Jan
23
3 minute read

I’m in the process of writing a bigger piece on the role of trig formulas but I was looking for an example of their use during a geometry proof and came to this one.  The initial trig that is used is fairly straightforward but takes advantage of the 10-20-40 ratios and the general use of double angle formulas plus roots of unity symmetry.

Example 1

**Part 1: **find expressions for side lengths.

  • Triangle ABC:  Let BC = 1 then AC = $ \cos{40} $ and $ AB = \sin{40} $

  • Triangle ACE:   $ AE = \cos{40} \cdot \frac{\sin{10}}{\cos{10}} $

  • Triangle ADE:   $ AD = \cos{40} \cdot \frac{\sin{10}}{\cos{10}} \cdot \frac{\cos{20}}{\sin{20}} $

  • Combining in Triangle ABD:  (Note: symmetry between denominator and numerator)

\[\tan{ADB} = \frac{AB}{AD} = \frac{\cos{10}\sin{20}\sin{40}}{\sin{10}\cos{20}\cos{40}}\]

**Part 2: **simplify the expression

  • Note: since cos(90-x) = sin(x) and vice versa we can make the top a bit regular
\[\tan{ADB} = \frac{\sin{20}\sin{40}\sin{80}}{\sin{10}\cos{20}\cos{40}}\]
  •       If m is the  floor of n / 2

         Therefore $ (\sin{20}\cdot \sin{40} \cdot \sin{60} \cdot \sin{80})^2 = \frac{9}{2^8} $ and $ \sin{20}\cdot \sin{40} \cdot \sin{60} \cdot \sin{80} = \frac{3}{16} $

  • $ \sin{60} = \frac{\sqrt{3}}{2} $ We combine to get the numerator is $ \frac{\sqrt{3}}{8} $  For the denominator we multiple by $ \frac{\cos{10}}{\cos{10}} $ and apply the double sine angle formula 3 times to find it equals $ \frac{1}{8} $

**Part 3: **Putting it all together

  • Combining we get $ \tan{ADB} = \sqrt{3} $ This is the well know ratio for tan 60.      
  • From there angle addition gets us that x + FEB + EBF = 180 or x + 100 + 30 = 180 and x = 50.

Alternate: $ \sin{x}\sin{(60-x)}\sin{(60+x)} = \sin{x}\cdot (\frac{\sqrt{3}}{2}\cos{x} - \frac{1}{2}\sin{x})(\frac{\sqrt{3}}{2}\cos{x} + \frac{1}{2}\sin{x}) $

$ = \frac{1}{4}\sin{x}(3\cos^2{x} - \sin^2{x}) $

$ = \frac{1}{4}(3 (1 - \sin^2{x})\sin{x} - \sin^3{x})  $

$ = \frac{1}{4}\sin{x}(3 \sin{x} - 4\sin^3{x}) $

$ = \frac{1}{4}\sin{3x} $

And likewise $ \cos{x}\cos{(60-x)}\cos{(60+x)} = \frac{1}{4}\cos{3x} $

If x = 20 we get the exact denominator and numerator of our expression above and can jump from

here to  $ \tan{ADB} = \frac{\sin{20}\sin{(60-20)}\sin{(60+20)}}{\cos{20}\cos{(60-20)}\cos{(60+20)}} = \tan{3\cdot 20} $

Additional Link: https://www.cut-the-knot.org/arithmetic/algebra/sin3x.shtml (a completely different approach to deriving above worth looking at)

Example 2

!

This is an example of a class of Algebra problems more easily attacked via a trigonometric substitution. If you try to naively unwind by squaring the sides enough to remove all the radicals you’ll end up with a not so easily solved degree 6 equation. In particular, the $ \sqrt{1-x^2} $ makes things difficult but it also strongly suggests the fundamental unit circle relationship: $ x^2 + y^2 = 1 $ so $ y = \sqrt{1 - x^2} $ or in trig terms $ \sqrt{1 - \cos^2{\theta}} = \sin{\theta} $

So let’s go ahead and subsitute $ x = \cos(\theta) $ which gets us:

\[\sqrt{\frac{1 + 2\cos{\theta}\sin{\theta}}{2}} + 2\cos^2{\theta} = 1\]

Enter the trig identies:   We can replace the 2 sin cos term with the double sine angle formula and likewise the 2cos^1 - 1 term is reducible with the double cos angle formula.

\[\sqrt{\frac{1 + \sin{2\theta}}{2}} - cos{2\theta} = 0\]

Squaring things now we arrive at:

$ \frac{1 + \sin{2\theta}}{2} = \cos^2{2\theta} $ $ 1 + \sin{2\theta} = 2\cos^2{2\theta} $ $ 1 + \sin{2\theta} = 2 - 2\sin^2{2\theta} $ $ 2\sin^2{2\theta} + \sin{2\theta} - 1 = 0 $

This is a quadratic equation which we can factor: $ (2\sin{2\theta} - 1)(\sin{2\theta} + 1) = 0 $

First case: $ \sin{2\theta} = \frac{1}{2} $ implying $ \theta = \frac{\pi}{12} $ or $ \frac{5 \pi}{12} $

So to find the cos value there’s a refresher on the 15-75-90 ratios here:

revisiting-the-15-75-90

But that’s not really necessary because we can run the double angle formulas in reverse

  • $\cos{2\theta} = \pm \frac{\sqrt{3}}{2} $
  • Then  $ \pm \frac{\sqrt{3}}{2}  = 2\cos^2{\theta} - 1 $ to get $x = \cos{\theta} = \frac{\sqrt{2 \pm \sqrt{3%}{2} $
  • After rechecking the potential solutions in the original equation only $ \frac{\sqrt{2 -\sqrt{3%}{2} $ works.

Second case: For the second easier quadratic root: $ \sin{2\theta} = -1 $ implying $ \theta = \frac{3 \pi}{4} $ Therefore $ x = \cos{\frac{3\pi}{4}} = -\frac{\sqrt{2}}{2} $

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