Approximation

**Punch** · August 29th 2011, 07:54 AM

Show that for small $x$ , $\frac{sinx-1}{cos(x-\frac{\pi}{4})} \approx \sqrt2(-1+2x-\frac{5}{2}x^2)$

$\frac{sinx-1}{cos(x-\frac{\pi}{4})}\approx\frac{x-1}{1-\frac{(x-\frac{\pi}{4})^2}{2}}$

**Siron** · August 29th 2011, 08:55 AM

Maybe you can do something with limits ( $x \to 0$ ), or isn't that the intention of this exercice? ...

**Punch** · August 29th 2011, 11:05 PM

But if I do that, then x will not be shown in the approximated answer

**Prove It** · August 30th 2011, 01:43 AM

Originally Posted by Punch

Show that for small $x$ , $\frac{sinx-1}{cos(x-\frac{\pi}{4})} \approx \sqrt2(-1+2x-\frac{5}{2}x^2)$

$\frac{sinx-1}{cos(x-\frac{\pi}{4})}\approx\frac{x-1}{1-\frac{(x-\frac{\pi}{4})^2}{2}}$

For starters, $\displaystyle \begin{align*} \cos{\left(x - \frac{\pi}{4}\right)} &= \cos{(x)}\cos{\left(\frac{\pi}{4}\right)} + \sin{(x)}\sin{\left(\frac{\pi}{4}\right)} \\ &= \frac{1}{\sqrt{2}}\cos{(x)} + \frac{1}{\sqrt{2}}\sin{(x)} \\ &= \frac{\cos{(x)} + \sin{(x)}}{\sqrt{2}} \end{align*}$

So

$\displaystyle \begin{align*} \frac{\sin{(x)} - 1}{\cos{\left(x - \frac{\pi}{4}\right)}} &= \frac{\sin{(x)} - 1}{\frac{\cos{(x)} + \sin{(x)}}{\sqrt{2}}} \\ &= \sqrt{2}\left[\frac{\sin{(x)} - 1}{\cos{(x)} + \sin{(x)}}\right] \end{align*}$

Now I would advise substituting the MacLaurin Series for $\displaystyle \sin{(x)}$ and $\displaystyle \cos{(x)}$ and seeing what you can do...

**CaptainBlack** · August 30th 2011, 02:42 AM

Originally Posted by Punch

Show that for small $x$ , $\frac{sinx-1}{cos(x-\frac{\pi}{4})} \approx \sqrt2(-1+2x-\frac{5}{2}x^2)$

$\frac{sinx-1}{cos(x-\frac{\pi}{4})}\approx\frac{x-1}{1-\frac{(x-\frac{\pi}{4})^2}{2}}$

Expand $\sin(x)-1$ and $1/\cos(x-\pi/4)$ about $x=0$ , as far as the term in $x^2$ (if there is one) and multiply out and truncate at the $x^2$ term.

CB

**Punch** · August 30th 2011, 03:24 AM

Originally Posted by Prove It

For starters, $\displaystyle \begin{align*} \cos{\left(x - \frac{\pi}{4}\right)} &= \cos{(x)}\cos{\left(\frac{\pi}{4}\right)} + \sin{(x)}\sin{\left(\frac{\pi}{4}\right)} \\ &= \frac{1}{\sqrt{2}}\cos{(x)} + \frac{1}{\sqrt{2}}\sin{(x)} \\ &= \frac{\cos{(x)} + \sin{(x)}}{\sqrt{2}} \end{align*}$

So

$\displaystyle \begin{align*} \frac{\sin{(x)} - 1}{\cos{\left(x - \frac{\pi}{4}\right)}} &= \frac{\sin{(x)} - 1}{\frac{\cos{(x)} + \sin{(x)}}{\sqrt{2}}} \\ &= \sqrt{2}\left[\frac{\sin{(x)} - 1}{\cos{(x)} + \sin{(x)}}\right] \end{align*}$

Now I would advise substituting the MacLaurin Series for $\displaystyle \sin{(x)}$ and $\displaystyle \cos{(x)}$ and seeing what you can do...

Following up, $\sqrt2[\frac{sinx-1}{cosx+sinx}]=\sqrt2[\frac{x-1}{1-\frac{x^2}{2}+x}]$

but I cant convert it into the given form

**CaptainBlack** · August 30th 2011, 05:08 AM

Originally Posted by Punch

Following up, $\sqrt2[\frac{sinx-1}{cosx+sinx}]=\sqrt2[\frac{x-1}{1-\frac{x^2}{2}+x}]$

but I cant convert it into the given form

$\frac{1}{1-\frac{x^2}{2}+x}=\frac{1}{1-\left(\frac{x^2}{2}-x\right)}=1+\left(\frac{x^2}{2}-x\right)+\left(\frac{x^2}{2}-x\right)^2+..\\ \\ \\ \phantom{xxxxxxxxxxx}=1-x+\frac{3x^2}{2}+..$

CB

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