Alternative to solve this?

  • #1
160
3
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?
 

Answers and Replies

  • #2
lurflurf
Homework Helper
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use
cosh(a)=cos(i a)
then
cos(2a)-cos(2b)=-2sin(a-b)sin(a+b)
and so forth
 
  • #3
160
3
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?
 
  • #4
lurflurf
Homework Helper
2,432
132
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)
 
  • #5
160
3
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!
 

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