# cosx-sinx

• Jul 28th 2008, 01:20 PM
hoops2008
cosx-sinx
Hi Everbody lost my basic maths books was wondering how to

convert cosx-sinx into a single sine function
• Jul 28th 2008, 02:41 PM
TwistedOne151
Sum to product identity
You can use one of the sum-to-product identities.
For example,
use $\sin\theta=\cos\left(\frac{\pi}{2}-\theta\right)$
along with
$\cos{x}-\cos{y}=-2\sin\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right)$
To get
$\cos{x}-\sin{x}=\cos{x}-\cos\left(\frac{\pi}{2}-x\right)$
$=-2\sin\left(\frac{x+\left(\frac{\pi}{2}-x\right)}{2}\right)\sin\left(\frac{x-\left(\frac{\pi}{2}-x\right)}{2}\right)$
$=-2\sin\left(\frac{\pi}{4}\right)\sin\left(\frac{2x-\frac{\pi}{2}}{2}\right)$
$=-\sqrt{2}\sin\left(x-\frac{\pi}{4}\right)$
Or, using $\sin(\theta+\pi)=-\sin\theta$,
$\cos{x}-\sin{x}=\sqrt{2}\sin\left(x-\frac{\pi}{4}+\pi\right)$
$=\sqrt{2}\sin\left(x+\frac{3\pi}{4}\right)$.

--Kevin C.
• Jul 28th 2008, 07:38 PM
Soroban
Hello, hoops2008!

Another approach . . .

Quote:

Convert $\cos x -\sin x$ into a single sine function
Multiply by $\frac{\sqrt{2}}{\sqrt{2}}\!:\;\;\frac{\sqrt{2}}{\s qrt{2}}\,(\cos x - \sin x) \;=\;\sqrt{2}\left(\frac{1}{\sqrt{2}}\cos x - \frac{1}{\sqrt{2}}\sin x\right)$

Note that: . $\bigg\{\begin{array}{c}\sin\frac{\pi}{4} \:=\:\frac{1}{\sqrt{2}} \\ \cos\frac{\pi}{4} \:=\:\frac{1}{\sqrt{2}} \end{array}$

So we have: . $\sqrt{2}\,\left(\sin\frac{\pi}{4}\cos x - \cos\frac{\pi}{4}\sin x\right) \;=\;\sqrt{2}\,\sin\left(\frac{\pi}{4} - x\right)

$

• Jul 29th 2008, 01:26 PM
hoops2008
Thanks for the replys fellas. I loved doing this kind of maths a long time ago and i am now very rusty.