## Thursday, May 14, 2015

Prove $\tan^2 20^\circ+\tan^2 40^\circ+\tan^2 80^\circ=33$.

Good trigonometry problem is hard to come by, and when we, the math educator found one, we have to take advantage of it and make full use of it.

Of course, we can rest assured that this trigonometry problem can be tackled using the sum-to-product and product-to-sum identities in a really messy and tedious way. After all, any given math problem can be solved in the most traditional way, isn't it?

So here goes the old traditional way of proof:

$\tan^220^\circ+\tan^240^\circ+\tan^280^\circ$

$=(\sec^220^\circ-1)+(\sec^240^\circ-1)+(\sec^280^\circ-1)$

$=(\sec^220^\circ+\sec^240^\circ+\sec^280^\circ)-3$

$=\dfrac {1}{\cos^220^\circ}+\dfrac {1}{\cos^240^\circ}+\dfrac {1}{\cos^280^\circ}-3$

$=\dfrac {(\cos^240^\circ)(\cos^280^\circ)+(\cos^220^\circ)(\cos^280^\circ)+(\cos^220^\circ)(\cos^240^\circ)}{(\cos^220^\circ)(\cos^240^\circ)(\cos^280^\circ)}-3$

For the numerator:

$(\cos^240^\circ)(\cos^280^\circ)+(\cos^220^\circ)(\cos^280^\circ)+(\cos^220^\circ)(\cos^240^\circ)$

$=\left[(\cos40^\circ)(\cos80^\circ)\right]^2+[(\cos20^\circ)(\cos80^\circ)]^2+[(\cos20^\circ)(\cos40^\circ)]^2$

$=\left[\dfrac{1}{2}(\cos120^\circ+\cos40^\circ)\right]^2+\left[\dfrac{1}{2}(\cos100^\circ+\cos60^\circ)\right]^2+\left[\dfrac{1}{2}(\cos60^\circ+\cos20^\circ)\right]^2$

$=\dfrac{1}{4}[(-\dfrac{1}{2}+\cos40^\circ)]^2+\dfrac{1}{4}[(\cos100^\circ+\dfrac{1}{2})]^2+\dfrac{1}{4}[(\dfrac{1}{2}+\cos20^\circ)]^2$

$=\dfrac{1}{4}\left[(\dfrac{1}{4}-\cos40^\circ+\cos^240^\circ\right]+\dfrac{1}{4}\left[(\cos^2100^\circ+\cos100^\circ+\dfrac{1}{4})\right]+\dfrac{1}{4}\left[(\dfrac{1}{4}+\cos20^\circ)+\cos^220^\circ\right]$

$=\dfrac{1}{4}\left[\dfrac{1}{4}-\cos40^\circ+\cos^240^\circ+\cos^2100^\circ+\cos100^\circ+\frac{1}{4}+\dfrac{1}{4}+\cos20^\circ+\cos^220^\circ\right]$

$=\dfrac{1}{4}\left[\dfrac{3}{4}+\cos20^\circ-\cos40^\circ+\cos100^\circ+\cos^220^\circ+\cos^240^\circ+\cos^2100^\circ\right]$

$=\dfrac{1}{4}\left[\dfrac{3}{4}+\cos20^\circ+\cos100^\circ-\cos40^\circ+\cos^220^\circ+\cos^240^\circ+\cos^2100^\circ\right]$

$=\dfrac{1}{4}\left[\dfrac{3}{4}+2\cos60^\circ\cos40^\circ-\cos40^\circ+\cos^220^\circ+\cos^240^\circ+\cos^2100^\circ\right]$

$=\dfrac{1}{4}\left[\dfrac{3}{4}+\cos40^\circ-\cos40^\circ+\dfrac{\cos40^\circ+1}{2}+\dfrac{\cos80^\circ+1}{2}+\dfrac{\cos200^\circ+1}{2}\right]$

$=\dfrac{1}{4}\left[\dfrac{3}{4}+\dfrac{1}{2}\left[\cos40^\circ+1+\cos80^\circ+1+\cos200^\circ+1\right]\right]$

