Alright so this problem has to do quite a bit with trigonometric identities.To start, write the question as an equation.\[sinx - cosx = Asin(x+c)\]The first thing that comes to mind is that it's possible to rewrite the right side using the compound angle formula.\[sinx - cosx = Asin(x+c)\]\[sinx - cosx = Asinxcosc + Acosxsinc\]From here on it's possible to make a system of equations:\[Asinxcosc = sinx\]\[Acosc = 1\] and\[Acosxsinc = -cosx\]\[Asinc = -1\]Now what we're going to do here is make them look like some sort of pythagorean identity so as to simplify them down to one variable.\[Acosc = 1\]\[A^2\cos^2c = 1\]and the other equation,\[Asinc = -1\]\[A^2\sin^sc = 1\]Now the next part may not be quite obvious, but we can solve for A by adding both of the equations together.\[A^2\sin^2c + A^2\cos^2c = 2\]\[A^2(\sin^2c + \cos^2c) = 2\]\[A^2(1) = 2\]\[A=\sqrt{2}\]In the first step, we added together both equations which was in equality to 2. Next we just factored out the A squared so that we could use the pythagorean identity to simplify the equation, and solve for A. So now that we have A solved, we need to solve for c.
Rewrite the original equation with the A value that we solved.\[\sqrt{2}sinxcosc + \sqrt{2}cosxsinc = sinx - cosx\]Using the same two system of equations we can now solve for c:\[\sqrt{2}sinxcosc = sinx\]\[\sqrt{2}cosc = 1\]\[cosc = \frac{ 1 }{ \sqrt{2} }\]\[c = \frac{ \pi }{ 4 }, \frac{7\pi}{4}\]Now for the other equation:\[\sqrt{2}cosxsinc = -cosx\]\[\sqrt{2}sinc = -1\]\[sinc = \frac{-1}{\sqrt{2}}\]\[c = \frac{5\pi}{4}, \frac{7\pi}{4}\]Because c cannot equal two values, we will use the related angles for both equations. \[sinx - cosx = \sqrt{2}sinxcos\frac{7\pi}{4} + \sqrt{2}cosxsin\frac{7\pi}{4}\]\[sinx - cosx = \sqrt{2}\sin(x + \frac{7\pi}{4})\]Although this is right, it is one of the many ways you can express it because sine is periodic. Essentially, this is equivalent to:\[sinx - cosx = \sqrt{2}\sin(x-\frac{\pi}{4})\]