Hello

I would appreciate some help with the following question, which is, prove that:

$\displaystyle \cos\theta - \sin\theta = -\sqrt{2}\sin(\theta - 45^o)$

If I do it like this:

$\displaystyle \cos\theta - \sin\theta \equiv R\sin(\theta + \alpha)$

$\displaystyle \equiv R(\sin\theta\cos\alpha + \cos\theta\sin\alpha)$

$\displaystyle 1 = R\sin\alpha$

$\displaystyle -1 = R\cos\alpha$

$\displaystyle R = \sqrt{2}$

$\displaystyle \alpha = arctan (-1) = - 45^o$

I get:

$\displaystyle \sqrt{2}\sin(\theta - 45^o)$

I thought of doing it like this:

$\displaystyle \cos\theta - \sin\theta \equiv\sin(\theta - \alpha)$

$\displaystyle \equiv R(\sin\theta\cos\alpha - \cos\theta\sin\alpha)$

$\displaystyle \equiv -R(-\sin\theta\cos\alpha +\cos\theta\sin\alpha)$

but then $\displaystyle (-R)^2 = R^2$

So I can't seem to do it.

I did first have to show that:

$\displaystyle \cos\theta - \sin\theta = \sqrt{2}\cos(\theta + 45^o)$, which I was able to do.

Any help would be very much appreciated,

Thank you