Solving trig equations — general solutions, solutions in given range: Worked examples

Before anything, one reminder of the vocabulary we lean on, so no symbol is unearned:

The three machines, all in one place:

The scenario matrix

Every trig-equation question sits in one of these cells. Each example below is tagged with the cell it hits. Notice we split the sign of the value for each function, because that is what decides which quadrants the answers live in.

Cell	What makes it different	Where the danger is	Example
A. cos, positive value	$cos θ = c$ , $c > 0$	forgetting the second, reflected solution — quadrants I & IV	Ex 1
B. cos, negative value	$cos θ = c$ , $c < 0$	$α$ is obtuse ( $> 9 0^{\circ}$ ) — quadrants II & III	Ex 2
C. sin, positive value	$sin θ = s$ , $s > 0$	supplementary pair $α$ and $π - α$ — quadrants I & II	Ex 3
D. sin, negative value	$sin θ = s$ , $s < 0$	$α$ negative — quadrants III & IV	Ex 4
E. tan, positive value	$tan θ = t$ , $t > 0$	single family, quadrants I & III	Ex 5
F. tan, negative value	$tan θ = t$ , $t < 0$	single family, negative principal $α$ — quadrants II & IV	Ex 6
G. degenerate (axis) values	value is $0$ , $\pm 1$	solutions collapse — $\pm$ gives repeats	Ex 7
H. impossible value	$∣ c ∣ > 1$ for sin/cos	there are no solutions	Ex 8
I. multiple angle	$cos 2 θ$ , $sin 3 θ \dots$	must widen the interval before dividing	Ex 9
J. phase shift	$sin (θ + k)$	shift the interval, solve, shift back	Ex 10
K. quadratic-in-trig	e.g. $2 sin^{2} θ - sin θ - 1 = 0$	factor first, two sub-equations	Ex 11
L. word problem	modelling with a real period	translate words → equation → filter by time	Ex 12

Figure — Solving trig equations — general solutions, solutions in given range

Figure s01 — the scenario map: horizontal lines cut the cosine curve twice, twice, once, and never for $c = 0.5$ , $c = - 0.5$ , $c = - 1$ and $c = 1.4$ . This picture explains why the answer-count changes from cell to cell.

The figure above is the map: every value of $c$ on the horizontal line meets the cosine curve either twice per turn (cells A, B), once at the peaks/troughs (cell G), or never if it lies above $1$ or below $- 1$ (cell H). The next two figures do the same job for sine and tangent, so all three machines get a visual justification.

Figure s02 — the sine curve. A positive value crosses in quadrants I & II (the two crossings are supplementary: $α$ and $π - α$ ); a negative value crosses in quadrants III & IV. This is why sine uses $nπ + (- 1)^{n} α$ , not $\pm$ .

Figure s03 — the tangent curve. Between each pair of vertical dashed asymptotes ( $π$ apart) the curve climbs from $- \infty$ to $+ \infty$ , hitting every value exactly once. That single crossing per period is why tan needs only one family $nπ + α$ — no $\pm$ , no $(- 1)^{n}$ .

Ex 1 — Cell A · cosine, positive value

Worked example Statement

Solve $cos θ = \frac{2}{2}$ for $0^{\circ} \leq θ < 36 0^{\circ}$ .

Forecast: the value is positive and between $0$ and $1$ , so the horizontal line cuts the cosine curve twice per turn (quadrants I and IV). Guess how many answers you'll keep before reading on.

Steps.

Principal value: $α = cos^{- 1} (\frac{2}{2}) = 4 5^{\circ}$ . Why this step? $cos 4 5^{\circ} = \frac{2}{2}$ is a standard angle — this is our one calculator answer.
General solution: $θ = 36 0^{\circ} n \pm 4 5^{\circ}$ . Why this step? cosine is even, so if $4 5^{\circ}$ works so does $- 4 5^{\circ}$ — the $\pm$ captures both reflections; $36 0^{\circ} n$ adds full turns (the period).
$n = 0$ : $θ = 4 5^{\circ}$ (keep) or $- 4 5^{\circ}$ (out of range). $n = 1$ : $θ = 36 0^{\circ} - 4 5^{\circ} = 31 5^{\circ}$ (keep); $36 0^{\circ} + 4 5^{\circ} = 40 5^{\circ}$ (out). Why this step? we turn the crank and filter by the interval.

