<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] Let us first develope a general methodology by considering the general identity: 
$$\ \tag{1} f(x)=A \cos(x) + B \sin(x)=C\left(\dfrac{A}{C}\cos(x) + \dfrac{B}{C} \sin(x)\right)$$
with $C:=\sqrt{A^2+B^2}$. 
As $\left(\dfrac{A}{C}\right)^2 + \left(\dfrac{B}{C}\right)^2=1$, point $\left(\dfrac{A}{C},\dfrac{B}{C}\right)$ is on the unit circle.
Thus, there exist an $x_0$ such that 
$$\ \tag{2} \cos(x_0)=\dfrac{A}{C} \ \ and \ \ \sin(x_0)=\dfrac{B}{C}$$
Remark : $\tan(x_0)=\dfrac{B}{A}$ (unless $A=0$).
Plugging (2) into (1) gives $f(x) = C cos(x-x_0)$
which is thus maximal for $x=x_0$ with maximal value $C$.
With the numerical values at hand, using remark above, the maximum occurs for $x=x_0=$atan$\left( \dfrac{-7}{2}\right)\approx -1.2925...$ and its value is 

$$\sqrt{53}\approx7.2801...$$ 

as can be seen on figure below.

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