Questions: Inverse Trigonometric Functions

Question 1

What is the value of arcsin(sin(5π/4))?

• A. 5π/4
• B. −π/4
• C. 3π/4
• D. π/4

Answer: (B)

sin(5π/4) = −√2/2. Now arcsin must return a value in [−π/2, π/2]. The angle in that range with sine −√2/2 is −π/4. The answer is NOT 5π/4, even though sin(5π/4) was the input — arcsin only 'undoes' sine for angles already in [−π/2, π/2]. 5π/4 lies outside that range, so the function forgets which branch it came from, just as √(x²) gives |x|, not x.

Question 2

A student writes: 'sin⁻¹(x) = 1/sin(x), so arcsin(2) = 1/sin(2).' What is wrong with this reasoning?

• A. Nothing — sin⁻¹ and 1/sin(x) are the same by the exponent rule
• B. sin⁻¹(x) means the inverse function of sine, not the reciprocal; and arcsin(2) is undefined since 2 > 1
• C. The formula is correct but arcsin(2) should be evaluated in degrees, not radians
• D. sin⁻¹(x) = 1/sin(x) only for x in [0, 1]

Answer: (B)

The notation sin⁻¹(x) means the arcsine — the inverse function — not 1/sin(x). The reciprocal of sine is csc(x). Additionally, arcsin has domain [−1, 1] because sine only takes values in that range; arcsin(2) is simply undefined. This confusion between inverse functions and reciprocals is extremely common and leads to systematic errors.

Question 3

arccos(−√2/2) = 3π/4, which is outside the range [0, π/2], so arccos cannot be correct here.

• T. True
• F. False

Answer: False

The range of arccos is [0, π], not [0, π/2]. 3π/4 is in [0, π], so it is a valid output of arccos. Indeed, cos(3π/4) = −√2/2, confirming the result. The restricted domain for arccos covers the upper semicircle — all the way from 0 to π — precisely so that it can handle negative values, which correspond to angles in the second quadrant.

Question 4

The notation sin⁻¹(x) and the notation arcsin(x) always refer to the same function.

• T. True
• F. False

Answer: True

Both notations mean the inverse sine function — the function that takes a value in [−1, 1] and returns the unique angle in [−π/2, π/2] with that sine. They are completely interchangeable. The potential for confusion is that the '−1' in sin⁻¹ looks like an exponent, which would mean 1/sin(x), but in function notation f⁻¹ always means the inverse function.

Question 5

Why must the domain of sine be restricted before we can define arcsin, and which restriction is chosen?

Model answer: Sine is periodic and therefore not one-to-one — every horizontal line between −1 and 1 intersects the sine curve infinitely many times. An inverse function can only exist for a one-to-one (injective) function. We restrict sine to [−π/2, π/2], one arc from the bottom to the top of the curve, where it is strictly increasing and passes the horizontal line test. This interval is chosen by convention because it includes 0 and is centered at the origin.

The restriction is not arbitrary geometry — it is a logical necessity. Without restricting the domain, 'arcsin(1/2)' would have infinitely many valid answers (π/6, 5π/6, π/6 ± 2π, etc.), making it not a function. The standard restriction [−π/2, π/2] is simply the most natural and convenient single branch; arccos uses [0, π] and arctan uses (−π/2, π/2) for analogous reasons.