Solutions — Inverse Trigonometric Functions
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Q1 — Evaluating arcsin
arcsin(k) = the unique angle θ ∈ [−π/2, π/2] with sinθ = k.
(a) sin(π/6) = 1/2 and π/6 ∈ [−π/2, π/2] → arcsin(1/2) = π/6
(b) sin(−π/2) = −1 and −π/2 ∈ [−π/2, π/2] → arcsin(−1) = −π/2
(c) sin(π/3) = √3/2 and π/3 ∈ [−π/2, π/2] → arcsin(√3/2) = π/3
(d) sin(0) = 0 and 0 ∈ [−π/2, π/2] → arcsin(0) = 0
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Q2 — Evaluating arccos
arccos(k) = the unique angle θ ∈ [0, π] with cosθ = k.
(a) cos(0) = 1 and 0 ∈ [0, π] → arccos(1) = 0
(b) cos(2π/3) = −1/2 and 2π/3 ∈ [0, π] → arccos(−1/2) = 2π/3
Common error: arccos(−1/2) ≠ −π/3. The range of arccos is [0,π], not [−π/2, π/2].
(c) cos(π/4) = 1/√2 and π/4 ∈ [0, π] → arccos(1/√2) = π/4
(d) cos(π) = −1 and π ∈ [0, π] → arccos(−1) = π
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Q3 — Evaluating arctan
arctan(k) = the unique angle θ ∈ (−π/2, π/2) with tanθ = k.
(a) tan(π/4) = 1 and π/4 ∈ (−π/2, π/2) → arctan(1) = π/4
(b) tan(0) = 0 → arctan(0) = 0
(c) tan(−π/3) = −√3 and −π/3 ∈ (−π/2, π/2) → arctan(−√3) = −π/3
(d) tan(π/6) = 1/√3 and π/6 ∈ (−π/2, π/2) → arctan(1/√3) = π/6
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Q4 — Domain and Range
(a) arcsin(x):
Domain: [−1, 1] (restricted from sin, which has outputs in [−1, 1])
Range: [−π/2, π/2] (the restricted domain of sin we inverted)
(b) arccos(x):
Domain: [−1, 1]
Range: [0, π]
(c) arctan(x):
Domain: ℝ (all real numbers — tan is defined on its full restricted domain (−π/2, π/2))
Range: (−π/2, π/2) (open interval: tan approaches but never reaches ±∞)
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Q5 — Simplifying Compositions
(a) sin(arcsin(3/5)): Since 3/5 ∈ [−1, 1], sin(arcsin x) = x applies directly.
sin(arcsin(3/5)) = 3/5
(b) arcsin(sin(5π/6)): Check if 5π/6 ∈ [−π/2, π/2]. It is not (5π/6 ≈ 2.62 > π/2 ≈ 1.57).
sin(5π/6) = sin(π − 5π/6) = sin(π/6) = 1/2
arcsin(1/2) = π/6 (which is in [−π/2, π/2])
arcsin(sin(5π/6)) = π/6
(c) cos(arccos(−2/3)): Since −2/3 ∈ [−1, 1], cos(arccos x) = x applies.
cos(arccos(−2/3)) = −2/3
(d) arctan(tan(3π/4)): Check if 3π/4 ∈ (−π/2, π/2). It is not (3π/4 ≈ 2.36 > π/2).
tan(3π/4) = −1
arctan(−1) = −π/4 (which is in (−π/2, π/2))
arctan(tan(3π/4)) = −π/4
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Q6 — Solving an Equation Using arcsin
arcsin(2x − 1) = π/6
Apply sin to both sides (since arcsin outputs values in [−π/2, π/2] and π/6 is in this range, this is valid):
2x − 1 = sin(π/6) = 1/2
2x = 1 + 1/2 = 3/2
x = 3/4
Restriction check: for arcsin(2x−1) to be defined, we need −1 ≤ 2x−1 ≤ 1, i.e. 0 ≤ x ≤ 1. Our solution x = 3/4 satisfies this. ✓
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Q7 — Properties of arctan
(a) Let θ = arctan(x). Then tanθ = x and θ ∈ (−π/2, π/2).
Consider −θ: we have −θ ∈ (−π/2, π/2) and tan(−θ) = −tanθ = −x.
By the definition of arctan (which gives the unique angle in (−π/2, π/2)):
arctan(−x) = −θ = −arctan(x). Therefore arctan is an odd function. ✓
(b) arctan(1) + arctan(−1) = π/4 + (−π/4) = 0
Consistent with arctan being an odd function: arctan(−1) = −arctan(1) = −π/4.
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Q8 — Comparing arcsin and arccos
(a) Proof that arcsin(x) + arccos(x) = π/2:
Let α = arcsin(x). Then sinα = x and α ∈ [−π/2, π/2].
Note that π/2 − α ∈ [0, π] (since α ∈ [−π/2, π/2] implies π/2 − α ∈ [0, π]).
Furthermore, cos(π/2 − α) = sinα = x.
Since π/2 − α is the unique angle in [0, π] whose cosine equals x, by definition arccos(x) = π/2 − α.
Therefore arcsin(x) + arccos(x) = α + (π/2 − α) = π/2. ✓
(b) arcsin(1/2) = π/6, so arccos(1/2) = π/2 − π/6 = π/3
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Q9 — Evaluating a Composite Expression
Let θ = arcsin(5/13). Then sinθ = 5/13 and θ ∈ [−π/2, π/2].
Since θ ∈ [−π/2, π/2], cosθ ≥ 0 (cosine is non-negative in this range).
Apply the Pythagorean identity:
cos²θ = 1 − sin²θ = 1 − (5/13)² = 1 − 25/169 = 144/169
cosθ = 12/13 (taking the positive square root)
Therefore cos(arcsin(5/13)) = 12/13
This technique (drawing a right triangle with opposite = 5, hypotenuse = 13, adjacent = 12) is a useful shortcut.
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Q10 — Sketching and Interpreting arctan
(a) Key features of y = arctan(x):
Domain: ℝ (all real x)
Range: (−π/2, π/2) — open interval, asymptotes are never reached
Horizontal asymptotes: y = π/2 (as x → +∞) and y = −π/2 (as x → −∞)
Intercepts: Both x-intercept and y-intercept at the origin (0, 0), since arctan(0) = 0
Monotonicity: Strictly increasing throughout its entire domain
Symmetry: Odd function — the graph is symmetric about the origin
Concavity: Concave down for x > 0 (approaches asymptote); concave up for x < 0
(b) arctan(x) > 0 when x > 0.
Since arctan is strictly increasing with arctan(0) = 0, it follows that arctan(x) > arctan(0) = 0 if and only if x > 0. The odd symmetry also confirms that arctan(x) < 0 when x < 0.