Answers: $θ = 4 5^{\circ}, 31 5^{\circ}$ .

Verify: $cos 4 5^{\circ} = cos 31 5^{\circ} = 0.7071 \dots = \frac{2}{2}$ ✓, and $31 5^{\circ} = 36 0^{\circ} - 4 5^{\circ}$ is the reflection below the axis, exactly the second crossing in figure s01.

Ex 2 — Cell B · cosine, negative value (obtuse principal)

Worked example Statement

Solve $cos θ = - \frac{1}{2}$ for $0 \leq θ < 2 π$ (radians).

Forecast: the value is negative, so both crossings sit in the left half of the circle. In radian measure "left half" means angles between $\frac{π}{2}$ and $\frac{3 π}{2}$ — that is quadrants II and III. Predict which two angles.

Steps.

Principal value: $α = cos^{- 1} (- \frac{1}{2}) = \frac{2 π}{3}$ (that is $12 0^{\circ}$ ). Why this step? $cos^{- 1}$ of a negative number returns an angle between $\frac{π}{2}$ and $π$ — the calculator does the sign-handling for us.
General solution: $θ = 2 nπ \pm \frac{2 π}{3}$ . Why this step? same cosine machine — the $\pm$ still comes from even symmetry, no change of rule.
$n = 0$ : $θ = \frac{2 π}{3}$ (keep) or $- \frac{2 π}{3}$ (out). $n = 1$ : $θ = 2 π - \frac{2 π}{3} = \frac{4 π}{3}$ (keep); $2 π + \frac{2 π}{3}$ (out). Why this step? $\frac{2 π}{3}$ lies between $\frac{π}{2}$ and $π$ , so it is quadrant II; $\frac{4 π}{3}$ lies between $π$ and $\frac{3 π}{2}$ , so it is quadrant III — matching our forecast in radian measure.

Answers: $θ = \frac{2 π}{3}, \frac{4 π}{3}$ .

Verify: $cos \frac{2 π}{3} = cos \frac{4 π}{3} = - 0.5$ ✓. Both lie left of the vertical axis, exactly where a negative cosine must live.

Ex 3 — Cell C · sine, positive value

Worked example Statement

Solve $sin θ = \frac{3}{2}$ for $0 \leq θ < 2 π$ .

Forecast: sine positive means the two crossings are above the horizontal axis — quadrants I and II (see figure s02). They are supplementary: $α$ and $π - α$ . Guess how many.

Steps.

Principal value: $α = sin^{- 1} (\frac{3}{2}) = \frac{π}{3}$ . Why this step? $sin \frac{π}{3} = \frac{3}{2}$ is a standard angle — our one calculator answer, in quadrant I.
General solution: $θ = nπ + (- 1)^{n} \frac{π}{3}$ . Why this step? sine's two crossings per turn are supplementary, and $(- 1)^{n}$ encodes exactly that — not cosine's $\pm$ .
$n = 0$ : $θ = \frac{π}{3}$ (keep, quadrant I). $n = 1$ : $θ = π - \frac{π}{3} = \frac{2 π}{3}$ (keep, quadrant II). $n = 2$ : $θ = 2 π + \frac{π}{3}$ (out). Why this step? keep the $n$ landing inside $[0, 2 π)$ .

Answers: $θ = \frac{π}{3}, \frac{2 π}{3}$ .

Verify: $sin \frac{π}{3} = sin \frac{2 π}{3} = \frac{3}{2} = 0.8660 \dots$ ✓, both above the axis, and $\frac{2 π}{3} = π - \frac{π}{3}$ is the supplementary partner.

Ex 4 — Cell D · sine, negative value

Worked example Statement

Solve $sin θ = - \frac{1}{2}$ for $0 \leq θ < 2 π$ .

Forecast: sine negative means the two crossings are below the horizontal axis — quadrants III and IV (figure s02, lower half). The calculator's $α$ will be negative; watch how the $(- 1)^{n}$ machine still delivers the right positive answers.

Steps.

