Recall that
[tex]\cos(x\pm y)=\cos x\cos y\mp\sin x\sin y[/tex]
It follows that
[tex]\sin x\sin y=\dfrac{\cos(x-y)-\cos(x+y)}2[/tex]
Replace [tex]y[/tex] with [tex]6x[/tex] and you get
[tex]\sin6x\sin x=\dfrac{\cos(x-6x)-\cos(x+6x)}2=\dfrac{\cos5x-\cos7x}2[/tex]
(using the fact that [tex]\cos(-5x)=\cos5x[/tex])
