12.4 Solving Trig Equations; Law of Sines and Cosines

12.4 Solving Trig Equations; Law of Sines and Cosines

/practice/accuplacer-advanced-algebraPractice questions with detailed explanations

Solving sin θ = 1/2 on [0, 2π)

A trigonometric equation asks for the angles that make a trig statement true. AAF almost always restricts you to an interval, most often [0, 2π) or [0, 2π], sometimes [0°, 360°). The method does not change:

1. Isolate the trig function. 2 sin θ = 1 becomes sin θ = 1/2 before you think about angles.
2. Reference angle. On the unit circle, sin θ = 1/2 at a 30° = π/6 reference (from the 30-60-90 table in Right-Triangle Trigonometry and Special Triangles).
3. Quadrants. Sine is positive in QI and QII (the ASTC diagram in Unit Circle, Radians, and Arc Length).
4. List every solution in the interval. QI: θ = π/6. QII: θ = π − π/6 = 5π/6. Both lie in [0, 2π). There is no QIII or QIV solution because sine is negative there.

So sin θ = 1/2 on [0, 2π) has solutions π/6 and 5π/6. Stopping at π/6 is the highest-yield miss in this section. The extra 2π + π/6 = 13π/6 is outside [0, 2π), so do not list it unless the interval is larger.

If the interval is in degrees, the same two angles are 30° and 150°.

Nearby equations using the same method

Worked isolation: 2 sin θ − 1 = 0 on [0, 2π). Then 2 sin θ = 1, sin θ = 1/2, solutions π/6 and 5π/6.

Worked isolation with a negative: √2 cos θ = −1 on [0, 2π). Then cos θ = −1/√2 = −√2/2. Reference π/4, cosine negative in QII and QIII: θ = 3π/4 and θ = 5π/4.

Worked linear: sin θ + 1 = 0 gives sin θ = −1, so θ = 3π/2 only. Sine equals −1 at one point per period, the bottom of the wave from Trigonometric Graphs and Identities.

A factoring variant: sin θ (sin θ − 1) = 0 means sin θ = 0 or sin θ = 1. On [0, 2π) that is 0, π from the first factor and π/2 from the second — three solutions, and 2π is the same angle as 0 if the interval is half-open [0, 2π).

Do not divide by a trig function if it might be zero. From sin θ tan θ = sin θ, move everything to one side: sin θ tan θ − sin θ = 0, sin θ (tan θ − 1) = 0. Then sin θ = 0 or tan θ = 1. Dividing by sin θ would drop the solutions of sin θ = 0.

Law of sines

In any triangle ABC — right or not — sides a, b, c opposite angles A, B, C satisfy

a / sin A = b / sin B = c / sin C

The common ratio equals the circumdiameter 2R, which is a useful exact-value device when a 30° angle is present: sin 30° = 1/2, so 2R = a / (1/2) = 2a.

Use the law of sines when you know an angle and its opposite side, plus one more angle or one more side.

AAS and ASA — unique triangle

AAS (two angles and a non-included side) and ASA (two angles and the included side) both determine a unique triangle because two angles already give the third: C = 180° − A − B. Then the law of sines produces the remaining sides.

Worked AAS with exact values: Triangle ABC has A = 30°, B = 45°, and side a = 10 (opposite A). Then C = 105°. The sine law with the 30° pair is

10 / sin 30° = 10 / (1/2) = 20

so 2R = 20. Then

b = 20 · sin 45° = 20 · (√2/2) = 10√2

The third side uses sin 105°. Angle-sum sine gives sin 105° = sin(60° + 45°) = sin 60° cos 45° + cos 60° sin 45° = (√3/2)(√2/2) + (1/2)(√2/2) = (√6 + √2)/4. Therefore c = 20 · (√6 + √2)/4 = 5(√6 + √2). AAF may stop at the b = 10√2 step; the point of the example is that AAS plus a special angle produces an exact side, not a calculator mush.

Worked ASA setup: A = 50°, B = 60°, included side c = 12 (the side between A and B). Then C = 70°. The law of sines starts from c / sin C = 12 / sin 70°, so a / sin 50° = 12 / sin 70° and b / sin 60° = 12 / sin 70°. AAF often asks which proportion is set up correctly rather than a calculator length.