Principal value: $α = sin^{- 1} (- \frac{1}{2}) = - \frac{π}{6}$ . Why this step? $sin^{- 1}$ returns angles between $- \frac{π}{2}$ and $+ \frac{π}{2}$ , so a negative input gives a negative angle. That is fine — it is a valid base for the general formula.
General solution: $θ = nπ + (- 1)^{n} (- \frac{π}{6})$ . Why this step? sine's two solutions per turn are supplementary, which the $(- 1)^{n}$ encodes.
Crank it:
- $n = 1$ : $θ = π - (- \frac{π}{6}) = π + \frac{π}{6} = \frac{7 π}{6}$ (keep, quadrant III).
- $n = 2$ : $θ = 2 π + (- \frac{π}{6}) = \frac{11 π}{6}$ (keep, quadrant IV).
- $n = 0$ : $θ = - \frac{π}{6}$ (out); $n = 3$ : $3 π + \frac{π}{6}$ (out). Why this step? we pick the $n$ values that land inside $[0, 2 π)$ .

Answers: $θ = \frac{7 π}{6}, \frac{11 π}{6}$ .

Verify: $sin \frac{7 π}{6} = sin \frac{11 π}{6} = - 0.5$ ✓, both below the axis as forecast.

Ex 5 — Cell E · tangent, positive value

Worked example Statement

Solve $tan θ = 3$ for $0^{\circ} \leq θ < 36 0^{\circ}$ .

Forecast: tan repeats every $18 0^{\circ}$ , hitting each value once per period (figure s03), so expect answers a clean $18 0^{\circ}$ apart. A positive tan lives in quadrants I and III (where $sin$ and $cos$ share a sign). Predict them.

Steps.

Principal value: $α = tan^{- 1} (3) = 6 0^{\circ}$ . Why this step? $tan 6 0^{\circ} = 3$ — a standard angle, quadrant I.
General solution: $θ = 18 0^{\circ} n + 6 0^{\circ}$ . Why this step? tan travels alone — one base plus multiples of its period $18 0^{\circ}$ . No $\pm$ , no $(- 1)^{n}$ .
$n = 0$ : $6 0^{\circ}$ (keep, quadrant I). $n = 1$ : $18 0^{\circ} + 6 0^{\circ} = 24 0^{\circ}$ (keep, quadrant III). $n = 2$ : $42 0^{\circ}$ (out). Why this step? one solution per tan period lands in range.

Answers: $θ = 6 0^{\circ}, 24 0^{\circ}$ .

Verify: $tan 6 0^{\circ} = tan 24 0^{\circ} = 1.732 \dots = 3$ ✓, and $240 - 60 = 180$ , one full tan period apart. Both sit in quadrants I and III where tan is positive.

Ex 6 — Cell F · tangent, negative value (negative principal)

Worked example Statement

Solve $tan θ = - 3$ for $0^{\circ} \leq θ < 36 0^{\circ}$ .

Forecast: tan repeats every $18 0^{\circ}$ , once per period. A negative tan lives in quadrants II and IV. Because the value is negative, the calculator's principal $α$ will be negative — watch how adding $18 0^{\circ}$ steps rescues the in-range answers. Predict them.

Steps.

Principal value: $α = tan^{- 1} (- 3) = - 6 0^{\circ}$ . Why this step? $tan^{- 1}$ returns angles between $- 9 0^{\circ}$ and $+ 9 0^{\circ}$ ; a negative input gives a negative angle. This negative $α$ is a valid base — it just lands outside our interval, so we step forward.
General solution: $θ = 18 0^{\circ} n - 6 0^{\circ}$ . Why this step? tan travels alone — one base plus multiples of $18 0^{\circ}$ .
$n = 0$ : $- 6 0^{\circ}$ (out). $n = 1$ : $18 0^{\circ} - 6 0^{\circ} = 12 0^{\circ}$ (keep, quadrant II). $n = 2$ : $36 0^{\circ} - 6 0^{\circ} = 30 0^{\circ}$ (keep, quadrant IV). $n = 3$ : $48 0^{\circ}$ (out). Why this step? the negative principal drops out; the $18 0^{\circ}$ steps bring us to the two in-range answers.

Answers: $θ = 12 0^{\circ}, 30 0^{\circ}$ .

Verify: $tan 12 0^{\circ} = tan 30 0^{\circ} = - 1.732 \dots = - 3$ ✓, and $300 - 120 = 180$ , one full tan period apart. Both sit in quadrants II and IV where tan is negative.

Ex 7 — Cell G · degenerate axis values (all of them)

Worked example Statement

Solve each for $0 \leq θ < 2 π$ : (a) $cos θ = 1$ , (b) $cos θ = - 1$ , (c) $sin θ = 0$ , (d) $sin θ = 1$ , (e) $sin θ = - 1$ .

