Trigonometry

What you need to know

• Exact values: $\sin 30^\circ = \cos 60^\circ = \frac12$, $\sin 45^\circ = \cos 45^\circ = \frac{\sqrt2}{2}$, $\sin 60^\circ = \cos 30^\circ = \frac{\sqrt3}{2}$, $\tan 30^\circ = \frac{1}{\sqrt3}$, $\tan 60^\circ = \sqrt3$.
• Law of cosines: $c^2 = a^2 + b^2 - 2ab\cos C$, valid for any angle.
• Law of sines: $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$, with $R$ the circumradius.
• Area $= \frac12 ab\sin C$, which gives $R = \frac{abc}{4\cdot\text{Area}}$.
• $\sin^2 x + \cos^2 x = 1$. Trig is rare on the AMC 10 but often the fastest route.

How AMC 10 tests it

• A triangle given by three sides, or two sides with a $60^\circ$ or $120^\circ$ included angle; find the third side, area, or circumradius.
• A cosine that comes out rational (like $\frac12$ or $\frac{11}{16}$) signals that a cevian length or a $13$-$14$-$15$-type triangle is intended.
• Problems 20–25: a length in a circle or regular polygon needing $\sin 15^\circ$, $\cos 36^\circ$, or a double-angle identity.

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, get one cosine from the law of cosines, then its sine from $\sin^2 + \cos^2 = 1$ for areas and circumradii.
3. Prefer 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. By the law of cosines at $A$,
$$\cos A = \frac{5^2 + 8^2 - 7^2}{2\cdot 5\cdot 8} = \frac{40}{80} = \frac12,$$
so $A = 60^\circ$ and $\sin A = \frac{\sqrt3}{2}$. By the extended law of sines,
$$R = \frac{BC}{2\sin A} = \frac{7}{\sqrt3} = \frac{7\sqrt3}{3}.$$
The answer is $\boxed{\textbf{(B)}\ \dfrac{7\sqrt3}{3}}$.

Pitfalls

• Sign error in the law of cosines: the term is $-2ab\cos C$, and $\cos C < 0$ for an obtuse angle.
• Pairing a side with the wrong angle in the law of sines; each side goes with the angle opposite it.
• Forgetting that $\sin C$ does not determine $C$ uniquely (ambiguous case), while $\cos C$ does.