Question #b9406

Jan 10, 2018

$\sin \left(8 \theta\right) = \frac{u}{8 a}$

Explanation:

I'm not too sure of your question.

I take it you are asking:

$\sin \theta = \frac{u}{a}$; What is $\sin \left(8 \theta\right)$?

We can relate this back to transformations of graphs in general. We're going to be moving away from trigonometry for just a bit!

For example:

$\text{Let } f \left(x\right) = {x}^{2} - 9 x + 18$
How about for a moment we consider the roots of $f \left(x\right) = 0$
$\text{Let } f \left(x\right) = 0$
${x}^{2} - 9 x + 18 = 0$
$\left(x - 3\right) \left(x - 6\right) = 0$
So this equation has roots of $x = 3$ and $x = 6$
graph{y=x^2-9x+18 [-10, 10, -5, 5]}

Consider $f \left(2 x\right)$
$f \left(2 x\right) = {\left(2 x\right)}^{2} - 9 \left(2 x\right) + 18$
$= 4 {x}^{2} - 18 x + 18$
Now, consider the roots of $f \left(2 x\right) = 0$
$\text{Let } f \left(2 x\right) = 0$
$4 {x}^{2} - 18 x + 18 = 0$
$2 {x}^{2} - 9 x + 9 = 0$
$x = \frac{9 \pm \sqrt{{\left(- 9\right)}^{2} - 4 \times 2 \times 9}}{2 \times 2}$
$x = 3$ or $x = \frac{3}{2}$
graph{4x^2-18x+18 [-10, 10, -5, 5]}

Look at the roots.
$f \left(x\right)$ has roots $x = 6$ and $x = 3$
$f \left(2 x\right)$ has roos $x = \frac{6}{2}$ and $x = \frac{3}{2}$

This isn't a proof, but it's enough for us to infer that:

if an expression $f \left(x\right)$ has roots $x = \alpha , \beta , \gamma , \delta \ldots$, then $f \left(k x\right)$ has roots $x = \frac{\alpha}{k} , \frac{\beta}{k} , \frac{\gamma}{k} , \frac{\delta}{k} \ldots$

Back to our trig

$\text{Let } g \left(x\right) = \sin x$ (because I used $f \left(x\right)$ further up).

For some value $x = \theta , g \left(\theta\right) = \frac{u}{a}$
$g \left(8 x\right) = \sin \left(8 x\right)$
$\sin \left(8 \theta\right) = g \left(8 \theta\right) = \frac{\frac{u}{a}}{8}$
$\sin \left(8 \theta\right) = \frac{u}{8 a}$