**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) $?

i remembered them but there's still no way i could have gotten that =)

ReplyDeleteMaybe after a lot of substituting and messing around? =P

ReplyDeleteIf only problems on the IMO today could be like this... not saying I'm getting into the IMO... just saying

ReplyDelete