$-7\cos(x) + 24\sin(x) = -25\sin(x - 16.26..) Why?$ Im confused as to why the $x$ angle is $-16.26..$ and not $360-16.26$ as well as why its $-25$ and not $25$? 
I dont understand why its $-25sin(x-16.26..)$ and not $25sin(x+343.74..)$
Can someone please help me! thank you. 
 A: Here. I found this paper that explains why this method works, but I will try and summarize.
This formula is derived from the trig identity $$\cos(x-\alpha) = \cos(x)\cos(\alpha) + \sin(x)\sin(\alpha)$$
Basically,
$$R\cos(x-\alpha) = R\cos(x)\cos(\alpha) + R\sin(x)\sin(\alpha)$$
This means for an equation $b\sin(x)+a\cos(x)$,
$$a = R\cos(\alpha)$$ 
and
$$b = R\sin(\alpha)$$
Additionally, $a^2+b^2 = R^2(\cos^2(\alpha) + \sin^2(\alpha)) = R^2$, which means $R = \pm\sqrt{a^2+b^2}$. Moreover, $$\frac{b}{a} = \frac{R\sin(\alpha)}{R\cos(\alpha)} = \tan(\alpha)$$
This means $\alpha = \tan^{-1}(\frac{b}{a})$. Thus, your equation can be simplified.
$$-7\sin(x) + 24\cos(x) = -\sqrt{7^2+24^2}\cos(x-\tan^{-1}(\frac{-24}{7})) = -25\cos(x+1.28700221759\ldots)$$
$\cos(x) = \sin(x + \frac{\pi}{2})$ implies the following.\
$$-7\sin(x) + 24\cos(x) =-25\sin(x+1.28700221759\ldots + \frac{(4n-3)\pi}{2})$$
When $n$ is $-2$ you get the equation you had above.
A: It follows directly from the identity
$$\sin(x-\theta)=\sin x\cos\theta-\cos x\sin\theta$$
$$-25\sin(x-\theta)=-7\cos x+24\sin x\iff\sin(x-\theta)=-\frac{24}{25}\sin x+\frac7{25}\cos x$$
Let $\theta$ be some angle such that
$$\begin{cases}\cos\theta=-\frac{24}{25}\\[1ex]\sin\theta=\frac7{25}\end{cases}$$
Then
$$\tan\theta=\frac{\sin\theta}{\cos\theta}=-\frac7{24}\implies\theta=\tan^{-1}\left(-\frac7{24}\right)+n\pi\approx(-16.26+360n)^\circ$$
where $n$ is any integer.
