8.5 Law of Sines, Law of Cosines, and Trigonometric Equations

The Most Advanced AAF Area

Trigonometry: Solving trigonometric equations, using right triangle trigonometry including special triangles, evaluating equivalent trigonometric functions, graphing trigonometric relationships, determining arc length and radian measures, and using the law of sines and the law of cosines.

At 1 to 3 of the 20 questions, trigonometry is a small share — but because these items sit at the top of the difficulty range, they matter disproportionately for students seeking placement into precalculus or calculus.

Beyond Right Triangles

Right-triangle ratios (SOH-CAH-TOA) require a 90° angle. For oblique triangles, two laws take over.

Standard notation: angles $A$, $B$, $C$ with opposite sides $a$, $b$, $c$. Side $a$ is always opposite angle $A$.

The Law of Sines

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$

Use it when you have a complete angle-side pair — an angle and the side opposite it. That covers:

• AAS — two angles and a non-included side
• ASA — two angles and the included side (find the third angle first)
• SSA — two sides and a non-included angle (the ambiguous case)

In triangle ABC, $A = 40°$, $B = 75°$, and $a = 12$. Find $b$.

$\dfrac{12}{\sin 40°} = \dfrac{b}{\sin 75°}$ $b = \dfrac{12 \sin 75°}{\sin 40°} \approx \dfrac{12(0.966)}{0.643} \approx 18.0$

The Ambiguous Case

SSA can yield two valid triangles, one, or none, because $\sin\theta = \sin(180° - \theta)$. If you solve for an angle and get 35°, then 145° may also satisfy the equation — check whether the three angles still sum to 180°. If they exceed it, discard the obtuse option.

The Law of Cosines

$c^2 = a^2 + b^2 - 2ab\cos C$

Use it when no complete angle-side pair exists:

• SAS — two sides and the included angle
• SSS — three sides (solving for an angle)

Sides $a = 8$, $b = 11$, included angle $C = 40°$. Find $c$.

$c^2 = 64 + 121 - 2(8)(11)\cos 40°$ $c^2 = 185 - 176(0.766) \approx 185 - 134.8 = 50.2$ $c \approx 7.1$

For SSS, rearrange: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$

The Pythagorean Connection

When $C = 90°$, $\cos 90° = 0$ and the law of cosines becomes $c^2 = a^2 + b^2 - 0 = a^2 + b^2$

The Pythagorean theorem is a special case of the law of cosines. This is a useful sanity check: a right-triangle result computed with the law of cosines must agree with Pythagoras.

Choosing Between the Laws

The deciding question: do I have an angle together with the side opposite it? Yes → sines. No → cosines.

Special Triangles

Exact values worth knowing, since the test rewards recognizing them without a calculator.

45°-45°-90°: sides $1 : 1 : \sqrt{2}$ 30°-60°-90°: sides $1 : \sqrt{3} : 2$ (opposite 30°, 60°, 90°)

Solving Trigonometric Equations

Because sine and cosine are periodic, equations typically have multiple solutions in an interval. Missing the second one is the characteristic error.

Procedure

1. Isolate the trigonometric function.
2. Find the reference angle.
3. Identify all quadrants where the function has the required sign.
4. Keep the solutions inside the stated interval.

Signs by Quadrant

Remembered as ASTC — All, Sine, Tangent, Cosine positive in quadrants I, II, III, IV respectively.

Worked Example

Solve $2\sin\theta - 1 = 0$ for $0° \le \theta < 360°$.

$\sin\theta = \tfrac{1}{2}$

Reference angle: 30°. Sine is positive in quadrants I and II:

• Quadrant I: $\theta = 30°$
• Quadrant II: $\theta = 180° - 30° = 150°$

Solutions: 30° and 150°.

Solve $\cos\theta = -\tfrac{\sqrt{2}}{2}$ for $0° \le \theta < 360°$.

Reference angle 45°; cosine is negative in quadrants II and III:

• Quadrant II: $180° - 45° = 135°$
• Quadrant III: $180° + 45° = 225°$

Solutions: 135° and 225°.

Reference Angle Formulas

Radian Measure and Arc Length

$180° = \pi \text{ radians}$

Convert by multiplying by $\dfrac{\pi}{180°}$ or $\dfrac{180°}{\pi}$.

$60° = 60 \cdot \dfrac{\pi}{180} = \dfrac{\pi}{3}$

Arc length: $s = r\theta$, with $\theta$ in radians.

A circle of radius 9 subtends a central angle of $\tfrac{2\pi}{3}$. Arc length? $s = 9 \cdot \tfrac{2\pi}{3} = 6\pi$

Using degrees directly in $s = r\theta$ is the standard error; convert first.