How can the trig identity |cos(z)|^2 = cos^2x + sinh^2y be proven?

[mattmns]

I am asked to prove the following: (Note: z = x + iy)

|cos(z)|² = cos²x + sinh²y
---------------

So I started the following way:

|cos(z)|² = |cos(x+iy)|²
= |cos(x)cosh(y) - i(sin(x)sinh(y))|²
= cos²(x)cosh²(y) + sin²(x)sinh²(y) [after having square root squared removed]

once I got here I was stuck. I am just not seeing how we can get this to equal cos²x + sinh²y

Is there some silly trig identity I don't know? Or did I make a mistake? Any ideas?

Thanks!

 

[StatusX]

I think you're right and the given answer is wrong. Try plugging in a few values of z to check.

 

[mattmns]

I thought about it maybe being wrong too, never thought of pluging in values though, duh!

Take $z = \pi / 2$ and clearly we get something different on both sides. Guess I will talk to my professor about it since he wrote the homework himself. Thanks!

 

[mattmns]

Actually, my z = pi/2 is not a counterexample (since x = pi/2 and y = 0, we get 0 on both sides).

For the actual proof, I just need to continue where I left off, but change sin² to 1-cos², which supposedly gets what we want (I have yet to work it out).