Problem 48
Question
Which of the expressions are defined, and which are not? Give reasons for your answers. a. \(\cot ^{-1}(-1 / 2)\) b. \(\cos ^{-1}(-5)\)
Step-by-Step Solution
Verified Answer
Expression a is defined; expression b is not defined.
1Step 1: Determine the Valid Range for Each Function
The functions given here are inverse trigonometric functions. For \( \cot^{-1}(x) \), the valid input range is \(-\infty < x < \infty\). For \( \cos^{-1}(x) \), the valid input range is \(-1 \, \leq x \leq \, 1\).
2Step 2: Evaluate Expression a
The expression is \( \cot^{-1}(-1/2) \). Since the input \( -1/2 \) falls within the range \(-\infty < x < \infty\), the expression is defined.
3Step 3: Evaluate Expression b
The expression is \( \cos^{-1}(-5) \). Here, the input \( -5 \) does not fall within the range \(-1 \, \leq x \leq \, 1\). Thus, the expression is not defined.
Key Concepts
Understanding CotangentExploring ArccosineValid Input Range for Inverse Trigonometric Functions
Understanding Cotangent
The cotangent function, often represented as "cot," is the reciprocal of the tangent function in trigonometry. In a right triangle, the cotangent of an angle is the ratio of the length of the adjacent side to the length of the opposite side. The formula is: \[\cot(\theta) = \frac{\text{adjacent}}{\text{opposite}}\]
For the inverse cotangent function, represented as \( \cot^{-1} \), it reverses this back to an angle from the ratio. Unlike the tangent function, whose range is typically limited, \( \cot^{-1}(x) \) can handle any real number \( x \). This means that any fraction, positive or negative, or even whole numbers can be valid inputs for \( \cot^{-1} \). The output of \( \cot^{-1} \) is usually an angle in radians which falls within the range of \( (0, \pi) \).
For the inverse cotangent function, represented as \( \cot^{-1} \), it reverses this back to an angle from the ratio. Unlike the tangent function, whose range is typically limited, \( \cot^{-1}(x) \) can handle any real number \( x \). This means that any fraction, positive or negative, or even whole numbers can be valid inputs for \( \cot^{-1} \). The output of \( \cot^{-1} \) is usually an angle in radians which falls within the range of \( (0, \pi) \).
- The function does not have strict input restrictions, making it quite versatile.
- Given this flexibility, calculations involving \( \cot^{-1} \) often yield useful insights without limitation.
Exploring Arccosine
The arccosine function, denoted as "arccos," is the inverse of the cosine function. In a right triangle, the cosine of an angle is the ratio of the adjacent side to the hypotenuse, expressed as: \[\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}\]
To find the angle from a given cosine value, we use \( \cos^{-1} \). Unlike \( \cot^{-1} \), \( \cos^{-1}(x) \) only accepts values within a narrow range.
To find the angle from a given cosine value, we use \( \cos^{-1} \). Unlike \( \cot^{-1} \), \( \cos^{-1}(x) \) only accepts values within a narrow range.
- The valid input range for \( \cos^{-1}(x) \) is from \(-1\) to \(1\).
- This is because the cosine of any angle cannot exceed these limits, as the adjacent side cannot be longer than the hypotenuse.
Valid Input Range for Inverse Trigonometric Functions
When working with inverse trigonometric functions, understanding valid input ranges is crucial for determining whether expressions are defined. Each inverse trigonometric function has specific input constraints based on the real-world interpretation of angles.
- For \( \cot^{-1}(x) \), there are no input restrictions. The function can handle any real input.
- For \( \cos^{-1}(x) \), inputs are tightly restricted to the range of \(-1 \leq x \leq 1\).
Other exercises in this chapter
Problem 47
Find the first and second derivatives of the functions. $$r=\frac{(\theta-1)\left(\theta^{2}+\theta+1\right)}{\theta^{3}}$$
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Use logarithmic differentiation to find the derivative of \(y\) with respect to the given independent variable. $$y=\frac{1}{t(t+1)(t+2)}$$
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Write a differential formula that estimates the given change in volume or surface area. The change in the lateral surface area \(S=\pi r \sqrt{r^{2}+h^{2}}\) of
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Find the derivatives of the functions in Exercises \(23-50\). $$q=\cot \left(\frac{\sin t}{t}\right)$$
View solution