Hyperbolic Functions

Question

Hey everyone, I need help with questions on hyperbolic functions. I was able to do part (a).

I proved for $\sinh(3y)$ by doing this:

\begin{align*} \sinh(3y) &= \sinh(2y +y)\\ &= \sinh(2y)\cosh(y) + \cosh(2y)\sinh(y)\\ &= 2\sinh(y)\cosh(y)\cosh(y) + (\cosh^2(y)+\sinh^2(y))\sinh(y)\\ &= 2\sinh(y)(1+\sinh^2(y)) + (1+\sinh^2(y) + \sinh^2(y))\sinh(y)\\ &= 2\sinh(y) + 2\sinh^3(y) + \sinh(y) +2\sinh^3(y)\\ &= 4\sinh^3(y) + 3\sinh(y). \end{align*}

Therefore, $0 = 4\sinh^3(y) + 3\sinh(y) - \sinh(3y)$.

I have no clue what to do for part (b) and part (c) but I do see similarities between part (a) and part(b) as you can subtitute $x = \sinh(y)$.

But yeah, I'm stuck and help would be very much appreciated.

Answer 1

Hint 1: Set $\color{#C00000}{x=\sinh(y)}$. Since $0=4\sinh^3(y)+3\sinh(y)-\sinh(3y)$, we have $$ 4x^3+3x-\sinh(3y)=0 $$ and by hypothesis, $$ 4x^3+3x-2=0 $$ So, if $\color{#C00000}{\sinh(3y)=2}$, both equations match. Solve for $x$.

Hint 2: Set $c\,x=\sinh(y)$ for appropriate $c$.

Answer 2

In the identity that you proved, put $\sinh 3y=2$. Then if $x=\sinh y$, the identity says that $4x^3+3x-2=0$. So we are almost finished, we have shown this $x$ is a solution of the equation.

Note that $\sinh t$ is a strictly increasing function, and that $\sinh t$ is large negative when $t$ is large negative, and large positive when $t$ is large positive. So it has an inverse function $\sinh^{-1}$, defined everywhere. We have $\sinh 3y=2$ if and only if $3y=\sinh^{-1} 2$ if and only if $y=\frac{1}{3}\sinh^{1}2$. Finally, to get $x$, take the $\sinh$ of this.

Now start from $ax^3+bx+c=0$. We want to manipulate this equation so that it will look like $4t^3+3t-d=0$, so that we can use the same trick.

We will make the substitution $x=kt$ for some constant $k$. This yields the equation $ak^3t^3+bkt +c=0$. We want the lead coefficient to be $4$. So multiply through by $\frac{4}{ak^3}$. We get the equation $$4t^3+\frac{4b}{ak^2}t+\frac{4b}{ak^3}c=0.\tag{$1$}$$ We want the coefficient of $t$ to be $3$. So we need $\frac{4b}{ak^2}=3$. That gives $$k=2\sqrt{\frac{b}{3a}}.$$

To solve the cubic $(1)$ using the $\sinh$ method, let $\sinh 3y=-\frac{4b}{ak^3}c$, where $k$ is as just calculated.