How to find max integer value of $6\sin(x)-8\cos(x)$ without using derivative

Question

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?

Answer 1

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

$$x_{min} = - \alpha\;\simeq -0.6435010.643501 $$ $$x_{max} = \pi - \alpha \; \simeq 2.498091544796$$

Answer 2

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,

$$6\cos (\theta)-8 \sin (\theta)=10 \cos( \theta+\arctan (\frac{4}{3}))$$

Answer 3

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 $$-\sqrt{a^2+b^2}\leq a\sin x+b\cos x\leq \sqrt{a^2+b^2}$$