 + \cos \theta} \] 3. **Canceling the Common Factor:** Notice that \((1 + \cos \theta)\) in the numerator and the denominator are the same, so we can cancel this common factor: \[ \frac{(1 - \cos \theta)(1 + \cos \theta)}{1 + \cos \theta} = 1 - \cos \theta \] 4. **Conclusion:** Now we have the simplified form of the LHS: \[ 1 - \cos \theta \] Which is exactly the right-hand side (RHS) of the original equation: \[ 1 - \cos \theta \] Therefore, we have verified the identity \( \frac{\sin^2 \theta}{1 + \cos \theta} = 1 - \cos \theta \). By showing all the steps involved, we have algebraically demonstrated that
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