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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.

Willie Wong
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George Randall
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    The statement in part (a) is analogous to the circular trig identity for $\sin 3\theta$. They are giving you a big hint about the remaining parts: compare the equations in (a) and (b) or (c) term-for-term... – colormegone Apr 22 '13 at 14:23

2 Answers2

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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 $c$.

robjohn
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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.

André Nicolas
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  • i still don't understand part (c) – George Randall Apr 22 '13 at 16:13
  • Let's try to pin down what you don't understand. By making the change of variable $x=kt$ for suitable $k$, we transform $ax^2+bx+c=0$ into an equation of shape $4t^3+3t-d$, which we can handle like in the numerical example. OK so far? So do the substitution, and multiply by a constant to make the lead coefficient $4$. Now pick $k$ so that the coefficient of $t$ will be $3$. This can be done. If you indicate where you have difficulty, I can expand the answer to give detail. It could be useful to you to take concrete values of $a$, $b$, $c$ and calculate along the lines of the above answer. – André Nicolas Apr 22 '13 at 16:27