Algebra and Trigonometry 10th Edition

$-2 \sin 8x \sin x$
We know that $\cos a-\cos b=-2 \sin (\dfrac{a+b}{2}) \sin (\dfrac{a-b}{2})$ $\cos 9 x-\cos 7 x=-2 \sin (\dfrac{9x+7x}{2}) \sin (\dfrac{9x-7x}{2})$ This implies that $\cos 9 x-\cos 7 x=-2 \sin (\dfrac{16x}{2}) \sin (\dfrac{2x}{2})$ Therefore, $\cos 9 x-\cos 7 x=-2 \sin 8x \sin x$