Question

Given that $\cos hx=\frac{5}{4}$, determine the values of $\sin h2x$
Use the formula for $\cos h(2x+x)$ to determine the value of $\cos h3x$.

Answer 1

we have,

$\cos hx=\frac{5}{4}$

and,

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

$4e^{2x}-10ex+4=0$

$e^x=\frac{10{+}_{-}\sqrt{100-64}}{8}$

$e^x=2$ or $\frac{1}{2}$

now,

here we have to calculate $\cos 3hx$ by spliting that into $\cos h\left(2x+x\right)$

$\sin hx=\frac{3}{4}$ or $-\frac{3}{4}$

$\left(i\right)\sin h2x=2\cos hx\sin hx$

$\sin 2hx=\frac{15}{8}$ or $\frac{-15}{8}$

$\left(ii\right)\cos h2x=2(\cos hx{)}^2-1$

$\cos h2x=\frac{17}{8}$

$\cos h3x=\cos h(2x+x)=\cos h2x\cos hx+\sin h2x\sin hx$

Hence, $\cos 3hx=\frac{40}{32}$