Correct Answer - Option 4 : 15/8
Concept:
The exponential representation of the basic hyperbolic functions are:
\(cosh(x) = \frac{e^x+e^{-x}}{2}\)
\(sinh(x) = \frac{e^x-e^{-x}}{2}\)
n.ln (m) = ln(m)n
eln(x) = x
Calculation:
We know, \(sinh(x) = \frac{e^x-e^{-x}}{2}\)
Put x = ln 4
\(⇒ sin h(ln\ 4) = \frac{e^{ln {4}}-e^{-ln{4}}}{2}\)
\(⇒ sin h(ln\ 4) = \frac{e^{ln {(4)}}-e^{ln{(4)^{-1}}}}{2}\)
Since, n.ln (m) = ln(m)n
\(⇒ sinh (ln \ 4) = \frac{4 \ - \ 4^{-1}}{2}\)
\(⇒ \frac{\frac{16 - 1}{4}} {2}\)
⇒ 15/8
