Step-by-Step Solution

Topic: Inverse Trigonometric Functions

(I) ALTERNATIVE INTERVAL

The principal value branch is [-π/2, π/2].

Any interval where the sine function is one-one (monotonic) and covers the range [-1, 1] works.

Other intervals include:

  • • [π/2, 3π/2]
  • • [3π/2, 5π/2]
  • • [-3π/2, -π/2]
Example: [π/2, 3π/2]

(II) EVALUATE EXPRESSION

Expression: sin-1(-1/2) – sin-1(1)

Step 1: sin-1(-1/2)

Since sin(-π/6) = -1/2 and -π/6 ∈ [-π/2, π/2], the value is -π/6.

Step 2: sin-1(1)

Since sin(π/2) = 1 and π/2 ∈ [-π/2, π/2], the value is π/2.

Step 3: Subtract

(-π/6) – (π/2) = -π/6 – 3π/6 = -4π/6

Value = -2π/3

(III)(A) GRAPH OF sin-1x

To draw the graph:

  • • X-axis (Domain): from -1 to 1.
  • • Y-axis (Range): from -π/2 to π/2.
  • • Key Points: (-1, -π/2), (0, 0), (1, π/2).

The curve passes through origin with increasing slope.

Graph of arcsin(x)

Graph of y = sin-1(x)

(III)(B) DOMAIN & RANGE OF f(x) = 2 sin-1(1 – x)

1. Find Domain:

The argument of sin-1 must be in [-1, 1].

-1 ≤ 1 – x ≤ 1

Subtract 1 from all sides: -2 ≤ -x ≤ 0

Multiply by -1 (inequality flips): 2 ≥ x ≥ 0

Domain: [0, 2]

2. Find Range:

We know that for the domain, sin-1(1 – x) lies in [-π/2, π/2].

Multiply the entire range by 2:

2 * [-π/2, π/2] = [-π, π]

Range: [-π, π]

Domain: [0, 2]
Range: [-π, π]

⚠️ Silly Mistake Audit: Inequality Flipping

Watch the Negative Sign!
When solving the domain inequality -2 ≤ -x ≤ 0, a very common mistake is forgetting to flip the inequality signs when multiplying or dividing by a negative number.
Incorrect: -2 ≤ x ≤ 0 (This implies x is negative, which is wrong)
Correct: 2 ≥ x ≥ 0 (This implies x is between 0 and 2).

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