Chapter 6: Problem 6
Show using a counterexample that the following is not an identity: \(\sin (x-y)=\sin x-\sin y\).
Short Answer
Expert verified
The identity \ \sin(x-y) \eq \sin x - \sin y \ by using x = π/2 and y = π/3 as a counterexample.
Step by step solution
01
Select Specific Values for x and y
Choose specific numerical values for the variables x and y that will work to show that the given equation is not always true. Let's use x = π/4 and y = π/4.
02
Calculate the Left-Hand Side (LHS)
Substitute x = π/4 and y = π/4 into the left-hand side of the equation \ \ \( \text{LHS} = \sin(x-y) \) \ \ Calculate \( \text{LHS} = \sin( \frac{\pi}{4} - \frac{\pi}{4} ) = \sin(0) = 0 \) \
03
Calculate the Right-Hand Side (RHS)
Substitute x = π/4 and y = π/4 into the right-hand side of the equation \ \ \ \( \text{RHS} = \sin x - \sin y \) \ \ \ Calculate \( \text{RHS} = \sin\( \frac{ \pi }{4} \) - \sin\( \frac{ \pi }{4} \) = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = 0 \) \
04
Choose Another Set of Values for a Valid Counterexample
To properly show the identity fails, choose different values for x and y. Let's use x = π/2 and y = π/3.
05
Calculate LHS with New x and y
Substitute x = π/2 and y = π/3 into the left-hand side of the equation \( \text{LHS} = \sin( x - y ) \) \ Calculate \( \text{LHS} = \sin( \frac{ \pi }{2} - \frac{ \pi }{3} ) = \sin( \frac{ \pi }{6} ) = \frac{ 1 }{2} \) \
06
Calculate RHS with New x and y
Substitute x = π/2 and y = π/3 into the right-hand side of the equation \( \text{RHS} = \sin x - \sin y \) \ Calculate \( \text{RHS} = \sin( \frac{ \pi }{2} ) - \sin( \frac{ \pi }{3} ) = 1 - \frac{ \sqrt{3} }{2} eq \frac{ 1 }{2} \). So \text{RHS} \eq \text{LHS}\
07
Conclude
Since \( \sin(x-y) \eq \sin x - \sin y \) for x = π/2 and y = π/3, it can be concluded that \ \sin(x-y) \eq \sin x - \sin y\ for any arbitrary x and y.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Trigonometric Counterexample
A counterexample is a specific case for which a general assumption does not hold true. In trigonometry, it is often used to disprove identities. By showing that an equation fails for certain values, we can prove that it is not a universal identity. In this exercise, we use the specific values \(\frac{\text{π}}{2}\) and \(\frac{\text{π}}{3}\) to demonstrate that the equation \(\text{sin}(x - y) = \text{sin} x - \text{sin} y\) does not always hold true. The left-hand side (LHS) and right-hand side (RHS) calculations yield different results, proving the equation invalid for those values.
Trigonometric Identities
Trigonometric identities are equations involving trigonometric functions that are true for all values within their domains. Some common identities include:
• \(\text{sin}^2(x) + \text{cos}^2(x) = 1\)
• \(\text{tan}(x) = \frac{\text{sin}(x)}{\text{cos}(x)}\)
These identities help in simplifying trigonometric expressions and solving equations. However, not every trigonometric equation is an identity. For example, \(\text{sin}(x - y) = \text{sin} x - \text{sin} y\) is not an identity, as shown by the counterexample.
• \(\text{sin}^2(x) + \text{cos}^2(x) = 1\)
• \(\text{tan}(x) = \frac{\text{sin}(x)}{\text{cos}(x)}\)
These identities help in simplifying trigonometric expressions and solving equations. However, not every trigonometric equation is an identity. For example, \(\text{sin}(x - y) = \text{sin} x - \text{sin} y\) is not an identity, as shown by the counterexample.
Sine Function
The sine function, denoted as \(\text{sin}(x)\), is one of the primary trigonometric functions. It gives the y-coordinate of a point on the unit circle as the angle \(x\) (measured in radians) from the positive x-axis to that point. For any angle \(x\):
• \(\text{sin}(0) = 0\)
• \(\text{sin}(\frac{\text{π}}{2}) = 1\)
• \(\text{sin}(\text{π}) = 0\)
This periodic function has symmetry and repeats every \(2 \text{π}\). Understanding the sine function is crucial in solving trigonometric problems and verifying identities.
• \(\text{sin}(0) = 0\)
• \(\text{sin}(\frac{\text{π}}{2}) = 1\)
• \(\text{sin}(\text{π}) = 0\)
This periodic function has symmetry and repeats every \(2 \text{π}\). Understanding the sine function is crucial in solving trigonometric problems and verifying identities.
Identity Verification
To verify or disprove a trigonometric identity, we often compare the LHS and RHS of the equation for specific values. In this exercise, we evaluated both sides for:
• \(x = \frac{\text{π}}{4}, y = \frac{\text{π}}{4}\)
Both sides equaled zero, supporting the equation temporarily. Another test with:
• \(x = \frac{\text{π}}{2}, y = \frac{\text{π}}{3}\)
yielded different values on LHS and RHS, disproving the identity. This method ensures the reliability of identities across all valid values, solidifying the understanding of trigonometric functions and their properties.
• \(x = \frac{\text{π}}{4}, y = \frac{\text{π}}{4}\)
Both sides equaled zero, supporting the equation temporarily. Another test with:
• \(x = \frac{\text{π}}{2}, y = \frac{\text{π}}{3}\)
yielded different values on LHS and RHS, disproving the identity. This method ensures the reliability of identities across all valid values, solidifying the understanding of trigonometric functions and their properties.