Maximum how?

Let $f(x) = (7\cos{x} + 24\sin{x}) (7\sin{x} - 24\cos{x})$

$\begin{aligned} \Rightarrow f(x) & = 25^2\left(\frac{7}{25}\cos{x} + \frac{24}{25}\sin{x}\right) \left(\frac{7}{25}\sin{x} - \frac{24}{25}\cos{x}\right) \\ & = -625 \sin{\left(x+\tan^{-1}{\frac{7}{24}}\right)} \cos{\left(x+\tan^{-1}{\frac{7}{24}}\right)} \\& = - 312.5 \sin{\left( 2 \left[ x+\tan^{-1}{\frac{7}{24}}\right] \right)} \end{aligned}$

$\Rightarrow f(x)$ is maximum, when $\sin{\left( 2 \left[ x+\tan^{-1}{\frac{7}{24}}\right] \right)} = -1$ and $f_{max}(x) = \boxed{312.5}$

We can expand the expression:

$49\sin x \cos x-168\cos^2 x + 168\sin^2 x - 576\sin x \cos x \\ -527\sin x \cos x-168(\cos^2 x - \sin^2 x)$

Now, use $\sin(2x)=2\sin x\cos x$ and $\cos(2x)=\cos^2 x-\sin^2 x$ :

$-\dfrac{527}{2}\sin(2x)-168\cos(2x)$

Finally, using Cauchy-Schwarz inequality:

$-\dfrac{527}{2}\sin(2x)-168\cos(2x) \leq \sqrt{\left(\left(-\dfrac{527}{2}\right)^2+(-168)^2\right)(\sin^2(2x)+\cos^2(2x))} \\ \leq \sqrt{\dfrac{277729}{4}+28224} \\ \leq \sqrt{\dfrac{390625}{4}} \\ \leq \boxed{\dfrac{625}{2}}$

We can generalize it, by finding the maximun of $(a\cos x+b\sin x)(a\sin x-b\cos x)$ :

$a^2\sin x\cos x-ab\cos^2 x+ab\sin^2 x-b^2\sin x\cos x \\ (a^2-b^2)\sin x\cos x-ab(\cos^2 x-\sin^2 x) \\ \dfrac{a^2-b^2}{2}\sin(2x)-ab\cos(2x) \leq \sqrt{\left(\left(\dfrac{a^2-b^2}{2}\right)^2+(-ab)^2\right)(\sin^2(2x)+\cos^2(2x))} \\ \leq \sqrt{\dfrac{a^4-2a^2 b^2+b^4+4a^2 b^2}{4}} \\ \leq \dfrac{\sqrt{(a^2+b^2)^2}}{2}$

Since we want a maximum, we take the positive value:

$\leq \boxed{\dfrac{a^2+b^2}{2}}$

Here's a calculus approach.

Consider the general expression $f(x) = (a \cos x + b\sin x)(a \sin x - b \cos x)$ for real constants $a,b$ .

For simplicity sake, let $S = \sin x, C = \cos x, T = \tan x$ . Applying product rule shows that $f'(x) = (aC + bS)^2 - (aS - bC)^2$ .

At the turning point, $f'(x) = 0$ , then $aC + bS = \pm \ (aS - bC)$ , equivalently $T = \frac SC = \frac{a+b}{a-b}, -\frac{a-b}{a+b}$ .

Substitute this value into $f(x)$ ,

$\begin{aligned} f(x) &= &(a + bT)(aT - b) \cos^2 T \\ &= &(a+bT)(aT - b) \cdot \frac 1{\tan^2 x + 1} \\ &=& \frac{1}{T^2+1} (a+bT)(aT-b) \\ \end{aligned}$

With a little bit of simplification, we get the extremal points of $f(x)$ as $\pm \ \frac{a^2+b^2}{2}$ .

In this case, $a = 7, b = 24 \Rightarrow \text{max}(f(x))= \frac{a^2+b^2}2 = \boxed{312.5}$ .