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Not sure what identity I should be using here: My gut tells me to use the Sin sum formula: $\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)$, but can't figure out how to.

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I can't make sense of the question in its current form. Please try to rephrase the question and put it also in the body (not only in the title) but also use LaTex to format it. Thx –  nt.bas Aug 6 '12 at 3:17
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(1) What is Asin? The $\,arcsin\,=$ the inverse of sine? (2)You had a "c" in the LHS argument: what happened to it in the RHS?... –  DonAntonio Aug 6 '12 at 3:18
    
A is a constant. A*sin(b+c). –  Murf Aug 6 '12 at 3:22
    
Ah, so I guess the question is: "Find $A, c$ such that $\sin(x) - \cos(x) = A\sin(x+c).$" –  user2468 Aug 6 '12 at 3:23
    
@J.D., thanks a lot in Scott's name! –  DonAntonio Aug 6 '12 at 3:25
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2 Answers

up vote 2 down vote accepted

Yes, you want to use that form. You have: $$A\sin(x+y)=A\sin x \cos y + A\sin y \cos x=\sin x - \cos x$$

So that means that $$A\cos y=1, A\sin y=-1$$

squaring the equations and adding them, we see that $$A^2\cos^2 y+A^2\sin^2 y=2$$ $$A^2=2$$ $$A=\pm\sqrt{2}$$

Let's take $A=\sqrt{2}$. Putting this into our first two equations, this implies that:

$$\sqrt{2}\cos y=1$$ $$\cos y=\frac 1 {\sqrt 2}$$

And similarly $$\sin y=\frac{-1}{\sqrt 2}$$

$y=-\pi/4$ solves both of these. So our answer is

$$\sin x - \cos x = A\sin(x+c)=\sqrt{2}\sin(x-\frac{\pi}4)$$

Note that there are other solutions that will work, but this is probably the simplest.

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Thanks...very well explained. –  Murf Aug 6 '12 at 3:32
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We can always express $a\sin x+b\cos x$ in the form $R\sin(x+\theta)$ where $R \ge 0$

If $a\sin x+b\cos x=R\sin(x+\theta)$

Or if $a\sin x+b\cos x=R\sin x\cos\theta + R\cos x\sin\theta$

Comparing the coefficients of $\sin x$ and $\cos x$,

$a=R\cos\theta$ and $b=R\sin\theta$.

Squaring & adding we get, $R^2=a^2+b^2=>R=\sqrt{a^2+b^2}$ as $R\ge 0$

Diving we get $\frac{R\sin\theta}{R\cos\theta}=\frac{b}{a}$,

or $\tan\theta=\frac{b}{a}\Rightarrow\theta=\tan^{-1}(\frac{b}{a})$.

Here, $a=1$, $b=-1$, so $R=\sqrt2$

and $\theta=\tan^{-1}(\frac{-1}{1})=\tan^{-1}(-1)$

Now this demands a bit care as $\tan^{-1}(-1)=n\pi-\frac{\pi}{4}$
=> $\theta$ can lie in the 2nd or in the 4th quadrant.

Observe that here $\cos\theta=\frac{1}{\sqrt2}$ and $\sin\theta=-\frac{1}{\sqrt2}$

So, $\theta$ lies in the 4th quadrant.

So, $\theta=2m\pi-\frac{\pi}{4}$ where m is any integer.

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TeX-related comment: You should use $\cos\theta$ or $\cos x$ and not $\ cos\theta$ or $\ cos x$; \ only adds space. Compare: $\cos\theta$ or $\cos x$ and not $\ cos\theta$ or $\ cos x$. –  Martin Sleziak Aug 6 '12 at 6:37
    
Thanks for your point, I will follow that. –  lab bhattacharjee Aug 6 '12 at 7:38
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