Math Core

Many AIME "trig problems" have no triangle in sight. They are algebra problems where the variables happen to be $\sin$ , $\cos$ or $\tan$ of some angles. The winning moves are to use identities to telescope or collapse products, to rewrite trig equations as polynomial equations (then apply Vieta), and to treat $\sin x$ and $\cos x$ as two variables tied by $\sin^2 x + \cos^2 x = 1$ .

Identities to have ready

Type	Identity
Double angle	$\sin 2x = 2\sin x\cos x$ , $\cos 2x = 2\cos^2 x - 1 = 1 - 2\sin^2 x$
Triple angle	$\cos 3x = 4\cos^3 x - 3\cos x$ , $\sin 3x = 3\sin x - 4\sin^3 x$ , $\tan 3x = \dfrac{3\tan x - \tan^3 x}{1 - 3\tan^2 x}$
Addition	$\tan(A + B) = \dfrac{\tan A + \tan B}{1 - \tan A\tan B}$
Product to sum	$2\sin A\sin B = \cos(A - B) - \cos(A + B)$
Complementary	$\sin(90^\circ - x) = \cos x$

Four standard techniques

1. Collapse products with the double-angle formula. Multiply a product of cosines of doubling angles by $\sin x$ . Each $\sin\theta\cos\theta$ becomes $\dfrac12\sin 2\theta$ , and the product collapses:

\cos x\cos 2x\cos 4x\cdots\cos 2^{n-1}x = \frac{\sin 2^n x}{2^n\sin x}.

2. Combine $a\sin x + b\cos x$ into one wave. Since $(a, b)$ has length $R = \sqrt{a^2 + b^2}$ , you can write $a\sin x + b\cos x = R\sin(x + \varphi)$ for some angle $\varphi$ . So its maximum is $R$ and its minimum is $-R$ .

3. Use symmetric functions of $\sin x$ and $\cos x$ . If you know $s = \sin x + \cos x$ , then squaring gives $\sin x\cos x = \dfrac{s^2 - 1}{2}$ . Every symmetric expression in $\sin x$ and $\cos x$ follows, just as in Vieta.

4. Turn a trig equation into a polynomial. The values $\cos 20^\circ$ , $\cos 100^\circ$ , $\cos 140^\circ$ all satisfy $\cos 3\theta = \dfrac12$ . By the triple-angle formula, they are the three roots of

4c^3 - 3c = \frac12, \qquad \text{that is,} \qquad 8c^3 - 6c - 1 = 0.

Now Vieta's formulas give you any symmetric expression in these three cosines.

Trig values as roots

If the angles in a problem are all solutions of one equation like $\cos n\theta = c$ or $\tan n\theta = c$ , expand with a multiple-angle formula to get a polynomial whose roots are the trig values. Then use Vieta.

A related trick for tangents: if $A + B = 45^\circ$ , then $\tan A + \tan B = 1 - \tan A\tan B$ , which rearranges to $(1 + \tan A)(1 + \tan B) = 2$ .

Common mistake

When you turn angles into polynomial roots, make sure the angles give distinct values and that you have all the roots. For example, $\cos 3\theta = \dfrac12$ is solved by $\theta = 20^\circ, 100^\circ, 140^\circ$ but also by $220^\circ, 260^\circ, 340^\circ$ , which have the same cosines. The cubic has exactly three roots, so list three distinct values, not six.

Worked examples

Worked example: A collapsing product

Compute $\cos 20^\circ\cos 40^\circ\cos 80^\circ$ .

Multiply and divide by $\sin 20^\circ$ and apply $\sin\theta\cos\theta = \dfrac12\sin 2\theta$ three times:

\sin 20^\circ\cos 20^\circ\cos 40^\circ\cos 80^\circ = \tfrac12\sin 40^\circ\cos 40^\circ\cos 80^\circ = \tfrac14\sin 80^\circ\cos 80^\circ = \tfrac18\sin 160^\circ.

Since $\sin 160^\circ = \sin 20^\circ$ , the product is $\dfrac18$ .

Worked example: One wave

Find the maximum of $3\sin x + 4\cos x + 2$ .

$3\sin x + 4\cos x = 5\sin(x + \varphi)$ where $\cos\varphi = \dfrac35$ and $\sin\varphi = \dfrac45$ . Its maximum is $5$ , so the maximum of the expression is $7$ .

Worked example: Pairing complementary angles

Compute $\sin^2 1^\circ + \sin^2 2^\circ + \dots + \sin^2 89^\circ$ .

Pair $k^\circ$ with $(90 - k)^\circ$ : $\sin^2 k^\circ + \sin^2(90 - k)^\circ = \sin^2 k^\circ + \cos^2 k^\circ = 1$ . There are $44$ pairs ( $k = 1$ to $44$ ) plus the middle term $\sin^2 45^\circ = \dfrac12$ . The sum is $44 + \dfrac12 = \dfrac{89}{2}$ .

Worked example: Vieta on cosines

Compute $\sec 20^\circ + \sec 100^\circ + \sec 140^\circ$ .

The three cosines are the roots of $8c^3 - 6c - 1 = 0$ , so $e_1 = 0$ , $e_2 = -\dfrac68 = -\dfrac34$ and $e_3 = \dfrac18$ . Then

\frac1{c_1} + \frac1{c_2} + \frac1{c_3} = \frac{e_2}{e_3} = \frac{-3/4}{1/8} = -6.

Since $\cos 100^\circ = -\cos 80^\circ$ and $\cos 140^\circ = -\cos 40^\circ$ , this also proves the surprising identity $\sec 40^\circ + \sec 80^\circ - \sec 20^\circ = 6$ .