Question

Given that $\sin hx=\frac{5}{12}$, find the values of $\cos hx$
Determine the value of $x$ as a natural logarithm

Answer 1

$\sin hx=\frac{5}{12}$

$\frac{e^x-e^{-x}}{2}=\frac{5}{12}$

$12(e^x+e^{-x})=10$

$12(e^{2x}-1)=10e^x$

$12e^{2x}-10e^x-12=0$

$e^x=\frac{10{+}_{-}\sqrt{100+576}}{24}$

$e^x=\frac{3}{2}$ or $\frac{-2}{3}$

Since, Negative value should not be choosed because exponential functions have only positive value function

$e^x=\frac{3}{2}$

and

$e^{-x}=\frac{2}{3}$

now we have to find value of $\cos hx$

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

Hence,

$\cos hx=\frac{13}{12}$

and We know that,

$e^x=\frac{3}{2}$

So,

$x=log\left(\frac{3}{2}\right)$