Problem 56
Question
Show that the equation is not an Identity. $$\log \left(\frac{1}{\sin t}\right)=\frac{1}{\log \sin t}$$
Step-by-Step Solution
Verified Answer
The equation is not an identity as it fails for \( t = \frac{\pi}{4} \).
1Step 1: Review the Equation
We need to evaluate the equation \( \log \left(\frac{1}{\sin t}\right) = \frac{1}{\log \sin t} \) to determine if it's an identity. An identity holds true for all values in the domain of the variables involved.
2Step 2: Simplify the Left Side
The left side of the equation is \( \log \left(\frac{1}{\sin t}\right) \). Using the log property \( \log(\frac{1}{x}) = -\log x \), this can be simplified to \( -\log(\sin t) \).
3Step 3: Analyze the Right Side
The right side of the equation is \( \frac{1}{\log \sin t} \). It's simply an inverse of the logarithm of \( \sin t \), and it doesn't simplify further.
4Step 4: Test for Specific Values of t
To test if the equation is an identity, choose specific values for \( t \). Let's test \( t = \frac{\pi}{4} \) where \( \sin t = \frac{1}{\sqrt{2}} \).
5Step 5: Calculate Both Sides for t = \( \frac{\pi}{4} \)
Plugging \( t = \frac{\pi}{4} \) into the simplified forms: - Left Side \( = -\log(\frac{1}{\sqrt{2}}) = \log(\sqrt{2}) \)- Right Side \( = \frac{1}{\log(\frac{1}{\sqrt{2}})} = \frac{1}{-\log(\sqrt{2})} = -\frac{1}{\log(\sqrt{2})} \)These are not equal.
6Step 6: Conclusion
Since at least one instance was found where \( -\log(\sqrt{2}) eq -\frac{1}{\log(\sqrt{2})} \), the original equation is not an identity. An identity must hold for all permissible values of \( t \).
Key Concepts
Logarithm PropertiesTrigonometric FunctionsMathematical Identities
Logarithm Properties
Logarithms are a way to express powers and roots in a different form. They have unique properties that help with simplifying equations and expressions. One of the key properties of logarithms is the reciprocal property, which states that the logarithm of the reciprocal of a number is the negative logarithm of the number itself. This is expressed as:
In the exercise, this property was utilized to simplify the left side of the equation \( \log\left(\frac{1}{\sin t}\right) \) to \( -\log(\sin t) \). This process is an important step, showing why the left side of the equation does not match with the right side, which is \( \frac{1}{\log\sin t} \). The left involves a negation, whereas the right is simply the reciprocal of the logarithm.
Misunderstanding or misapplying logarithm properties can often lead to incorrect conclusions about mathematical relationships, such as erroneously assuming the equation is an identity.
- \( \log\left(\frac{1}{x}\right) = -\log x \).
In the exercise, this property was utilized to simplify the left side of the equation \( \log\left(\frac{1}{\sin t}\right) \) to \( -\log(\sin t) \). This process is an important step, showing why the left side of the equation does not match with the right side, which is \( \frac{1}{\log\sin t} \). The left involves a negation, whereas the right is simply the reciprocal of the logarithm.
Misunderstanding or misapplying logarithm properties can often lead to incorrect conclusions about mathematical relationships, such as erroneously assuming the equation is an identity.
Trigonometric Functions
Trigonometric functions are fundamental in trigonometry and essential for understanding relationships in right-angled triangles and the unit circle. The six main trigonometric functions include sine, cosine, tangent, cosecant, secant, and cotangent.
The exercise primarily focuses on the sine function, denoted as \( \sin t \). It is defined as the ratio of the length of the opposite side to the hypotenuse in a right-angled triangle. In the unit circle, it represents the y-coordinate of a point on the circle.
The exercise primarily focuses on the sine function, denoted as \( \sin t \). It is defined as the ratio of the length of the opposite side to the hypotenuse in a right-angled triangle. In the unit circle, it represents the y-coordinate of a point on the circle.
- Understanding sine helps solve the equation since its values range from -1 to 1.
- It is periodic with a period of \( 2\pi \), meaning it repeats every \( 2\pi \) radians.
Mathematical Identities
Mathematical identities are equations that are universally true for all values of the variable within their domain. Identifying whether an equation is an identity is crucial when simplifying or transforming expressions.
To test if an equation is an identity, it must satisfy the equation for all possible values of the variable. In our exercise, the given equation \( \log\left(\frac{1}{\sin t}\right) = \frac{1}{\log \sin t} \) was tested with a specific value of \( t \) to determine if it holds as an identity. This process revealed a discrepancy, proving it is not an identity.
It's important to understand that mathematical identities are foundational in solving equations because they help simplify complex relationships into universally valid forms. By comprehensively analyzing an equation across different values, we can confirm whether it is always true, which is essential for problem-solving and further applications in mathematics.
To test if an equation is an identity, it must satisfy the equation for all possible values of the variable. In our exercise, the given equation \( \log\left(\frac{1}{\sin t}\right) = \frac{1}{\log \sin t} \) was tested with a specific value of \( t \) to determine if it holds as an identity. This process revealed a discrepancy, proving it is not an identity.
It's important to understand that mathematical identities are foundational in solving equations because they help simplify complex relationships into universally valid forms. By comprehensively analyzing an equation across different values, we can confirm whether it is always true, which is essential for problem-solving and further applications in mathematics.
Other exercises in this chapter
Problem 55
Show that the equation is not an Identity. $$(\sin \theta+\cos \theta)^{2}=\sin ^{2} \theta+\cos ^{2} \theta$$
View solution Problem 55
Find the solutions of the equation that are in the interval \([0,2 \pi)\). $$2 \tan t-\sec ^{2} t=0$$
View solution Problem 56
Exer. \(55-56:\) (a) Compare the decimal approximations of both sides of equation ( 1). (b) Find the acute angle \(x\) such that equation ( 2) is an identity. (
View solution Problem 56
Find the solutions of the equation that are in the interval \([0,2 \pi)\). $$\tan \theta+\sec \theta=1$$
View solution