ATAR Notes: Forum
Uni Stuff => Science => Faculties => Mathematics => Topic started by: dcc on March 29, 2008, 02:45:56 am
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1 - sin10 = cos80
Start with the triple angle formula:

So using this for sin10 & sin30

Which we can turn into a cubic equation as so:

Letting 

Let 
^3 + p(u - v) + q = 0)
expanding out and collecting terms, we get (trust me):
) + (u - v)(p - 3uv) = 0)
Now to solve the equation, we have to solve (see here for the reasoning):

now going back to what we set q & p as:
 + \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, )
Now, finding u & v (ignoring the two other pair of solutions, as they are nonsensical in the context of this question)
)
)
now subbing these back into our equation for a (in terms of u + v)
 - \dfrac{1}{2}(\cos 100 - isin100))
since sin80 = sin100, we lose all the imaginary bits!
)
since cos(100) = -cos80
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Why would you do that? Everyone knows that by looking at a right-angled triangle!
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Obviously too late at night :P
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Since when was there a triple angle formula :(
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There's an n-th angle formula too, which is simply derived by expanding then equating the real and imaginary parts of,
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An exercise in uselessness. Good work, nonetheless. :P