<s>[INST] What is the maximal value of $2 \sin x - 7 \cos x$? What is the maximal value of $2 \sin x - 7 \cos x$?
How do I calculate this? Do I have to write out the $\sin$ and the $\cos$?
 [/INST] Hint: one can use formula
$$\sin x \cos y- \cos x \sin y = \sin(x-y),$$
and use such 'tricky' $y_0$, that
$$
\sin y_0 = \dfrac{7}{?}, \;\cos y_0 = \dfrac{2}{?}.
$$

 Let $y_0$ is such that $\sin y_0 = \dfrac{7}{\sqrt{7^2+2^2}} = \dfrac{7}{\sqrt{53}}$, and $\cos y_0 = \dfrac{2}{\sqrt{53}}$, namely $y_0 = \arcsin(7/\sqrt{53})$. 

Then $2\sin x - 7\cos x =\sqrt{53}\left(\dfrac{2}{\sqrt{53}}\sin x - \dfrac{7}{\sqrt{53}}\cos x\right)
 = \sqrt{53}\sin(x-y_0)$.
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