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

Verified step by step guidance

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.
2. Evaluate each option by considering the cosine or sine values at \(x = \pi\):
3. For option (a), check if \(\arccos(-1) = \pi\) because \(\cos \pi = -1\); this suggests \(x = \pi\) is a solution for (a).
4. For option (b), \(\arccos(1) = 0\) since \(\cos 0 = 1\), so \(x = \pi\) is not a solution here.
5. For option (c), \(\arcsin(-1) = -\frac{\pi}{2}\) because \(\sin(-\frac{\pi}{2}) = -1\), so \(x = \pi\) is not a solution here.

Key Concepts

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.

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.

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.