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.
Show in the form $A\sin(x+c)=\sin x - \cos x$?
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0And if you know the formula for $\sin x+\sin y$, see [Wikipedia](http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Product-to-sum_and_sum-to-product_identities), you could combine it with the fact that $-\cos x= -\sin(\pi/2-x)=\sin(x-\pi/2)$ – 2012-08-06
2 Answers
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.
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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1TeX-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 $\ cos x$. – 2012-08-06