How to find max integer value of $6\sin(x)-8\cos(x)$ without using derivative I have $6\sin(x)-8\cos(x)$ and want to find its maximum value. If we use derivative and assuming that it is equal to 0, we get $6\cos(x)+8\sin(x)=0$ which implies that $\tan(x)=-\cfrac{3}{4}$ and $x = 143^\circ$ and we can conclude that the maximum value is equal to $10$ . How can I find the maximum integer value of $6\sin(x)-8\cos(x)$ without using derivative?

Let $\dfrac{b}{a}=\tan\alpha$ then
$$\cos\alpha=\dfrac{1}{\sqrt{1+\tan^2\frac{\alpha}{2}}}=\dfrac{a}{\sqrt{a^2+b^2}}$$
so
\begin{eqnarray}
a\sin x+b\cos x
&=&
a(\sin x+\dfrac{b}{a}\cos x)\\
&=&
a(\sin x+\tan\alpha\cos x)\\
&=&
a(\sin x+\dfrac{\sin\alpha}{\cos\alpha}\cos x)\\
&=&
\dfrac{a}{\cos\alpha}(\sin x\cos\alpha+\sin\alpha\cos x)\\
&=&
\dfrac{a}{\cos\alpha}\sin(x+\alpha)\\
&=&
\sqrt{a^2+b^2}\sin(x+\alpha)
\end{eqnarray}
from
$$-1\leq\sin(x+\alpha)\leq1$$
we have
$$-\sqrt{a^2+b^2}\leq a\sin x+b\cos x\leq \sqrt{a^2+b^2}$$
