Trigonometry

What you need to know

• Exact values. In a right triangle, for an acute angle: sine $=\frac{\text{opposite}}{\text{hypotenuse}}$, cosine $=\frac{\text{adjacent}}{\text{hypotenuse}}$, tangent $=\frac{\text{opposite}}{\text{adjacent}}$. From the $30$-$60$-$90$ and $45$-$45$-$90$ triangles: $\sin 30^\circ = \cos 60^\circ = \frac12$, $\sin 45^\circ = \cos 45^\circ = \frac{\sqrt2}{2}$, $\sin 60^\circ = \cos 30^\circ = \frac{\sqrt3}{2}$, $\tan 30^\circ = \frac{1}{\sqrt3}$, $\tan 60^\circ = \sqrt3$.
• Law of Cosines. In any triangle, with $C$ the angle across from side $c$: $c^2 = a^2 + b^2 - 2ab\cos C$. It works for every angle, obtuse ones too. Example: sides $3$ and $5$ with a $120^\circ$ angle between them: $c^2 = 9 + 25 - 30\cdot\left(-\frac12\right) = 49$, so $c = 7$.
• Law of Sines. $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$, where each side goes with the angle across from it and $R$ is the circumradius (the radius of the circle through all three corners). Example: an equilateral triangle with side $6$ has $2R = \frac{6}{\sin 60^\circ} = 4\sqrt3$, so $R = 2\sqrt3$.
• Area. Area $= \frac12 ab\sin C$. Putting in $\sin C = \frac{c}{2R}$ from the Law of Sines gives $R = \frac{abc}{4\cdot\text{Area}}$.
• Pythagorean identity. $\sin^2 x + \cos^2 x = 1$, where $\sin^2 x$ means $(\sin x)^2$. Example: if $\cos C = \frac35$ in a triangle, then $\sin C = \sqrt{1 - \frac{9}{25}} = \frac45$.
• Trigonometry is rare on the AMC 10, but it is often the fastest route.

How AMC 10 tests it

• A triangle given by its three sides, or by two sides with a $60^\circ$ or $120^\circ$ angle between them; find the third side, the area, or the circumradius.
• A cosine that comes out as a fraction (like $\frac12$ or $\frac{11}{16}$) is a sign that a cevian length or a $13$-$14$-$15$-type triangle is intended.
• Problems 20–25: a length in a circle or regular polygon that needs $\sin 15^\circ = \frac{\sqrt6-\sqrt2}{4}$, $\cos 36^\circ = \frac{1+\sqrt5}{4}$, or a double-angle identity such as $\sin 2x = 2\sin x\cos x$.

Standard approaches

1. With a $30^\circ$, $45^\circ$, or $60^\circ$ angle, drop an altitude and use special right triangles.
2. Given three sides, find one cosine with the Law of Cosines. Then get its sine from $\sin^2 + \cos^2 = 1$, and use it for the area or the circumradius.
3. Use the Law of Cosines when the answer is a side length. Use the Law of Sines for $R$, or when two angles are known.

Worked example

Triangle $ABC$ has $AB = 5$, $AC = 8$, and $BC = 7$. What is the radius of the circle passing through $A$, $B$, and $C$?

(A) $\dfrac72$ (B) $\dfrac{7\sqrt3}{3}$ (C) $4$ (D) $\dfrac{7\sqrt2}{2}$ (E) $5$

Solution. Step 1. Use the Law of Cosines at $A$. The side across from $A$ is $BC = 7$, and the sides next to $A$ are $5$ and $8$:
$$\cos A = \frac{5^2 + 8^2 - 7^2}{2\cdot 5\cdot 8} = \frac{40}{80} = \frac12.$$
So $A = 60^\circ$ and $\sin A = \frac{\sqrt3}{2}$.

Step 2. The Law of Sines with the $2R$ part (the extended Law of Sines) gives $\frac{BC}{\sin A} = 2R$, so
$$R = \frac{BC}{2\sin A} = \frac{7}{\sqrt3} = \frac{7\sqrt3}{3}.$$
The answer is $\boxed{\textbf{(B)}\ \dfrac{7\sqrt3}{3}}$.

Pitfalls

• Sign mistakes in the Law of Cosines. The term is $-2ab\cos C$, and $\cos C$ is negative for an obtuse angle, so the term becomes positive. Example: for $C = 120^\circ$, $-2ab\cos C = +ab$.
• Pairing a side with the wrong angle in the Law of Sines. Each side goes with the angle across from it.
• Forgetting that $\sin C$ does not pin down $C$ (the ambiguous case), while $\cos C$ does. Example: $\sin C = \frac12$ allows $C = 30^\circ$ or $150^\circ$, but $\cos C = \frac12$ means $C = 60^\circ$.