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.
it helps solve equations for speed and frequency
Unlike equations (or inequalities), identities are always true. It is, therefore, not possible to solve them to obtain values of the variable(s).
By using trigonometry or using Pythagoras' theorem for a right angle triangle.
lol hahaha
With trigonometry by using the cosine rule
With trigonometry by using the cosine rule
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You can solve for sin18 by making use of trigonometry.
Tell me the equations first.
by calculating the angles
There are people who use this web site that can and will solve equations.
The answer depends on the nature of the equations.