Showing $\pm |2x(\cos(az)+ i y \sin(az))| = \pm |2x| \sqrt{y^2 + (1-y^2)\cos^2(az)}$

Question

I'm trying to figure how why the following equality is true:

$$\pm |2x(\cos(az)+ i y \sin(az))| = \pm |2x| \sqrt{y^2 + (1-y^2)\cos^2(az)}$$

my mathematica === trial says false, I'm not sure how the equality holds. Every other mathematica command I know: ExpToTrig, TrigReduce, TrigExpand, TrigToExp, Simplify, etc don't seem to yield anything fruitful.

Is there any named identity between imaginary expressions (left hand side) and squared inside a square root (right hand side) that I'm not aware of?

Thanks

Answer 1

Provided $a$, $z$ and $y$ are all real, \begin{align*} &\pm \left| {2x(\cos (az) + iy\sin (az))} \right| = \pm \left| {2x} \right|\left| {\cos (az) + iy\sin (az)} \right| \\ & = \pm \left| {2x} \right|\sqrt {\cos ^2 (az) + (y\sin (az))^2 } = \pm \left| {2x} \right|\sqrt {\cos ^2 (az) + y^2 \sin ^2 (az)} \\& = \pm \left| {2x} \right|\sqrt {\cos ^2 (az) + y^2 (1 - \cos ^2 (az))} = \pm \left| {2x} \right|\sqrt {y^2 + (1 - y^2 )\cos ^2 (az)}. \end{align*}