Question

Solve : If $coshx=\frac{5}{4}$ then cosh (3x)=

Answer 1

$\cosh x=\frac{e^x+e^{-x}}{2}=\frac{5}{4}$

$\cosh \left(3x\right)=\frac{e^{3x}+e^{-3x}}{2}=?$

$e^x+e^{-x}=\frac{5}{2}$

${(e^x+e^{-x})}^3=\frac{125}{8}$

$e^{3x}+e^{-3x}+3e^x{e}^{-x}(e^x+e^{-x})=\frac{125}{8}$

$e^{3x}+e^{-3x}+3\left(\frac{5}{2}\right)=\frac{125}{8}$

$e^{3x}+e^{-3x}=\frac{125}{8}-\frac{15}{2}=\frac{65}{8}$

$\cosh \left(3x\right)=\frac{e^{3x}+e^{-3x}}{2}=\frac{65}{16}$