Question

Imaginary part of ${\cos h}(\alpha + i\beta) - {\cos h}(\alpha - i\beta) =$

• A. $2\sinh\alpha\sinh\beta$
• B. $2{\sin h}\alpha\sin\beta$
• C. $\cosh\alpha\cos\beta$
• D. $2\cos\alpha{\cos h}\beta$

Answer 1

Answer: (B)

$2{\sin h}\alpha\sin\beta$

Answer 2

${\cos h}(\alpha + i\beta) - {\cos h}(\alpha - i\beta) = 2{\sin h}\alpha{\sin h}(i\beta)$

$= 2{\sin h}\alpha.i\sin\beta = 2i{\sin h}{\sin\beta}$

∴ Imaginary part $= 2{\sin h}\alpha\sin\beta$.