$=\dfrac{1}{4}\left[\dfrac{3}{4}+\dfrac{1}{2}\left[3+\cos40^\circ+\cos200^\circ+\cos80^\circ\right]\right]$

$=\dfrac{1}{4}\left[\dfrac{3}{4}+\dfrac{1}{2}\left[3+2\cos120^\circ\cos80^\circ+\cos80^\circ\right]\right]$

$=\dfrac{1}{4}\left[\dfrac{3}{4}+\dfrac{1}{2}\left[3-\cos80^\circ+\cos80^\circ\right]\right]$

$=\dfrac{1}{4}\left[\dfrac{3}{4}+\dfrac{1}{2}\left[3\right]\right]$

$=\dfrac{9}{16}$

For the denominator:

$(\cos^220^\circ)(\cos^240^\circ)(\cos^280^\circ)$

$=\dfrac{1}{8}\left[(\cos40^\circ+1)(\cos80^\circ+1)(\cos160^\circ+1)\right]$

$=\dfrac{1}{8}\left[(\cos40^\circ\cos80^\circ+\cos40^\circ+\cos80^\circ+1)(\cos160^\circ+1)\right]$

$=\dfrac{1}{8}\left[(\dfrac{1}{2}(\cos120^\circ+\cos40^\circ)+2\cos60^\circ\cos20^\circ+1)(\cos160^\circ+1)\right]$

$=\dfrac{1}{8}\left[(\dfrac{1}{2}(\cos40^\circ-\dfrac{1}{2})+\cos20^\circ+1)(\cos160^\circ+1)\right]$

$=\dfrac{1}{8}\left[(\dfrac{\cos40^\circ}{2}-\dfrac{1}{4}+\cos20^\circ+1)(\cos(180-20)^\circ+1)\right]$

$=\dfrac{1}{8}\left[(\dfrac{\cos40^\circ}{2}-\dfrac{1}{4}+\cos20^\circ+1)(-\cos20^\circ+1)\right]$

$=\dfrac{1}{8}\left[(\dfrac{\cos40^\circ}{2}+\dfrac{3}{4}+cos20^\circ)(-\cos20^\circ+1)\right]$

$=\frac{1}{8}\left[\dfrac{-\cos20^\circ\cos40^\circ}{2}+\dfrac{\cos40^\circ}{2}-\dfrac{3\cos20^\circ}{4}+\dfrac{3}{4}-\cos^220^\circ+\cos20^\circ\right]$

$=\dfrac{1}{8}\left[\dfrac{-\dfrac{1}{2}(\cos60^\circ+\cos20^\circ)}{2}+\dfrac{\cos40^\circ}{2}+\dfrac{\cos20^\circ}{4}+\dfrac{3}{4}-\dfrac{\cos40^\circ+1}{2}\right]$

$=\dfrac{1}{8}\left[-\dfrac{(\dfrac{1}{2}+\cos20^\circ)}{4}+\dfrac{\cos40^\circ}{2}+\dfrac{cos20}{4}+\dfrac{3}{4}-\dfrac{\cos40^\circ}{2}-\dfrac{1}{2}\right]$

$=\dfrac{1}{8}\left[-\dfrac{1}{8}-\dfrac{\cos20^\circ}{4}+\dfrac{\cos40^\circ}{2}+\dfrac{\cos20^\circ}{4}+\dfrac{3}{4}-\dfrac{\cos40^\circ}{2}-\dfrac{1}{2}\right]$

$=\dfrac{1}{8}\left[-\dfrac{1}{8}+\dfrac{3}{4}+\dfrac{1}{2}\right]$

$=\dfrac{1}{64}$

Therefore, we have:

$\tan^220^\circ+\tan^240^\circ+\tan^280^\circ=\frac {\dfrac {9}{16}}{\dfrac {1}{64}}-3=36-3=33$(Q.E.D.)

Good news is, we can tackle this problem fairly easily if we are feed with the right trigonometric identities. I will show that easy proof in my next blog post, see ya!

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