a cos x + b sinx

**Furyan** · July 19th 2012, 04:11 PM

Hello

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

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

If I do it like this:

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

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

$1 = R\sin\alpha$

$-1 = R\cos\alpha$

$R = \sqrt{2}$

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

I get:

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

I thought of doing it like this:

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

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

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

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

So I can't seem to do it.

I did first have to show that:

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

Any help would be very much appreciated,

Thank you

**emakarov** · July 19th 2012, 04:43 PM

Originally Posted by Furyan

prove that:

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

Rewrite the right-hand side using the formula $\sin(\theta-\alpha)=\sin\theta\cos\alpha - \cos\theta\sin\alpha$ .

**Kmath** · July 19th 2012, 04:52 PM

Why you do not try this,

$\cos \theta -\sin \theta=\sqrt{2}(\cos \theta \frac{1}{\sqrt{2}}-\sin\theta\frac{1}{\sqrt{2}})$
$=-\sqrt{2}(\sin\theta\cos 45 -\cos\theta\sin45)=-\sqrt{2}\sin{(\theta-45)}$

**emakarov** · July 19th 2012, 05:03 PM

Originally Posted by Furyan

If I do it like this:

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

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

$1 = R\sin\alpha$

$-1 = R\cos\alpha$

$R = \sqrt{2}$

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

I get:

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

The error here is that $\tan(\alpha)=-1$ does not imply that $\alpha=-45^\circ$ . In this case, $\sin\alpha=1/\sqrt{2}$ and $\cos\alpha=-1/\sqrt{2}$ , so $\alpha=135^\circ$ . Thus,

$\cos\theta-\sin\theta=\sqrt{2}\sin(\theta+135)=\sqrt{2}\sin(1 80-\theta-135)=\sqrt{2}\sin(45-\theta)$ $=-\sqrt{2}\sin(\theta-45)$ .

**Furyan** · July 19th 2012, 05:23 PM

Thank you both for taking the time to reply.

(There's nothing in the chapter I'm working through about either of the methods you have shown me, but I will study your posts in detail to see what I can learn from them.

Thank you very much for you efforts.)

Edit: Ah, actually there is an example showing how to determine which quadrant the angle is in. I never learnt that, I'm not sure how I got by without it. It seems pretty important, in fact, looking back it's one of the first things I should have learnt in trigonometry.

**Furyan** · July 19th 2012, 05:42 PM

Thank you,

(I will see if I can get:

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

just now seeing that $\alpha= 135$ is beyond me, but I guess I need to know that.)

Thank you very much indeed for your help.

Edit: I think I understand now. Since $\sin\alpha > 0$ and both $\cos\alpha$ and $\tan\alpha < 0, \alpha$ is in the second quadrant, $90^o \leq \alpha \leq 180^o$ . Thanks for pointing that out. I'm glad I know that now.

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