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There are two ways to solve for the double angle formulas in trigonometry. The first is to use the angle addition formulas for sine and cosine.
* sin(a + b) = sin(a)cos(b) + cos(a)sin(b) * cos(a + b) = cos(a)cos(b) - sin(a)sin(b) if a = b, then
* sin(2a) = sin(a)cos(a) + cos(a)sin(a) = 2sin(a)cos(a) * cos(2a) = cos2(a) - sin2(b) The cooler way to solve for the double angle formulas is to use Euler's identity. eix = cos(x) + i*sin(x). Yes, that is "i" as in imaginary number.
we we put 2x in for x, we get
* e2ix = cos(2x) + i*sin(2x) This is the same as
* (eix)2 = cos(2x) + i*sin(2x) We can substitute our original equation back in for eix.
* (cos(x) + i*sin(x))2 = cos(2x) + i*sin(2x) We can distribute the squared term.
* cos2(x) + i*sin(x)cos(x) + i*sin(x)cos(x) + (i*sin(x))2 = cos(2x) + i*sin(2x) And simplify. Because i is SQRT(-1), the i squared term becomes negative.
* cos2(x) + 2i*sin(x)cos(x) - sin2(x) = cos(2x) + i*sin(2x) * cos2(x) - sin2(x) + 2i*sin(x)cos(x) = cos(2x) + i*sin(2x) Now you can plainly see both formulas in the equation arranged quite nicely. I don't yet know how to get rid of the i, but I'm working on it.
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