Lesson 5: Inverse Trigonometric Functions and Solving Trig Equations

Solving a trig equation backwards — recovering the angle from a value — requires inverse functions, and inverse functions require restricted domains. This lesson builds the three inverse trig functions properly, then turns to the real skill: finding *all* solutions of a trig equation, on an interval or in general, without losing any or inventing extras.

Why inverses need restricted domains

Sine is not one-to-one: sin(π/6) = sin(5π/6) = 1/2, so "the angle whose sine is 1/2" is ambiguous. The fix, as with sqrt(x) in Lesson 1, is to restrict to an interval where the function *is* one-to-one and still hits every value once:

These ranges are conventions chosen for continuity with calculus (they are the ranges your calculator uses). Every inverse-trig answer you ever give must land in the stated range — that constraint decides several examples below.

Evaluating compositions

Worked examples. arcsin(sin(5π/4)): sin(5π/4) = −sqrt(2)/2, and the angle in [−π/2, π/2] with that sine is −π/4. So arcsin(sin(5π/4)) = −π/4, not 5π/4 — the composition f^−1(f(x)) equals x only when x lies in the restricted range. Similarly arccos(cos(7π/6)) = arccos(−sqrt(3)/2) = 5π/6 (the angle in [0, π] with cosine −sqrt(3)/2), and arctan(tan(3π/4)) = arctan(−1) = −π/4. For mixed compositions, draw the triangle: sin(arccos(3/5)) — arccos(3/5) is an angle in [0, π] with adjacent 3, hypotenuse 5, opposite 4, so the sine is 4/5 (positive, as the range demands).

Solving equations: basic angles and general solutions

Worked example. Solve sin x = sqrt(3)/2. The reference angle is π/3, and sine is positive in quadrants I and II, so on [0, 2π) the solutions are x = π/3 and x = 2π/3. The general solution adds every full turn: x = π/3 + 2πn or x = 2π/3 + 2πn, for any integer n. Similarly 2 cos x + 1 = 0 gives cos x = −1/2, so x = 2π/3 or 4π/3 on [0, 2π). Tangent has period π, not 2π, so tan x = 1 has general solution x = π/4 + πn — one family, not two. Matching the family structure to the period of the function is essential when equations are used inside models, where n counts revolutions or cycles.

Solving harder equations: quadratics and double angles

Worked example — quadratic in sin x. Solve 2 sin^2 x − sin x − 1 = 0 on [0, 2π). Factor: (2 sin x + 1)(sin x − 1) = 0. sin x = 1 gives x = π/2. sin x = −1/2 gives x = 7π/6 and 11π/6. Three solutions total. Worked example — double angle. Solve cos 2x = sin x on [0, 2π). Use cos 2x = 1 − 2 sin^2 x: 1 − 2 sin^2 x = sin x → 2 sin^2 x + sin x − 1 = 0 → (2 sin x − 1)(sin x + 1) = 0. sin x = 1/2 gives x = π/6, 5π/6; sin x = −1 gives x = 3π/2. The strategy is uniform: use an identity to reduce everything to one trig function, then solve the resulting polynomial equation — but never divide both sides by a trig expression, which silently discards the solutions where that expression is zero.

Modelling: recovering an angle from a ratio

Surveyors measure ratios and need angles. A building casts a shadow: from a point 60 m from its base, the elevation to the top satisfies tan θ = 45/60 = 0.75 (the building is 45 m tall), so θ = arctan(0.75) ≈ 36.87°. A wheelchair ramp rising 3 m over a 12 m horizontal run has angle arctan(3/12) = arctan(0.25) ≈ 14.04° — check this against building codes and you see why arctan is the working tool: regulations state maximum *slopes*, but builders set out *angles*. Note the calculator discipline: confirm whether the problem wants degrees or radians before pressing the inverse key, and sanity-check the size (14° is a steep ramp; 0.245 rad is the same statement).

Common mistakes

• Assuming arcsin(sin x) = x. It holds only on [−π/2, π/2]. arcsin(sin(2π/3)) = π/3, not 2π/3 — the range check is the whole lesson in miniature.
• Losing a family of solutions. sin x = 1/2 has two solutions per period, in quadrants I and II. Students who stop at arcsin get π/6 and miss 5π/6 — typically half the marks.
• Dividing by sin x or cos x. In sin x cos x = sin x, dividing by sin x yields cos x = 1 but loses sin x = 0. Factor instead: sin x (cos x − 1) = 0, and keep both families.
• Writing the general solution with the wrong period. Sine and cosine repeat every 2π; tangent every π. Adding 2πn to a tangent solution is harmless but incomplete — the πn family contains solutions the 2πn family misses (for tan x = 1, x = 5π/4).
• Mode errors. arctan(0.75) is 0.644 in radians and 36.87 in degrees. Both are "the answer" to different questions; state your unit.

