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Ch. 6 - Inverse Circular Functions and Trigonometric Equations
Lial - Trigonometry 12th Edition
Lial12th EditionTrigonometryISBN: 9780136552161당신이 사용하는 게 아니라요?교과서 변경
7장, 문제 5

Which one of the following equations has solution π?
a. arccos (―1) = x
b. arccos 1 = x
c. arcsin (―1) = x

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1
Recall the definition of the inverse cosine function, arccos, which returns an angle \(x\) in the range \(0 \leq x \leq \pi\) such that \(\cos x\) equals the given value.
Evaluate each option by considering the cosine or sine values at \(x = \pi\):
For option (a), check if \(\arccos(-1) = \pi\) because \(\cos \pi = -1\); this suggests \(x = \pi\) is a solution for (a).
For option (b), \(\arccos(1) = 0\) since \(\cos 0 = 1\), so \(x = \pi\) is not a solution here.
For option (c), \(\arcsin(-1) = -\frac{\pi}{2}\) because \(\sin(-\frac{\pi}{2}) = -1\), so \(x = \pi\) is not a solution here.

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Inverse Trigonometric Functions

Inverse trigonometric functions, such as arccos and arcsin, return the angle whose trigonometric value matches the given input. They are used to find angles from known sine or cosine values, with outputs restricted to specific principal value ranges.
추천 영상:
4:28
Introduction to Inverse Trig Functions

Range and Domain of arccos and arcsin

The arccos function has a domain of [-1, 1] and a range of [0, π], meaning it outputs angles between 0 and π radians. The arcsin function also has a domain of [-1, 1] but a range of [-π/2, π/2], so it outputs angles between -π/2 and π/2 radians.
추천 영상:
4:22
Domain and Range of Function Transformations

Evaluating arccos and arcsin at Specific Values

Evaluating arccos(-1) yields π because cos(π) = -1, arccos(1) yields 0 since cos(0) = 1, and arcsin(-1) yields -π/2 because sin(-π/2) = -1. Understanding these specific values helps identify which equation has π as a solution.
추천 영상:
7:28
Evaluate Composite Functions - Values Not on Unit Circle