- 1,011

- 6

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

|cos(z)|

---------------

So I started the following way:

|cos(z)|

= |cos(x)cosh(y) - i(sin(x)sinh(y))|

= cos

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

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

Thanks!

|cos(z)|

^{2}= cos^{2}x + sinh^{2}y---------------

So I started the following way:

|cos(z)|

^{2}= |cos(x+iy)|^{2}= |cos(x)cosh(y) - i(sin(x)sinh(y))|

^{2}= cos

^{2}(x)cosh^{2}(y) + sin^{2}(x)sinh^{2}(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 cos

^{2}x + sinh^{2}yIs there some silly trig identity I don't know? Or did I make a mistake? Any ideas?

Thanks!

Last edited: