Question

Write \(2{e^{2x}} + 3{e^{ - 2x}}\) in terms of \(\sinh 2x\)and\(\cosh 2x\).

Short answer

The expression in terms of \(\sin 2x\) and \(\cos 2x\)is;\(2{e^{2x}} + 3{e^{ - 3x}} = - \sinh 2x + 5\cosh 2x\).

Step-by-step solution

Definition of Hyperbolic function

The formulas for the hyperbolic function as shown below:

\(\begin{aligned}\sinh x = \frac{{{e^x} - {e^{ - x}}}}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\mathop{\rm csch}\nolimits} x = \frac{1}{{\sinh x}}\\\cosh x = \frac{{{e^x} + {e^{ - x}}}}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,{\mathop{\rm sech}\nolimits} x = \frac{1}{{\cosh x}}\\\tanh x = \frac{{\sinh x}}{{\cosh x}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\coth x = \frac{{\cosh x}}{{\sinh x}}\end{aligned}\)

Write \(2{e^x} + 3{e^{ - 2x}}\) in terms of \(\sinh 2x\) and \(\cosh 2x\)

Consider the expression as\(2{e^{2x}} + 3{e^{ - 2x}} = a\sinh 2x + b\cosh 2x\)……(1)

Write the above expression in terms of \({e^x}\) and \({e^{ - x}}\) as shown below:

\(\begin{aligned}2{e^{2x}} + 3{e^{ - 2x}} &= a\left( {\frac{{{e^{2x}} - {e^{ - 2x}}}}{2}} \right) + b\left( {\frac{{{e^{2x}} + {e^{ - 2x}}}}{2}} \right)\\2{e^{2x}} + 3{e^{ - 2x}} &= a\left( {\frac{{{e^{2x}}}}{2} - \frac{{{e^{ - 2x}}}}{2}} \right) + b\left( {\frac{{{e^{2x}}}}{2} + \frac{{{e^{ - 2x}}}}{2}} \right)\\2{e^{2x}} + 3{e^{ - 2x}} &= \frac{a}{2}{e^{2x}} - \frac{a}{2}{e^{ - 2x}} + \frac{b}{2}{e^{2x}} + \frac{b}{2}{e^{ - 2x}}\\2{e^{2x}} + 3{e^{ - 2x}} &= \frac{{a + b}}{2}{e^{2x}} + \frac{{ - a + b}}{2}{e^{ - 2x}}\end{aligned}\)

Compare the terms of \({e^{2x}}\) and \({e^{ - 2x}}\)as shown below:

\(\begin{aligned}\frac{{a + b}}{2} = 2\\a + b = 4\end{aligned}\)……(2)

And,

\(\begin{aligned}\frac{{ - a + b}}{2} = 3\\ - a + b = 6\end{aligned}\)……(3)

Solve the equation (2) and (3) to obtain as shown below:

\(\begin{aligned}2b = 10\\b = 5\end{aligned}\)

Substitute \(b = 5\) in equation (2) as shown below:

\(\begin{aligned}a + 5 &= 4\\a &= 5 - 4\\a &= - 1\end{aligned}\)

Substitute \(a = - 1\) and \(b = 5\) in equation (1) to obtain as shown below:

\(2{e^{2x}} + 3{e^{ - 3x}} = - \sinh 2x + 5\cosh 2x\)

Hence, the expression in terms of \(\sin 2x\) and \(\cos 2x\) is \(2{e^{2x}} + 3{e^{ - 3x}} = - \sinh 2x + 5\cosh 2x\).