Friday, February 9, 2007

Don't You Wish You Remembered Those Trig Identities. Topic: Trigonometry. Level: AMC/AIME.

Problem: (1962 IMO - #4) Solve the equation $\cos^2{(x)}+\cos^2{(2x)}+\cos^2{(3x)} = 1$.

Solution: Subtract the $\cos^2{(3x)}$ from both sides, and replace the LHS by $\sin^2{(3x)}$, so we now have

$\cos^2{(x)}+\cos^2{(2x)} = \sin^2{(3x)}$.

Recalling the sine angle addition identity, we write

$\sin{(3x)} = \sin{(2x+x)} = \sin{(2x)}\cos{(x)}+\sin{(x)}\cos{(2x)}$.

So our equation is

$\cos^2{(x)}+\cos^2{(2x)} = [\sin{(2x)}\cos{(x)}+\sin{(x)}\cos{(2x)}]^2$.

Expanding, collecting terms, and simplifying using $1-\sin^2{(2x)} = \cos^2{(2x)}$ and $1-\sin^2{(x)} = \cos^2{(x)}$, we get

$2\cos^2{(x)} \cos^2{(2x)} = 2\sin{(x)}\cos{(x)}\sin{(2x)}\cos{(2x)}$.

Divide by $2$, rearrange, and factor:

$\cos{(x)}\cos{(2x)}[\cos{(x)}\cos{(2x)}-\sin{(x)}\sin{(2x)}] = 0$.

Substitute using the double-angle identities $\cos{(2x)} = 1-2\sin^2{(x)}$ and $\sin{(2x)} = 2\sin{(x)}\cos{(x)}$ to finally find

$\cos^2{(x)} \cos{(2x)} [1-4\sin^2{(x)}] = 0$.

Solving yields

$x = \frac{\pi}{2}+k\pi \mbox{ ; } \frac{\pi}{4}+\frac{k \pi}{2} \mbox{ ; } \frac{\pi}{6}+k\pi \mbox{ ; } \frac{5 \pi}{6}+k \pi$

for all integers $k$. QED.

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Comment: Pretty standard and slightly ugly manipulation of trig identities, but overall not too bad. Considering you get like an hour or more to do this, I don't feel too bad about it. Other solutions taking into account symmetry or using DeMoivre's and stuff are also out there, but this seemed much more straightforward and easier to develop.

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Practice Problem: (2007 AMC 12A - #24) For each integer $n > 1$, let $F(n)$ be the number of solutions of the equation $\sin{(x)} = \sin{(nx)}$ on the interval $[0, \pi]$. What is $\displaystyle \sum_{n=2}^{2007} F(n)$?