Restrict trigonometric functions so their inverses exist, and use inverse functions to solve trigonometric equations from modeling contexts

Learning goal: Understand that restricting a trigonometric function to a domain on which it is always increasing or always decreasing allows its inverse to be constructed; use the inverse functions to solve trigonometric equations that arise in modeling contexts; evaluate the solutions using technology; and interpret them in terms of the context.

Scope note. F.TF.6 and F.TF.7 are marked (+) in Ohio's Learning Standards for mathematics, which Ohio uses to indicate standards "for students who are planning on taking advanced courses". They are taught here in full, as extension content for students on that path.

Why sine has no inverse as it stands. A function has an inverse only when every output comes from exactly one input, so that "undoing" the function gives a single answer. Sine fails this badly: sin(π/6) = 1/2, but so do sin(5π/6), sin(13π/6), sin(−7π/6) and infinitely many others, because of the symmetry and periodicity of the last lesson. Asked "which angle has sine 1/2?", sine has no single answer, so there is no function that undoes it on its whole domain. On a graph, the horizontal line y = 1/2 crosses the sine curve infinitely often.

The fix: restrict the domain. On an interval where a function is always increasing, two different inputs always give two different outputs (the larger input gives the larger output), so every output is hit at most once and the restricted function has an inverse. The same holds on an interval where it is always decreasing. Sine is always increasing on [−π/2, π/2], rising from −1 to 1 and taking every value in between exactly once. Restricting sine to that interval gives a function with an inverse, written arcsin or sin⁻¹: arcsin y is the one number in [−π/2, π/2] whose sine is y, defined for −1 ≤ y ≤ 1. Cosine is not increasing or decreasing on [−π/2, π/2] (it rises and then falls), so a different interval is chosen: cosine is always decreasing on [0, π], falling from 1 to −1, and arccos y is the one number in [0, π] whose cosine is y. Tangent is always increasing on the open interval (−π/2, π/2), where it runs through every real number, so arctan y is defined for every real y and lies in (−π/2, π/2). These intervals are conventions, but not arbitrary ones: each is an interval of length π that starts at 0 or is centered on 0, on which the function is one-directional and takes every value in its range exactly once.

Exact evaluations. arcsin(1/2) = π/6, because sin(π/6) = 1/2 and π/6 is in [−π/2, π/2]; the angle 5π/6 also has sine 1/2, but it is outside the interval, so it is not the answer. arcsin(−√2/2) = −π/4. arccos(−1/2) = 2π/3, since cos(2π/3) = −1/2 and 2π/3 lies in [0, π]. arctan(1) = π/4 and arctan(−√3) = −π/3. With technology, arcsin(0.8) ≈ 0.9273, arccos(0.3) ≈ 1.2661 and arctan(2.5) ≈ 1.1903, all in radian mode.

From one solution to all solutions. The inverse function returns one solution, the one in the restricted interval. A modeling problem usually needs the others too, and the unit circle supplies them. Write a for the given value and n for an integer. If sin θ = a (with −1 ≤ a ≤ 1) has the solution θ₁ = arcsin a, then π − θ₁ is also a solution (same sine, by the y-axis reflection), and adding 2πn to either gives all the rest. If cos θ = a (with −1 ≤ a ≤ 1) has the solution θ₁ = arccos a, then −θ₁ is also a solution (cosine is even), plus 2πn. If tan θ = a has the solution θ₁ = arctan a, all solutions are θ₁ + nπ, because the period of tangent is π.

A fully worked model: the Ferris wheel. From the previous lesson, a rider's height on a wheel of radius 20 m whose axle is 22 m above the ground, turning once every 4 minutes from the bottom, is h(t) = 22 − 20 cos(πt/2) meters. When during the first two turns is the rider exactly 32 m above the ground? Set 22 − 20 cos(πt/2) = 32, so −20 cos(πt/2) = 10 and cos(πt/2) = −1/2. Let u = πt/2. The inverse function gives one value: u = arccos(−1/2) = 2π/3. Cosine is even, so u = −2π/3 is also a solution, and all solutions are u = ±2π/3 + 2πk. Convert back: t = 2u/π, so t = 4/3 + 4k or t = −4/3 + 4k. The values in [0, 8) are t = 4/3 ≈ 1.33 min, t = −4/3 + 4 = 8/3 ≈ 2.67 min, t = 16/3 ≈ 5.33 min and t = 20/3 ≈ 6.67 min. Interpretation: in each 4-minute turn the rider passes 32 m twice, once on the way up (after 1 min 20 s) and once on the way down (after 2 min 40 s), symmetric about the top at t = 2. Check: h(4/3) = 22 − 20 cos(2π/3) = 22 + 10 = 32 ✓.

