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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.

Q: How do you solve double angle equations for trigonometry?

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you use the the 3 trigonometry functions , sin=opposite divided by hypotenuse cos=adjacent divided by hypotenuse tan=opposite divided by adjacent these are used to work out angles and side lengths in right angle triangles only!!! sine,cosine,tangent :)

Probably you should start by looking up the double-angle formulas, reducing the "4a" to some combination of "2a".

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You need to be able to solve logarithms and be very good at algebra. In college, you have to be able to do college level algebra before you can take trig.

Angle of elevation: tangent angle = opposite/adjacent and by rearranging the given formula will help to solve the problem

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Tell me the equations first.

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