Forecast: each value is a peak, trough, or axis-crossing — the horizontal line just touches the curve rather than slicing it. So the $\pm$ or $(- 1)^{n}$ branches should collapse, giving fewer answers than a generic value. Predict the count for each.

Steps.

(a) $cos θ = 1$ . $α = cos^{- 1} (1) = 0$ . General: $θ = 2 nπ \pm 0 = 2 nπ$ . In $[0, 2 π)$ only $θ = 0$ . Why this step? $+ 0$ and $- 0$ are the same, so the $\pm$ collapses — the peak is touched once per turn.
(b) $cos θ = - 1$ . $α = cos^{- 1} (- 1) = π$ . General: $θ = 2 nπ \pm π$ . The two branches differ by $2 π$ (a full period), so they repeat. In $[0, 2 π)$ only $θ = π$ . Why this step? the trough is a single touch; $\pm π$ gives one distinct angle per turn.
(c) $sin θ = 0$ . $α = sin^{- 1} (0) = 0$ . General: $θ = nπ + (- 1)^{n} \cdot 0 = nπ$ . In $[0, 2 π)$ : $θ = 0, π$ . Why this step? sine crosses zero twice per turn (going up at $0$ , coming down at $π$ ) — the $(- 1)^{n}$ does nothing to a zero, and $nπ$ lists both.
(d) $sin θ = 1$ . $α = sin^{- 1} (1) = \frac{π}{2}$ . General: $θ = nπ + (- 1)^{n} \frac{π}{2}$ . $n = 0 \Rightarrow \frac{π}{2}$ ; $n = 1 \Rightarrow π - \frac{π}{2} = \frac{π}{2}$ (repeat). Only $θ = \frac{π}{2}$ . Why this step? the sine peak is a single touch; the supplementary partner of $\frac{π}{2}$ is itself.
(e) $sin θ = - 1$ . $α = - \frac{π}{2}$ . General: $θ = nπ + (- 1)^{n} (- \frac{π}{2})$ . $n = 1 \Rightarrow π + \frac{π}{2} = \frac{3 π}{2}$ . Only $θ = \frac{3 π}{2}$ . Why this step? the sine trough, single touch, at the bottom of the circle.

Answers: (a) $0$ ; (b) $π$ ; (c) $0, π$ ; (d) $\frac{π}{2}$ ; (e) $\frac{3 π}{2}$ .

Verify: $cos 0 = 1$ , $cos π = - 1$ , $sin 0 = sin π = 0$ , $sin \frac{π}{2} = 1$ , $sin \frac{3 π}{2} = - 1$ ✓. Peaks/troughs give one answer per turn; the zero-crossing gives two — exactly the collapse we forecast.

Ex 8 — Cell H · impossible value (no solutions)

Worked example Statement

Solve $sin θ = 1.4$ for $θ \in R$ .

Forecast: sine and cosine never leave the band $[- 1, 1]$ (look at figure s02 — the wave never rises above the top of its range). So what can we say about $sin θ = 1.4$ ?

Steps.

Note the range of sine: $- 1 \leq sin θ \leq 1$ for every real $θ$ . Why this step? it comes straight from the unit circle — a coordinate can never exceed the radius $1$ . See Unit Circle and Radian Measure.
Compare: $1.4 > 1$ . Why this step? the target lies outside the achievable band.
Conclusion: no solutions. There is no $α$ because $sin^{- 1} (1.4)$ is undefined. Why this step? the machine needs a valid principal value; there is none.

Answer: No real $θ$ satisfies the equation — the solution set is empty.

Verify: the maximum of $sin θ$ is $1$ , and $1.4$ exceeds it, so the equation cannot hold ✓. (Always sanity-check $∣ c ∣ \leq 1$ before reaching for the calculator.)

Ex 9 — Cell I · multiple angle (widen, then divide)

Worked example Statement

Solve $sin 3 θ = \frac{2}{2}$ for $0 \leq θ < π$ .

Forecast: the angle inside is $3 θ$ . As $θ$ sweeps $[0, π)$ , the inside sweeps three times as far. Guess how many answers this produces.

Steps.

