- #1
mattmns
- 1,128
- 6
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!
|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!
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