Can Direct Use of cosh(2y) and cos(2x) Prove the Trigonometric Identity?

[MrWarlock616]

I need to prove that if $\sin(x+iy)=cos(\alpha)+i\sin(\alpha)$ , then $\cosh(2y)-\cos(2x)=2$

expanding sin(x+iy) and equating real and imaginary parts, we get:
$\cos(\alpha)=\sin(x)\cosh(y)$ -- (1)
and $\sin(\alpha)=\cos(x)\sinh(y)$ -- (2)

if we use $\cos^2(\alpha)+\sin^2(\alpha)=1$ and put the values from (1) and (2) , we can get the required expression. My question is that, is there an alternative to do this? Can I prove it by directly using the values of cosh2y and cos2x [from (1) and (2)]? I tried but it seems impossible. Why is it so?

 

[lurflurf]

use
cosh(a)=cos(i a)
then
cos(2a)-cos(2b)=-2sin(a-b)sin(a+b)
and so forth

 

[MrWarlock616]

use
cosh(a)=cos(i a)
then
cos(2a)-cos(2b)=-2sin(a-b)sin(a+b)
and so forth

ok..what is a and b?

 

[lurflurf]

Whatever you want (those are identities), in this case perhaps
cosh(2y)=cos(2i y)
cos(2iy)-cos(2x)=-2sin(iy-x)sin(iy+x)

 

[MrWarlock616]

Whatever you want (those are identities), in this case perhaps
cosh(2y)=cos(2i y)
cos(2iy)-cos(2x)=-2sin(iy-x)sin(iy+x)

Ok that is brilliant. thanks!