Substitute $ϕ = 3 θ$ and widen the interval: $0 \leq θ < π \Rightarrow 0 \leq ϕ < 3 π$ . Why this step? tripling $θ$ triples the interval. Skip this and you lose two-thirds of the answers.
Principal value: $α = sin^{- 1} (\frac{2}{2}) = \frac{π}{4}$ .
General solution in $ϕ$ : $ϕ = nπ + (- 1)^{n} \frac{π}{4}$ (sine machine).
Crank until $ϕ \geq 3 π$ :
- $n = 0$ : $ϕ = \frac{π}{4}$
- $n = 1$ : $ϕ = π - \frac{π}{4} = \frac{3 π}{4}$
- $n = 2$ : $ϕ = 2 π + \frac{π}{4} = \frac{9 π}{4}$
- $n = 3$ : $ϕ = 3 π - \frac{π}{4} = \frac{11 π}{4}$
- $n = 4$ : $ϕ = 4 π + \frac{π}{4}$ (out, $\geq 3 π$ ). Why this step? we collect every $ϕ$ in $[0, 3 π)$ .
Divide each by $3$ : $θ = \frac{π}{12}, \frac{π}{4}, \frac{3 π}{4}, \frac{11 π}{12}$ . Why this step? $θ = ϕ /3$ undoes the substitution.

Answers: $θ = \frac{π}{12}, \frac{π}{4}, \frac{3 π}{4}, \frac{11 π}{12}$ — four solutions, as tripling suggested.

Verify: for each, $sin 3 θ = 0.7071 \dots = \frac{2}{2}$ ✓. Uses Compound and Double Angle Formulae thinking (a multiple angle) — the widen-then-divide routine from the parent note.

Ex 10 — Cell J · phase shift

Worked example Statement

Solve $cos (θ - \frac{π}{6}) = \frac{1}{2}$ for $0 \leq θ < 2 π$ .

Forecast: the machine solves for the bracket; the $- \frac{π}{6}$ just slides everything along. Substituting a new letter for the bracket keeps our rules identical.

Steps.

Let $u = θ - \frac{π}{6}$ and shift the interval: $0 \leq θ < 2 π \Rightarrow - \frac{π}{6} \leq u < 2 π - \frac{π}{6}$ , i.e. $- \frac{π}{6} \leq u < \frac{11 π}{6}$ . Why this step? subtracting $\frac{π}{6}$ from $θ$ shifts the whole window down by $\frac{π}{6}$ — solve where $u$ actually lives.
Principal value: $α = cos^{- 1} (\frac{1}{2}) = \frac{π}{3}$ .
General solution in $u$ : $u = 2 nπ \pm \frac{π}{3}$ (cosine machine).
Crank inside $[- \frac{π}{6}, \frac{11 π}{6})$ :
- $n = 0$ : $u = \frac{π}{3}$ (keep) or $u = - \frac{π}{3}$ (out, below $- \frac{π}{6}$ ).
- $n = 1$ : $u = 2 π - \frac{π}{3} = \frac{5 π}{3}$ (keep, since $\frac{5 π}{3} < \frac{11 π}{6}$ ). Why this step? only these $u$ fall inside the shifted window.
Undo the shift, $θ = u + \frac{π}{6}$ :
- $θ = \frac{π}{3} + \frac{π}{6} = \frac{π}{2}$
- $θ = \frac{5 π}{3} + \frac{π}{6} = \frac{11 π}{6}$ Why this step? return to the original variable.

Answers: $θ = \frac{π}{2}, \frac{11 π}{6}$ .

Verify: $cos (\frac{π}{2} - \frac{π}{6}) = cos \frac{π}{3} = \frac{1}{2}$ ✓, and $cos (\frac{11 π}{6} - \frac{π}{6}) = cos \frac{5 π}{3} = \frac{1}{2}$ ✓ (both brackets reduce to an angle whose cosine is $\frac{1}{2}$ ).

Ex 11 — Cell K · quadratic in a trig function

Worked example Statement

Solve $2 sin^{2} θ - sin θ - 1 = 0$ for $0^{\circ} \leq θ < 36 0^{\circ}$ .

Forecast: this is a quadratic where the unknown is $sin θ$ . Factor it, get two simple equations, then solve each with our machines. Guess: could it mix a generic case and a degenerate case?

Steps.