Exam Experience

1. arcsin(sin(2π/3)) equals

• (A) 2π/3
• (B) π/3
• (C) π/6
• (D) −π/3

Answer: (B). sin(2π/3) = sqrt(3)/2, and arcsin must return an angle in [−π/2, π/2]; the angle there with sine sqrt(3)/2 is π/3. (A) ignores the range restriction — the exact error this lesson targets.

2. The complete solution set of 2 sin x − 1 = 0, for all real x, is

• (A) x = π/6 + 2πn
• (B) x = 5π/6 + 2πn
• (C) x = π/6 + 2πn or x = 5π/6 + 2πn
• (D) x = π/6 + πn

Answer: (C). sin x = 1/2 in quadrants I and II gives two families. (D) is wrong: π/6 + πn includes 7π/6, where sine is −1/2. Neither single family in (A) or (B) is complete.

3. How many solutions does sin x = −sqrt(2)/2 have on [0, 2π)?

• (A) 1
• (B) 2
• (C) 3
• (D) 4

Answer: (B). Sine equals −sqrt(2)/2 at x = 5π/4 and x = 7π/4 (quadrants III and IV) — two solutions. The endpoints 0 and 2π are not roots, and the interval includes each interior solution once.

Constructed response. Solve cos 2x + cos x = 0 for 0 ≤ x < 2π. (a) Rewrite the equation using a double-angle identity so that only cos x appears. (b) Factor and find all solutions. (c) Which of the solutions also satisfy sin x > 0?

Scoring guide. (a) 1 point: cos 2x = 2 cos^2 x − 1, giving 2 cos^2 x + cos x − 1 = 0. (b) 3 points: factor as (2 cos x − 1)(cos x + 1) = 0 — 1 point; cos x = 1/2 gives x = π/3, 5π/3 — 1 point; cos x = −1 gives x = π — 1 point. (c) 1 point: sin x > 0 in quadrants I–II, so among {π/3, π, 5π/3} only x = π/3 qualifies (sin π = 0 is not > 0, and sin(5π/3) < 0).

Going deeper: solution counting and an identity among inverses

Counting solutions on longer intervals. Each period of sine contributes the same solution pattern, so extending the interval multiplies the count. sin x = 1/2 on [0, 2π) has 2 solutions (π/6, 5π/6); on [0, 4π) it has 4 (add 2π: 13π/6, 17π/6); on [0, 2nπ), 2n. But boundary placement matters: on [0, π], sin x = 1 has just one solution (π/2, the peak itself, shared by both "families"), and sin x = 0 has three solutions on [0, 2π] (0, π, 2π) but only two on [0, 2π). Peaks and endpoints are where counting arguments fail — list the solutions, don't just extrapolate the pattern.

Equations needing an identity first. Solve sin 2x = cos x on [0, 2π). Expand: 2 sin x cos x = cos x. Do not divide by cos x (it is zero at π/2 and 3π/2, both solutions). Factor instead: cos x (2 sin x − 1) = 0. cos x = 0 gives x = π/2, 3π/2; sin x = 1/2 gives x = π/6, 5π/6. Four solutions — and the two found by the factored-out term are exactly the two that division would have destroyed.

A graceful identity: arcsin u + arccos u = π/2 for all u in [−1, 1]. Proof: let θ = arcsin u, so sin θ = u with θ ∈ [−π/2, π/2]; then cos(π/2 − θ) = sin θ = u, and π/2 − θ ∈ [0, π] — the arccos range — so arccos u = π/2 − θ. Hence the sum is π/2. The two inverse functions are complementary in the literal sense, which is why their graphs are mirror images summing to a constant.

Workshop: tangent equations and inverse evaluations

Tangent's four-solution pattern. Solve tan^2 x − 1 = 0 on [0, 2π). tan x = ±1: tangent is 1 at π/4 and 5π/4, and −1 at 3π/4 and 7π/4 — four solutions, one per quadrant, because tangent's period π means each sign appears twice per 2π interval. Sketching tan's graph (Lesson 4's X-section) makes the count obvious before the algebra starts: the horizontal lines y = 1 and y = −1 each cross the tangent curve twice on [0, 2π).

Exact inverse evaluations, mixed. arctan(sqrt(3)) = π/3 (the angle in (−π/2, π/2) with tangent sqrt(3)); arccos(−1/2) = 2π/3; and cos(arcsin(2/5)): the triangle for arcsin(2/5) has opposite 2, hypotenuse 5, adjacent sqrt(25 − 4) = sqrt(21), so the cosine is sqrt(21)/5. Mixed compositions always route through the triangle — draw it, label from the *inside* function's meaning, read off the *outside* function.