1. ## Trigonometric Function

Hello, I need help on this question:

Find the value, or values, of x for which $4sinhx - 3coshx = 5$ giving your answer, or answers, to 3 significant figures.

2. Originally Posted by Air
Hello, I need help on this question:

Find the value, or values, of x for which $4sinhx - 3coshx = 5$ giving your answer, or answers, to 3 significant figures.

There are a couple of ways. Here's one that uses the basic definitions:
$4sinh(x) - 3cosh(x) = 5$

$4 \cdot \frac{e^x - e^{-x}}{2} - 3 \cdot \frac{e^x + e^{-x}}{2} = 5$

Now for a little grouping work:
$2(e^x - e^{-x}) - \frac{3}{2}(e^x + e^{-x}) = 5$

$\left ( 2 - \frac{3}{2} \right ) e^x + \left (-2 - \frac{3}{2} \right ) = 5$

$\frac{1}{2}e^x - \frac{7}{2}e^{-x} = 5$

Multiply both sides by $2e^x$:
$e^{2x} - 7 = 10e^x$

$e^{2x} - 10e^x - 7 = 0$

This is a quadratic in $e^x$ so you can finish this. If that is hard for you to see then:

Let $y = e^x$. The equation becomes:
$y^2 - 10y - 7 = 0$

Solve this with the quadratic formula (or completing the square) then take the natural log of it to get x.

For reference I got $x \approx 2.3662$.

-Dan

3. Hello, Air!

Solve: . $4\sinh x - 3\cosh x \:= \:5$ . to 3 significant figures.
I would use the defintions of $\sinh x$ and $\cosh x$

. . $4\overbrace{\left(\frac{e^x - e^{-x}}{2}\right)}^{\sinh x} \:-\; 3\overbrace{\left(\frac{e^x + e^{-x}}{2}\right)}^{\cosh x} \;=\;5$

Multiply by 2: . $4(e^x-e^{-x}) - 3(e^x+e^{-x}) \:=\:10$

This simplifies to: . $e^x - 10 - 7e^{-x} \:=\:0$

Multiply by $e^x\!:\;\;e^{2x} - 10e^x - 7 \:=\:0 \quad\Rightarrow\quad(e^x)^2 - 10(e^x) - 7 \:=\:0$ . . . a quadratic

Quadratic Formula: . $e^x \;=\;\frac{10 \pm\sqrt{10^2 - 4(1)(-7)}}{2(1)} \;=\;\frac{10 \pm\sqrt{128}}{2}\;=\;\frac{10 \pm8\sqrt{2}}{2}$

Therefore: . $e^x \:=\:5 + 4\sqrt{2}\quad\Rightarrow\quad x \;=\;\ln(5 + 4\sqrt{2}) \;\approx\;2.37$

4. Originally Posted by Soroban
Hello, Air!

I would use the defintions of $\sinh x$ and $\cosh x$

. . $4\overbrace{\left(\frac{e^x - e^{-x}}{2}\right)}^{\sinh x} \:-\; 3\overbrace{\left(\frac{e^x + e^{-x}}{2}\right)}^{\cosh x} \;=\;5$

Multiply by 2: . $4(e^x-e^{-x}) - 3(e^x+e^{-x}) \:=\:10$

This simplifies to: . $e^x - 10 - 7e^{-x} \:=\:0$

Multiply by $e^x\!:\;\;e^{2x} - 10e^x - 7 \:=\:0 \quad\Rightarrow\quad(e^x)^2 - 10(e^x) - 7 \:=\:0$ . . . a quadratic

Quadratic Formula: . $e^x \;=\;\frac{10 \pm\sqrt{10^2 - 4(1)(-7)}}{2(1)} \;=\;\frac{10 \pm\sqrt{128}}{2}\;=\;\frac{10 \pm8\sqrt{2}}{2}$

Therefore: . $e^x \:=\:5 + 4\sqrt{2}\quad\Rightarrow\quad x \;=\;\ln(5 + 4\sqrt{2}) \;\approx\;2.37$

Edit: Too fast for me, Dan!

5. Originally Posted by Soroban
Edit: Too fast for me, Dan!
Yeah, but you make fewer mistakes than I do. I'm always happy when I see that I got the same answer you did.

-Dan