SSA — the ambiguous case (trap)

SSA (two sides and a non-included angle) is not a congruence shortcut. Given acute A, side a opposite it, and side b adjacent to it:

• If a ≤ b sin A, you can have no triangle (or exactly one right triangle when equality holds).
• If b sin A < a < b, two triangles are possible (the ambiguous case).
• If a ≥ b, one triangle.

AAF will not make you grind both supplementary candidates on every item, but it will punish the assumption that SSA behaves like AAS. If a stem gives only A, a, and b with a < b, do not announce a unique triangle. That is the trap. SAS, SSS, ASA, and AAS are unique; SSA is the exception.

Law of cosines

When the known pair is SAS (two sides and the included angle) or SSS (three sides), the law of sines has no complete ratio to start from. Use the law of cosines:

c^2 = a^2 + b^2 − 2ab cos C

The other two letters cycle the same way: a^2 = b^2 + c^2 − 2bc cos A and b^2 = a^2 + c^2 − 2ac cos B. If C = 90°, cos 90° = 0 and the formula collapses to Pythagoras. That is why the law of cosines is the non-right generalization of a^2 + b^2 = c^2.

SAS — included angle known

Worked numeric non-right triangle: Sides 7 and 10 enclose a 60° angle. Find the side opposite that 60°.

c^2 = 7^2 + 10^2 − 2 · 7 · 10 · cos 60°

cos 60° = 1/2, so

c^2 = 49 + 100 − 140 · (1/2) = 149 − 70 = 79

c = √79

Do not “simplify” √79 into √80 or 9. 79 is prime, so √79 is the exact length. A cousin that does clean up: sides 8 and 5 enclosing 60° give c^2 = 64 + 25 − 80 · (1/2) = 89 − 40 = 49, so c = 7. Use whichever form the options match.

If the included angle is obtuse, cosine is negative, so −2ab cos C becomes a plus and the opposite side is longer than in the acute case. Worked: sides 6 and 9 enclose 120°. cos 120° = −1/2. Then c^2 = 36 + 81 − 2 · 6 · 9 · (−1/2) = 117 + 54 = 171, c = √171 = 3√19.

SSS — three sides, find an angle

Solve the law of cosines for the cosine:

cos C = (a^2 + b^2 − c^2) / (2ab)

Worked: Sides 7, 8, 5 with c = 7 the side opposite C. Then cos C = (8^2 + 5^2 − 7^2) / (2 · 8 · 5) = (64 + 25 − 49) / 80 = 40/80 = 1/2, so C = 60°. That is the inverse of the SAS example above — same triangle, other direction.

If the right-hand side is negative, C is obtuse. If it equals 0, C is 90°. If it is greater than 1 or less than −1, the side lengths do not make a triangle (AAF usually stays inside a real triangle).

Which law, which case

After SAS produces a side, you may finish remaining angles with the sine law. Check that an obtuse angle is not replaced by its acute supplementary twin: if the law of cosines already said C is obtuse, do not let sin C = sin(180° − C) talk you into the acute one.

AAF traps for equations and triangle laws

• One solution only. sin θ = 1/2 has two solutions per period. Graphs in 12.3 already showed the wave hitting 1/2 twice.
• Wrong interval unit. Options in radians mean you report π/6, not 30°, unless a stem mixed them and an option matches the other unit.
• Forgetting isolation. 2 sin θ = 1 is not sin θ = 2 or θ = 1/2.
• SSA treated as unique. Two sides and a non-included angle are not SAS. SAS includes the angle between the two sides.
• Cosine law with the wrong angle. The angle in c^2 = a^2 + b^2 − 2ab cos C must be the one opposite side c, which is the included angle of sides a and b.
• Degree measure inside s = rθ leftover. This section does not use arc length, but a mixed CAT item might. Keep radian conversion in the same toolkit.
• Calculator cosine for 60°. cos 60° = 1/2 exactly. Do not replace it with 0.5 in a way that then gets rounded out of √79.

/practice/accuplacer-advanced-algebraPractice questions with detailed explanations