Skip to main content
Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 7 - Transcendental FunctionsProblem 7.6.6b
Chapter 7, Problem 7.6.6b

Use reference triangles in an appropriate quadrant to find the angles in Exercises 1–8.
6. b. arccsc(-2/√3)

Verified step by step guidance
1
Recall that \( \arccsc(x) \) is the inverse cosecant function, which gives an angle \( \theta \) such that \( \csc(\theta) = x \). Here, we want to find \( \theta = \arccsc\left(-\frac{2}{\sqrt{3}}\right) \).
Rewrite the cosecant in terms of sine: since \( \csc(\theta) = \frac{1}{\sin(\theta)} \), we have \( \sin(\theta) = \frac{1}{\csc(\theta)} = \frac{1}{-\frac{2}{\sqrt{3}}} = -\frac{\sqrt{3}}{2} \).
Determine the reference angle by considering the positive value of sine: \( \sin(\theta_{ref}) = \frac{\sqrt{3}}{2} \). From the unit circle, this corresponds to an angle of \( \frac{\pi}{3} \) radians (or 60 degrees).
Since the sine value is negative and cosecant is negative, identify the quadrant where sine is negative. Sine is negative in Quadrants III and IV. The principal range of \( \arccsc(x) \) is \( [-\frac{\pi}{2}, 0) \cup (0, \frac{\pi}{2}] \) excluding zero, so the angle must be in Quadrant IV (negative angle between 0 and \( -\frac{\pi}{2} \)).
Therefore, the angle \( \theta \) is the negative of the reference angle, so \( \theta = -\frac{\pi}{3} \). This is the angle in the appropriate quadrant corresponding to \( \arccsc\left(-\frac{2}{\sqrt{3}}\right) \).

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
2m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Inverse Trigonometric Functions

Inverse trigonometric functions, like arccsc, return the angle whose trigonometric ratio equals a given value. For arccsc(x), it gives the angle whose cosecant is x. Understanding their domains and ranges is essential to find correct angle values.
Recommended video:
06:35
Derivatives of Other Inverse Trigonometric Functions

Reference Triangles

Reference triangles are right triangles used to relate trigonometric ratios to angles in different quadrants. By constructing a triangle with known side ratios, you can determine the reference angle and then adjust for the correct quadrant.
Recommended video:
6:04
Introduction to Trigonometric Functions

Quadrant Sign Rules

Trigonometric functions have specific signs in each quadrant. Since arccsc(-2/√3) is negative, the angle lies where cosecant is negative (quadrants II or IV). Knowing these sign rules helps identify the correct quadrant for the angle.
Recommended video:
Guided course
7:39
Introduction to Exponent Rules
Related Practice
Textbook Question

Evaluate the integrals in Exercises 67–74 in terms of

b. natural logarithms.

69. ∫(from 5/4 to 2)dx/(1-x²)

Textbook Question

91. [Technology Exercise] 91. The continuous extension of to (sin x)^x to [0, π]

b. Verify your conclusion in part (a) by finding lim(x→0⁺)f(x) with l’Hôpital’s Rule.

1
views
Textbook Question

In Exercises 67–72, you will explore some functions and their inverses together with their derivatives and tangent line approximations at specified points. Perform the following steps using your CAS:


b. Solve the equation y=f(x) for x as a function of y, and name the resulting inverse function g.

67. y= √(3x-2), 2/3 ≤ x ≤ 4, x_0=3

Textbook Question

Use reference triangles in an appropriate quadrant to find the angles in Exercises 1–8.

4. b. arcsin(-1/√2)

Textbook Question

Verify the integration formulas in Exercises 37–40.

37. b. ∫sech(x)dx = sin⁻¹(tanh x) + C

1
views
Textbook Question

3. Use the properties of logarithms to write the expressions in Exercises 3 and 4 as a single term.

b. ln(3x² - 9x) + ln(1/3x)