sin* (t) – cos*(t) – cos²(t) = 1 sin²(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°.
Related questions
Question
![**Objective:**
Show that
\[
\frac{\sin^4(t) - \cos^4(t)}{\sin^2(t) - \cos^2(t)} = 1
\]
**Explanation:**
We start by recognizing the expression \(\sin^4(t) - \cos^4(t)\) as a difference of squares:
\[
\sin^4(t) - \cos^4(t) = (\sin^2(t))^2 - (\cos^2(t))^2 = (\sin^2(t) - \cos^2(t))(\sin^2(t) + \cos^2(t))
\]
Since \(\sin^2(t) + \cos^2(t) = 1\) (using the Pythagorean identity), the expression simplifies to:
\[
\sin^4(t) - \cos^4(t) = (\sin^2(t) - \cos^2(t)) \cdot 1
\]
Thus, we can rewrite the original expression as:
\[
\frac{(\sin^2(t) - \cos^2(t)) \cdot 1}{\sin^2(t) - \cos^2(t)}
\]
This simplifies to 1, given that \(\sin^2(t) - \cos^2(t) \neq 0\):
\[
\frac{\sin^2(t) - \cos^2(t)}{\sin^2(t) - \cos^2(t)} = 1
\]
Therefore, the equality is proven.
---
In this mathematical proof, recognizing and applying the identity for the difference of squares and the Pythagorean identity are key steps in simplifying the expression.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5c5734d7-71be-4626-8cb5-227ca6c7fb69%2F0c4bf571-fdd4-479e-b021-bbc4fc90af7a%2F6maouho_processed.png&w=3840&q=75)
Transcribed Image Text:**Objective:**
Show that
\[
\frac{\sin^4(t) - \cos^4(t)}{\sin^2(t) - \cos^2(t)} = 1
\]
**Explanation:**
We start by recognizing the expression \(\sin^4(t) - \cos^4(t)\) as a difference of squares:
\[
\sin^4(t) - \cos^4(t) = (\sin^2(t))^2 - (\cos^2(t))^2 = (\sin^2(t) - \cos^2(t))(\sin^2(t) + \cos^2(t))
\]
Since \(\sin^2(t) + \cos^2(t) = 1\) (using the Pythagorean identity), the expression simplifies to:
\[
\sin^4(t) - \cos^4(t) = (\sin^2(t) - \cos^2(t)) \cdot 1
\]
Thus, we can rewrite the original expression as:
\[
\frac{(\sin^2(t) - \cos^2(t)) \cdot 1}{\sin^2(t) - \cos^2(t)}
\]
This simplifies to 1, given that \(\sin^2(t) - \cos^2(t) \neq 0\):
\[
\frac{\sin^2(t) - \cos^2(t)}{\sin^2(t) - \cos^2(t)} = 1
\]
Therefore, the equality is proven.
---
In this mathematical proof, recognizing and applying the identity for the difference of squares and the Pythagorean identity are key steps in simplifying the expression.
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