Technology and interpretation. In the tide model h(t) = 3.1 + 2.1 cos(2π(t − 3)/12.4), when is the depth exactly 4.0 m? Then cos(2π(t − 3)/12.4) = 0.9/2.1 ≈ 0.4286. A calculator gives arccos(0.4286) ≈ 1.1279, so 2π(t − 3)/12.4 = ±1.1279 + 2πk, giving t − 3 = ±1.1279 × 12.4/(2π) ≈ ±2.226 and t ≈ 0.774 or 5.226, then every 12.4 hours after: 13.174 and 17.626. In clock time these are about 12:46 a.m., 5:14 a.m., 1:10 p.m. and 5:38 p.m. Interpretation: a boat needing 4.0 m of water has it from 12:46 to 5:14 in the morning and from 1:10 to 5:38 in the afternoon, since the depth is above 4.0 m between each pair of crossings that surrounds a high tide.

The misconception. Students treat sin⁻¹ as 1/sin, or believe arcsin(sin θ) = θ for every θ. Neither is true. sin⁻¹ is the inverse function on the restricted domain, not a reciprocal, and arcsin(sin(5π/6)) = arcsin(1/2) = π/6, not 5π/6, because 5π/6 is outside [−π/2, π/2]. The inverse returns the representative in the restricted interval; the context and the unit circle supply the rest.

Worked, exact inverse values: Evaluate arccos(√3/2), arcsin(−1) and arctan(−1). arccos(√3/2) = π/6, because cos(π/6) = √3/2 and π/6 is in [0, π]. arcsin(−1) = −π/2, since sin(−π/2) = −1 and −π/2 is the left endpoint of [−π/2, π/2]. arctan(−1) = −π/4, since tan(−π/4) = −1 and −π/4 is in (−π/2, π/2); the angle 3π/4 also has tangent −1 but lies outside the interval.

Worked, a spring with technology: A weight on a spring has displacement y(t) = 6 sin(3t) cm, t in seconds. When is it first 3 cm above rest, and when next? Solve 6 sin(3t) = 3, so sin(3t) = 1/2. Let u = 3t: u = arcsin(1/2) = π/6, and by the y-axis reflection u = π − π/6 = 5π/6, then + 2πk. So t = π/18 ≈ 0.1745 s and t = 5π/18 ≈ 0.8727 s, and again every 2π/3 ≈ 2.094 s. Interpretation: the weight passes 3 cm on the way up at 0.17 s and on the way down at 0.87 s, symmetric about the top of the bounce at t = π/6 ≈ 0.52 s. Check: 6 sin(3 × 0.8727) = 6 sin(2.618) = 6 × 0.5 = 3 ✓.

Worked, why the restriction is needed: Explain why arcsin cannot be defined using the interval [0, π] instead. On [0, π], sine rises from 0 to 1 and then falls back to 0, so it is not always increasing or always decreasing; the value 1/2 occurs twice, at π/6 and 5π/6, and an inverse would have to return two answers. On [−π/2, π/2] sine is always increasing, every value from −1 to 1 occurs exactly once, and the inverse is a genuine function. The interval [0, π] is instead used for arccos, because cosine is always decreasing there.

Worked, the daylight model: With L(d) = 12.2 + 3.1 sin(2π(d − 81)/365) hours of daylight on day d (an invented model), on which days is there exactly 14 hours of daylight? sin(2π(d − 81)/365) = 1.8/3.1 ≈ 0.5806. arcsin(0.5806) ≈ 0.6195, and the second solution in one cycle is π − 0.6195 ≈ 2.5221. So 2π(d − 81)/365 ≈ 0.6195 or 2.5221, giving d − 81 ≈ 35.99 or 146.51, so d ≈ 117 and d ≈ 228. Interpretation: daylight passes 14 hours going up around day 117 and going down around day 228, symmetric about the model's longest day, day 81 + 365/4 = 172.25, and there are 14 or more hours between those two days.

Representation task: Draw y = sin x for −2π ≤ x ≤ 2π and the horizontal line y = 0.6. Mark every crossing in that range and record their x-values with a calculator: arcsin(0.6) ≈ 0.6435, then π − 0.6435 ≈ 2.4981, and each of those plus or minus 2π. Shade the interval [−π/2, π/2] on the x-axis, note that only one crossing lies inside it, and record why that one is the value of arcsin(0.6). Repeat for y = cos x with the line y = −0.4 and the interval [0, π], recording arccos(−0.4) ≈ 1.9823 and the second solution −1.9823 + 2π ≈ 4.3009.

Independent performance: Take the periodic model you built in the previous lesson (or the Ferris wheel h(t) = 22 − 20 cos(πt/2)) and pose a question of the form "when does the quantity equal a given value?" Solve it in writing: isolate the trigonometric function, apply the appropriate inverse function, state the restricted interval the inverse uses, find the second solution in one cycle from the unit circle, add multiples of the period, and list every solution in a stated time window, evaluated with a calculator to two decimal places. Then interpret each solution in a sentence about the situation and check one by substitution. Success criteria: the inverse-function step names the restricted interval correctly; the second solution is found by the correct symmetry (π − u for sine, −u for cosine, u + π for tangent); all solutions in the window are listed and none outside it; and the interpretation says what happens physically at each time.