# a cos x + b sinx

• Jul 19th 2012, 04:11 PM
Furyan
a cos x + b sinx
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
• Jul 19th 2012, 04:43 PM
emakarov
Re: a cos x + b sinx
Quote:

Originally Posted by Furyan
prove that:

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

Rewrite the right-hand side using the formula $\displaystyle \sin(\theta-\alpha)=\sin\theta\cos\alpha - \cos\theta\sin\alpha$.
• Jul 19th 2012, 04:52 PM
Kmath
Re: a cos x + b sinx
Why you do not try this,

$\displaystyle \cos \theta -\sin \theta=\sqrt{2}(\cos \theta \frac{1}{\sqrt{2}}-\sin\theta\frac{1}{\sqrt{2}})$
$\displaystyle =-\sqrt{2}(\sin\theta\cos 45 -\cos\theta\sin45)=-\sqrt{2}\sin{(\theta-45)}$
• Jul 19th 2012, 05:03 PM
emakarov
Re: a cos x + b sinx
Quote:

Originally Posted by Furyan
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:

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

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

$\displaystyle \cos\theta-\sin\theta=\sqrt{2}\sin(\theta+135)=\sqrt{2}\sin(1 80-\theta-135)=\sqrt{2}\sin(45-\theta)$ $\displaystyle =-\sqrt{2}\sin(\theta-45)$.
• Jul 19th 2012, 05:23 PM
Furyan
Re: a cos x + b sinx
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. (Doh)
• Jul 19th 2012, 05:42 PM
Furyan
Re: a cos x + b sinx
(Bow)Thank you,

(I will see if I can get:

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

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

Thank you very much indeed for your help. (Bow)

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