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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 7.6.3c
Chapter 7, Problem 7.6.3c

Use reference triangles in an appropriate quadrant to find the angles in Exercises 1–8.
3. c. sin^(-1)(-√3/2)

Verified step by step guidance
1
Recognize that the problem asks for the angle \( \theta \) such that \( \sin(\theta) = -\frac{\sqrt{3}}{2} \). This means we are looking for the inverse sine (arcsin) of \( -\frac{\sqrt{3}}{2} \).
Recall that the sine function is negative in the third and fourth quadrants. Since the range of \( \sin^{-1} \) (arcsin) is restricted to \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \), the angle must lie in either the fourth quadrant (negative angles) or the first quadrant (positive angles, but sine positive there). So here, the angle will be in the fourth quadrant (between \( -\frac{\pi}{2} \) and 0).
Identify the reference angle whose sine is \( \frac{\sqrt{3}}{2} \). From common special angles, \( \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \). So the reference angle is \( \frac{\pi}{3} \).
Since the sine value is negative and the angle must be in the fourth quadrant (due to the range of arcsin), the angle is \( -\frac{\pi}{3} \). This is the angle whose sine is \( -\frac{\sqrt{3}}{2} \) within the principal range of arcsin.
Express the final answer as \( \theta = \sin^{-1}\left(-\frac{\sqrt{3}}{2}\right) = -\frac{\pi}{3} \).

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Key Concepts

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

Inverse Sine Function (sin⁻¹ or arcsin)

The inverse sine function returns the angle whose sine value is a given number. Its output is restricted to the range [-π/2, π/2] (or [-90°, 90°]) to ensure it is a function. Understanding this helps find the principal angle corresponding to a sine value.
Recommended video:
4:03
Inverse Sine

Reference Triangles and Quadrants

Reference triangles are right triangles used to find angle measures based on trigonometric ratios. Knowing the quadrant where the angle lies is essential because sine values are positive or negative depending on the quadrant, affecting the angle's actual measure.
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Trig Values in Quadrants II, III, & IV

Sign of Sine in Different Quadrants

Sine is positive in the first and second quadrants and negative in the third and fourth quadrants. Since sin⁻¹(-√3/2) is negative, the angle must lie in either the fourth or first quadrant (considering the principal range), guiding the correct angle determination.
Recommended video:
4:03
Inverse Sine
Related Practice
Textbook Question

86. This exercise explores the difference between

lim(x→∞)(1 + 1/x²)^x

and

lim(x→∞)(1 + 1/x)^x = e

c. Confirm your estimate of lim(x→∞)f(x) by calculating it with l’Hôpital’s Rule.

Textbook Question

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

7. c. arcsec(-2)

Textbook Question

4. Which of the following functions grow faster than x² as x→∞? Which grow at the same rate as x²? Which grow slower?

c. x²e^(-x)

Textbook Question

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

5. c. arccos(√3/2)

Textbook Question

In Exercises 1–4, solve for t.

1. c. e^((ln 0.2)t) = 0.4

Textbook Question

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

c. 3ln ∛(t² - 1) - ln(t+1)