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?
 A: Write 
$$f = 6 \sin(x) - 8 \cos(x) = \sqrt{6^2 + 8^2} (\frac {6}{\sqrt{6^2 + 8^2}} \sin(x) + \frac {8}{\sqrt{6^2 + 8^2}} \cos(x) ) $$
and now let
$$\frac {6}{\sqrt{6^2 + 8^2}} = \sin(\alpha)$$
then (remember $\sin(z)^2 + \cos(z)^2 = 1$ for any $z$)
$$\frac {8}{\sqrt{6^2 + 8^2}} = \cos(\alpha)$$
And we can write
$$f = 10 \;(\;\sin(\alpha)\sin(x) - \cos(\alpha) \cos(x)\; ) = - 10 \cos(x+\alpha) $$
where we have used the trigonometric addition theorem.     
Hence the maximum is obviously $10$.
EDIT
As a bonus let us determine the values of $x$ where $f$ has its maximum (and its minimum).
From the previous result we have to look for the extrema of $\cos(z)$ which leads us to
$$f\to -10:  x =2 n \pi - \alpha $$ 
$$f\to +10:  x=(2 n+1) \pi - \alpha$$
Where
$$\alpha = \arctan(\frac{3}{4})\;\simeq 0.6435010.643501 $$
and $n$ integer.
The extrema closest to the origin are at
$$x_{min} = - \alpha\;\simeq -0.6435010.643501 $$
$$x_{max} = \pi - \alpha \; \simeq 2.498091544796$$
A: Suppose,
$$a\cos(\theta)-b\sin (\theta)=R \cos (\theta+y)$$
Expanding out the right hand side using the trigonometric angle addition theorem we have,
$$R\cos (\theta)\cos(y)-R\sin (\theta)\sin(y)$$
Than all we need to make sure is that,
$$a=R\cos (y)$$
$$b=R \sin (y)$$
Squaring both equations and adding gives,
$$R^2=a^2+b^2$$
But we need to chose $y$ so both the previous conditions are satisfied.
In the case $a=6$ and $b=8$ then $R^2=6^2+8^2=100$. Let us chose $R=10$.
Then we need to satisfy,
$$6=10\cos (y)$$
$$8=10 \sin (y)$$
Dividing the second equation by the third first $\tan (y)=\frac{8}{6}=\frac{4}{3}$. It is easy to check that $\arctan (\frac{4}{3}) \in [0,\frac{\pi}{2}]$ solves both equations, so it is a valid one. And we have,
$$6\cos (\theta)-8 \sin (\theta)=10 \cos( \theta+\arctan (\frac{4}{3}))$$
A: 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
$$-\sqrt{a^2+b^2}\leq a\sin x+b\cos x\leq \sqrt{a^2+b^2}$$
