If sinh x = frac{12}{5},then sinh 3x + cosh 3 | vedclass

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(A) We know that $\sinh x = \frac{e^x - e^{-x}}{2}$ and $\cosh x = \frac{e^x + e^{-x}}{2}$.Given $\sinh x = \frac{12}{5}$,we have $\frac{e^x - e^{-x}}

Vedclass · Accepted Answer

$125$

Vedclass · Answer

(A) We know that $\sinh x = \frac{e^x - e^{-x}}{2}$ and $\cosh x = \frac{e^x + e^{-x}}{2}$.Given $\sinh x = \frac{12}{5}$,we have $\frac{e^x - e^{-x}}{2} = \frac{12}{5}$,which implies $e^x - e^{-x} = \frac{24}{5}$.Let $e^x = t$. Then $t - \frac{1}{t} = \frac{24}{5}$.Multiplying by $5t$,we get $5t^2 - 24t - 5 = 0$.Factoring the quadratic equation: $(5t + 1)(t - 5) = 0$.Since $e^x = t$ must be positive,$t = 5$,so $e^x = 5$.We need to find $\sinh 3x + \cosh 3x$.Using the definitions,$\sinh 3x + \cosh 3x = \frac{e^{3x} - e^{-3x}}{2} + \frac{e^{3x} + e^{-3x}}{2} = e^{3x}$.Since $e^x = 5$,$e^{3x} = (e^x)^3 = 5^3 = 125$.

If $\sinh x = \frac{12}{5}$,then $\sinh 3x + \cosh 3x = $

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