Question

The imaginary part of $ sin\text{ }h(x+iy) $ is

• A. $ cos\text{ }x\text{ }sinh\text{ }y $
• B. $ -cosh\text{ }x\text{ }sin\text{ }y $
• C. $ cosh\text{ }x\text{ }sin\text{ }y $
• D. None of these

Answer 1

Answer: (C)

$ \sinh (x+iy)=\frac{1}{i}\sin [i(x+iy)] $
$ [\because \sinh (\theta )=i\sin i\theta ] $
$ =-i\sin (ix-y) $
$ =-i[\sin (ix)\cos y-\cos (ix)\sin y] $
$ =-i[i\sin x.\cos y-\cosh x\sin y] $
$ =-{{i}^{2}}\sin x.\cos y+i\cosh x\sin y $
$ =\sinh x\cos y+i\cosh x\sin y $
$ \therefore $ Imaginary part of $ \sinh (x+iy)=\cosh x\sin y $