Inverse Trigonometric Functions — Important Questions

[-π/2, π/2]

Step 1: The principal value branch of sin⁻¹ is the specific restricted domain of sine that makes it one-one and onto. Step 2: We restrict sine to [-π/2, π/2] where it is increasing and one-one with range [-1, 1]. Step 3: Therefore, sin⁻¹: [-1, 1] → [-π/2, π/2]. The range is the closed interval [-π/2, π/2]. Step 4: Note the endpoints are INCLUDED (closed brackets) because sin(-π/2) = -1 and sin(π/2) = 1 are valid. Step 5: [0, π] is the range of cos⁻¹(x), and [-1, 1] is the DOMAIN of sin⁻¹(x). Common mistake: Confusing ranges of sin⁻¹ and cos⁻¹.

2π/3

Step 1: We need y such that cos(y) = -1/2, where y lies in [0, π]. Step 2: We know cos(π/3) = 1/2. Since cosine is negative, we need an angle in the second quadrant. Step 3: cos(π - π/3) = -cos(π/3) = -1/2. So cos(2π/3) = -1/2. Step 4: Check: 2π/3 lies in [0, π], so it is valid. Step 5: Therefore, cos⁻¹(-1/2) = 2π/3. Common mistake: Some students write -π/3, but the range of cos⁻¹ is [0, π], so negative values are NOT allowed.

-π/6

Step 1: We need y such that sin(y) = -1/2, where y lies in [-π/2, π/2]. Step 2: We know sin(π/6) = 1/2. Since we need sin(y) = -1/2, and sin is an odd function, sin(-π/6) = -sin(π/6) = -1/2. Step 3: Check: -π/6 lies in [-π/2, π/2], so it is valid. Step 4: Therefore, sin⁻¹(-1/2) = -π/6. Step 5: Common mistake: Writing 5π/6 — this is incorrect because 5π/6 is NOT in the principal value branch [-π/2, π/2] for sin⁻¹.

(-π/2, π/2)

Step 1: The tangent function is restricted to the open interval (-π/2, π/2) to make it one-one and onto. Step 2: Note: This interval is OPEN (parentheses, not brackets) because tan(π/2) and tan(-π/2) are UNDEFINED. Step 3: The tangent function approaches ±∞ as x → ±π/2, so these endpoints are excluded. Step 4: Therefore, tan⁻¹: ℝ → (-π/2, π/2), and the range is the open interval (-π/2, π/2). Step 5: Common mistake: Writing [-π/2, π/2] with closed brackets — this is wrong because ±π/2 are not in the range of tan⁻¹.