Problem 17
Question
Calculate the value of the given inverse trigonometric function at the given point. $$ \operatorname{arcsec}(\sec (-5 \pi / 6)) $$
Step-by-Step Solution
Verified Answer
The result of \( \operatorname{arcsec}(\sec(-5\pi/6)) \) is \( 5\pi/6 \).
1Step 1: Simplify the Function Argument
First, calculate the value of \( \sec(-5\pi/6) \). Recall that \( \sec(\theta) = \frac{1}{\cos(\theta)} \). Hence, \( \sec(-5\pi/6) = \frac{1}{\cos(-5\pi/6)} \). Because cosine is an even function, \( \cos(-5\pi/6) = \cos(5\pi/6) \), and \( \cos(5\pi/6) = -\frac{\sqrt{3}}{2} \). Therefore, \( \sec(-5\pi/6) = -\frac{2}{\sqrt{3}} \), which simplifies to \( -\frac{2\sqrt{3}}{3} \) after rationalizing the denominator.
2Step 2: Use the Definition of Arcsec
The expression \( \operatorname{arcsec}(x) \) is the angle \( \theta \) such that \( \sec(\theta) = x \). Here, we need \( \theta \) for which \( \sec(\theta) = -\frac{2\sqrt{3}}{3} \). This angle \( \theta \) must lie in the range \([0, \pi) \) except \( \frac{\pi}{2} \).
3Step 3: Find the Angle in the Range
Since secant is negative and \( \sec(\theta) = -\frac{2\sqrt{3}}{3} \), \( \cos(\theta) = -\sqrt{3}/2 \). The reference angle that corresponds to \( \cos(\theta) = \sqrt{3}/2 \) is \( \pi/6 \). So, the angles where \( \cos(\theta) = -\sqrt{3}/2 \) are in the second quadrant as \( 5\pi/6 \) because cosines of angles are negative in the second quadrant.
4Step 4: Verify the Solution
Finally, let's verify by checking that \( \sec(5\pi/6) = -\frac{2\sqrt{3}}{3} \), which matches our requirement. Therefore, the angle that satisfies the equation is \( \theta = 5\pi/6 \).
Key Concepts
arcsecsecant functioncosine functionunit circle
arcsec
The function \( \operatorname{arcsec} \), known as the inverse secant, finds an angle whose secant is a given number. If you know \( \sec(\theta) = x \), then \( \operatorname{arcsec}(x) = \theta \). This inverse function helps us transform values back into angles, effectively reversing the secant function process.
\[ \operatorname{arcsec}(x) = \theta \text { where } \sec(\theta) = x \]
The range of \( \operatorname{arcsec}(x) \) is crucial. Typically, the angles are restricted to \([0, \pi) \) excluding \( \frac{\pi}{2} \), because secant is undefined there. These restrictions ensure that each input has exactly one output, maintaining it as a proper function.
Understanding the behavior of arcsec is vital when working with inverse trigonometric operations, particularly if you're computing it manually or verifying solutions.
\[ \operatorname{arcsec}(x) = \theta \text { where } \sec(\theta) = x \]
The range of \( \operatorname{arcsec}(x) \) is crucial. Typically, the angles are restricted to \([0, \pi) \) excluding \( \frac{\pi}{2} \), because secant is undefined there. These restrictions ensure that each input has exactly one output, maintaining it as a proper function.
Understanding the behavior of arcsec is vital when working with inverse trigonometric operations, particularly if you're computing it manually or verifying solutions.
secant function
The secant function, denoted as \( \sec(\theta) \), is one of the six fundamental trigonometric functions. It is defined as the reciprocal of the cosine function:
\[ \sec(\theta) = \frac{1}{\cos(\theta)} \]
This function is particularly useful for solving trigonometry problems involving right triangles and circles.
Secant is closely connected to the unit circle, offering insight into how trigonometric functions operate using geometry. Realizing the domains and periodic nature of the secant function is central to mastering its use in trigonometrics.
\[ \sec(\theta) = \frac{1}{\cos(\theta)} \]
This function is particularly useful for solving trigonometry problems involving right triangles and circles.
- When \( \cos(\theta) \) approaches zero, \( \sec(\theta) \) becomes undefined, occurring at angles \( \frac{\pi}{2} + k\pi \), \( k \in \mathbb{Z} \).
- Secant is periodic with a period of \( 2\pi \), meaning its values repeat every \( 2\pi \) radians.
Secant is closely connected to the unit circle, offering insight into how trigonometric functions operate using geometry. Realizing the domains and periodic nature of the secant function is central to mastering its use in trigonometrics.
cosine function
The cosine function, expressed as \( \cos(\theta) \), is another cornerstone of trigonometry. It signifies the ratio of the adjacent side over the hypotenuse in a right triangle and features prominently within the unit circle.
Understanding these properties not only aids in solving trigonometric equations but also provides foundational insights for more advanced applications like engineering and physics.
- \( \cos(\theta) = 1 \) at \( \theta = 0 \) and \( \cos(\theta) = -1 \) at \( \theta = \pi \).
- It is an even function, meaning \( \cos(-\theta) = \cos(\theta) \), contributing to symmetric behavior about the \( y\)-axis.
- Its period is \( 2\pi \), reflecting how values return to the start every \( 2\pi \).
Understanding these properties not only aids in solving trigonometric equations but also provides foundational insights for more advanced applications like engineering and physics.
unit circle
A central tool in trigonometry, the unit circle helps illustrate how trigonometric functions interact with angles and coordinates. It is a circle with a radius of 1 centered at the origin of a coordinate plane.
Utilizing the unit circle simplifies understanding the cyclical nature of trig functions and enables easy calculation of function values at key angles like \( 0, \frac{\pi}{2}, \pi, \) and \( \frac{3\pi}{2} \). This visualization is key when learning or teaching trigonometry, making it a go-to resource for both students and educators.
- Angles in the unit circle are measured from the positive \( x\)-axis, moving counterclockwise.
- The coordinates of any point on the circle correspond to \( (\cos(\theta), \sin(\theta)) \) for an angle \( \theta \).
- Symmetries in the unit circle allow one to easily determine the signs of trigonometric functions based on the quadrant.
Utilizing the unit circle simplifies understanding the cyclical nature of trig functions and enables easy calculation of function values at key angles like \( 0, \frac{\pi}{2}, \pi, \) and \( \frac{3\pi}{2} \). This visualization is key when learning or teaching trigonometry, making it a go-to resource for both students and educators.
Other exercises in this chapter
Problem 16
Describes the position of a moving body at time \(t\). Determine whether, at time \(t=4,\) the body is moving forward, backward, or neither. $$ p(t)=1 / t $$
View solution Problem 16
Find a point \(x\) where \(f^{\prime}(x)=6\). $$ f(x)=12 \sin (x) $$
View solution Problem 17
Use the method of increments to estimate the value of \(f(x)\) at the given value of \(x\) using the known value \(f(c)\) $$ f(x)=e^{x}, c=0, x=-0.17 $$
View solution Problem 17
Use implicit differentiation to find the tangent line to the given curve at the given point \(P_{0}\). \(\sin ^{2}(x)+\cos ^{2}(y)=5 / 4 \quad P_{0}=(\pi / 3, \
View solution