Author Topic: cool things dcc proves by accident (Read 2217 times) Tweet Share

dcc · « **on:** March 29, 2008, 02:45:56 am »

1 - sin10 = cos80

Start with the triple angle formula:
$\sin 3\theta = 3\sin \theta - 4\sin^3 \theta$

So using this for sin10 & sin30
$\sin 30 = 3\sin 10 - 4\sin^3 10$

Which we can turn into a cubic equation as so:
$0 = \sin^3 10 - \dfrac{3}{4}\sin 10 + \dfrac{1}{8}$

Letting $a = \sin 10, p = -\dfrac{3}{4}, q = \dfrac{1}{8}$

$a^3 + pa + q = 0$

Let $a = u - v$

$(u - v)^3 + p(u - v) + q = 0$

expanding out and collecting terms, we get (trust me):

$(q - (v^3 - u^3)) + (u - v)(p - 3uv) = 0$

Now to solve the equation, we have to solve (see here for the reasoning):

$u^3 = \dfrac{-9q + \sqrt{12p^3 + 81q^2}}{18}\mbox{ and }v^3 = \dfrac{9q + \sqrt{12p^3 + 81q^2}}{18}$

now going back to what we set q & p as:

$u^3 = \dfrac{-9(\dfrac{1}{8}) + \sqrt{12(-\dfrac{3}{4})^3 + 81(\dfrac{1}{8})^2}}{18} = \dfrac{-\dfrac{9}{8} + \sqrt{\dfrac{243}{64}i^2}}{18} = \dfrac{-\dfrac{9}{8} + \dfrac{9\sqrt{3}i}{8}}{18} = -\dfrac{1}{16} + \dfrac{\sqrt{3}}{16}i = \dfrac{1}{8}(\cos 120 +i\sin 120)$

similiarly, $v^3 = \dfrac{1}{16} + \dfrac{\sqrt{3}}{16}i = \dfrac{1}{8}(\cos 60 + i\sin 60)$

Now, finding u & v (ignoring the two other pair of solutions, as they are nonsensical in the context of this question)

$u_{1} = \dfrac{1}{2}(\cos 80 - i\sin 80)$

$v_{1} = \dfrac{1}{2}(\cos 100 - i\sin 100)$

now subbing these back into our equation for a (in terms of u + v)

$\sin 10 = \dfrac{1}{2}(\cos 80 - isin80) - \dfrac{1}{2}(\cos 100 - isin100)$

since sin80 = sin100, we lose all the imaginary bits!

$\sin 10 = \dfrac{1}{2}(\cos 80 - \cos 100)$

since cos(100) = -cos80

$\sin 10 = \dfrac{1}{2}(\cos 80 + \cos 80) = \boxed{\cos 80}$

Collin Li · « **Reply #1 on:** March 29, 2008, 03:08:23 am »

Why would you do that? Everyone knows that by looking at a right-angled triangle!

Ahmad · « **Reply #2 on:** March 29, 2008, 10:07:10 am »

Obviously too late at night

cara.mel · « **Reply #3 on:** March 29, 2008, 02:50:53 pm »

Since when was there a triple angle formula

Ahmad · « **Reply #4 on:** March 29, 2008, 03:20:12 pm »

There's an n-th angle formula too, which is simply derived by expanding then equating the real and imaginary parts of,

$\cos n\theta + i \sin n\theta = (\cos \theta + i \sin \theta)^n$

enwiabe · « **Reply #5 on:** March 29, 2008, 03:29:00 pm »

An exercise in uselessness. Good work, nonetheless.

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Author Topic: cool things dcc proves by accident (Read 2217 times) Tweet Share

dcc

cool things dcc proves by accident

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