# $f(x) = (\cos x - \sin x) (17 \cos x -7 \sin x)$

$f(x) = (\cos x - \sin x) (17 \cos x -7 \sin x)$

Determine the greatest and least values of $\frac{39}{f(x)+14}$ and state a value of x at which greatest values occurs.

Do I just use a graphing calculator for this? Is there a way I could do this without a graphing calculator?

• May be computing the derivative and looking at the solutions of $\Big(\frac{39}{f(x)+14}\Big)'=0$ – Claude Leibovici Mar 23 '15 at 9:16

you can do this without calculus. here is a way. i will use $t$ for $x.$ we have \begin{align}y &= 14 + (\cos t - \sin t)(17 \cos t - 7 \sin t) \\&= 14 + 17 \cos^2 t-24 \sin t \cos t+7 \sin ^2 t\\ &=14 + \frac{17}{2}(1 + \cos 2t)-12 \sin 2t+\frac 72(1-\cos 2t)\\ &=26+5\cos 2t-12\sin 2t\\ &=26 + 13\cos(2t+\phi), \text{ where} \cos \phi = \frac 5{13}, \sin \phi = \frac{12}{13} \end{align} so the maximum value of $y$ is $39$ and minimum value of $y$ is $13.$

therefore $$1 \le \frac {39}{ 14 + (\cos t - \sin t)(17 \cos t - 7 \sin t)} \le 3.$$

$$f(x) = (\cos x - \sin x) (17 \cos x -7 \sin x)\\=17c^2-24sc+7s^2=7+5(1+c_2)-12s_2=12+5c_2-12s_2$$ And as we know $|as+bc|\le\sqrt{a^2+b^2}$, so: $$-1=12-13\le f(x)\le 12+13=25$$

*$s:=\sin,c:=\cos,s_n=s(nx)$

HINT:

$$f(x)=17\cos^2x+7\sin^2x-24\sin x\cos x=\dfrac{17(1+\cos2x)+7(1-\cos2x)-24\sin2x}2$$

$$=12+5\cos2x-12\sin2x=12+\sqrt{12^2+5^2}\cos\left(2x+\arctan\dfrac{12}5\right)$$

Now for real $x,-1\le\cos\left(2x+\arctan\dfrac{12}5\right)\le1$