Trigonometric inequality to prove Prove the following inequality for each $x$:
$$|\sin x+7\cos x| \leq \sqrt{50}$$
My remark: It seems to me that the solution uses the fact that $1^2+7^2=50$.
Thanks in advance!

One of the ways is to use the addition formula: $R \sin (a+x) = (R\cos a) \sin x + (R\sin a) \cos x$. We let $R \cos a = 1$ and $R \sin a = 7$. Hence $50 = 1^2 + 7^2 = R^2 (\cos^2 + \sin^2) = R^2$ and $R = \sqrt{50}$. 
$|\sin x + 7\cos x| = |\sqrt{50} \sin (a+x)| \leq \sqrt{50}$