Let $x = sin θ$ : $2 x^{2} - x - 1 = 0$ . Why this step? naming $sin θ$ as $x$ reveals the familiar quadratic shape.
Factor: $(2 x + 1) (x - 1) = 0 \Rightarrow x = - \frac{1}{2}$ or $x = 1$ . Why this step? factoring splits it into two solvable trig equations.
Branch $sin θ = 1$ (degenerate, cell G): $θ = 9 0^{\circ}$ only — the single peak. Why this step? $sin = 1$ is the top of the wave, one touch per turn.
Branch $sin θ = - \frac{1}{2}$ (cell D): from Ex 4's logic in degrees, $θ = 21 0^{\circ}, 33 0^{\circ}$ (quadrants III and IV). Why this step? apply the sine machine to the negative value.

Answers: $θ = 9 0^{\circ}, 21 0^{\circ}, 33 0^{\circ}$ .

Verify: $2 sin^{2} θ - sin θ - 1 = 0$ at each: $sin 9 0^{\circ} = 1 \Rightarrow 2 - 1 - 1 = 0$ ✓; $sin 21 0^{\circ} = sin 33 0^{\circ} = - 0.5 \Rightarrow 2 (0.25) + 0.5 - 1 = 0$ ✓. Draws on Trigonometric Identities / factoring to reduce to $sin θ = c$ .

Ex 12 — Cell L · real-world word problem

Worked example Statement

A Ferris wheel seat's height above the ground is modelled by $h (t) = 8 - 6 cos (\frac{π}{15} t) metres,$ where $t$ is time in seconds. Find the first two times $t \geq 0$ at which the seat is exactly $11$ m high.

Forecast: setting $h = 11$ gives a cosine equation with a scaled time inside. The period is $\frac{2 π}{π /15} = 30$ s, so the seat passes each height twice per loop. Predict roughly when.

Steps.

Set $h = 11$ : $8 - 6 cos (\frac{π}{15} t) = 11 \Rightarrow cos (\frac{π}{15} t) = - \frac{1}{2}$ . Why this step? rearrange into $cos (angle) = number$ , the standard form.
Let $ϕ = \frac{π}{15} t$ . Principal value: $α = cos^{- 1} (- \frac{1}{2}) = \frac{2 π}{3}$ (cell B, a negative value → obtuse $α$ ).
General solution: $ϕ = 2 nπ \pm \frac{2 π}{3}$ . The first two positive values are $ϕ = \frac{2 π}{3}$ and $ϕ = 2 π - \frac{2 π}{3} = \frac{4 π}{3}$ . Why this step? we want the earliest times, so the two smallest positive $ϕ$ .
Convert back: $t = \frac{15}{π} ϕ$ .
- $ϕ = \frac{2 π}{3} \Rightarrow t = \frac{15}{π} \cdot \frac{2 π}{3} = 10$ s.
- $ϕ = \frac{4 π}{3} \Rightarrow t = \frac{15}{π} \cdot \frac{4 π}{3} = 20$ s. Why this step? $t = ϕ \cdot 15/ π$ inverts the scaling.

Answers: $t = 10$ s and $t = 20$ s.

Verify (with units): $h (10) = 8 - 6 cos \frac{2 π}{3} = 8 - 6 (- 0.5) = 11$ m ✓; $h (20) = 8 - 6 cos \frac{4 π}{3} = 11$ m ✓. Both times are positive seconds inside the first $30$ s loop — physically sensible.

Recall

Recall Which cell needs the interval widened before you solve?

Cell I (multiple angle). Substitute $ϕ = k θ$ , multiply the interval limits by $k$ , solve, then divide every answer by $k$ .

Recall Why does

$cos θ = - 1$ give half as many answers as $cos θ = 0.3$ in the same range? Because $- 1$ is a trough — the curve only touches it once per turn, so the two $\pm$ branches coincide (they differ by a full period). A generic value crosses twice per turn.

Recall What is the very first thing to check for

$sin θ = c$ or $cos θ = c$ ? Whether $∣ c ∣ \leq 1$ . If not, there are no solutions (cell H) and you stop immediately.

Connections

Parent topic — the three machines these examples exercise.
Unit Circle and Radian Measure — why $∣ sin ∣, ∣ cos ∣ \leq 1$ (cell H) and where the $\pm$ comes from.
Graphs of Trigonometric Functions — the crossing-count picture behind the scenario matrix.
Inverse Trigonometric Functions — the principal value $α$ for negative and axis inputs.
Trigonometric Identities — reduces cell K quadratics to $sin / cos / tan = c$ .
Compound and Double Angle Formulae — the machinery behind cells I and J.