Using complex numbers to find trig identities

[Miike012]

I can find for example Tan(2x) by using Euler's formula for example

Let the complex number Z be equal to 1 + itan(x)

Then if I calculate Z² which is equal to 1 + itan(2x) I can find the identity for tan(2x) by the following...

Z² =(Z)² = (1 + itan(x))² = 1 + (2i)tan(x) -tan(x)² = 1 -tan(x)² + i(2tan(x))

now tan(2x) = Im(Z²)/Re(Z²) = 2tan(x)/(1 - tan²(x)).

QUESTION:
Is there a method (still using complex numbers) to find half angle identities?

 

[Infrared]

Sure. Say you want to find $\sin(x/2)$. $$\sin(x/2)= \frac{e^\frac{ix}{2}-e^{-\frac{ix}{2}}}{2i}$$ Square both sides and simplify using $\frac{e^{ix}+e^{-ix}}{2}=cos(x)$ to get $$\sin^2(x/2)=\frac{1-cos(x)}{2}$$

 

[Miike012]

Sure. Say you want to find $\sin(x/2)$. $$\sin(x/2)= \frac{e^\frac{ix}{2}-e^{-\frac{ix}{2}}}{2i}$$ Square both sides and simplify using $\frac{e^{ix}+e^{-ix}}{2}=cos(x)$ to get $$\sin^2(x/2)=\frac{1-cos(x)}{2}$$

would it work the same for sin(x/n) where n>2

 

[Infrared]

I don't think so. If you were to expand $$(\frac{e^\frac{ix}{n}-e^{-\frac{ix}{n}}}{2i})^n$$ you would get a lot of terms that look like $e^{aix/n}$ that don't look so easy to simplify.

 

[Miike012]

I don't think so. If you were to expand $$(\frac{e^\frac{ix}{n}-e^{-\frac{ix}{n}}}{2i})^n$$ you would get a lot of terms that look like $e^{aix/b}$ where a and b are integers less than n that don't look so easy to simplify.

Thanks for your help

 

[epenguin]

If for any function f, you have a formula f(2x) = g[f(x)]

Then f(x) = g[f(x/2)], and f(x/2) = g^-1[f(x)]

Which you can obtain provided you can invert g, which in this case you can - it is solving a quadratic.

You will surely not obtain any different results by whatever other method.