Question

I don't understand number 5

Answer 1

If you expand the desired form, you get

$a\sin(bx+c)=a(\sin bx\cos c+\cos bx\sin c)$

and if we match this up with the given form, we can see that $b=4$, and we have to find $a$ and $c$ such that

$\begin{cases}a\cos c=-6\\a\sin c=-2\end{cases}$

If we divide the second equation by the first, we have

$\dfrac{a\sin c}{a\cos c}=\dfrac{-2}{-6}\iff\tan c=\dfrac13$

If we assume $-\dfrac\pi2<c<\dfrac\pi2$, then we can take the inverse tangent of both sides to get $c=\tan^{-1}\left(\dfrac13\right)$. If we do this, then we find

$a\sin\left(\tan^{-1}\left(\dfrac13\right)\right)=\dfrac a{\sqrt{10}}=-2\implies a=-2\sqrt{10}$

So we can rewrite

$y=-6\sin4x-2\cos4x$

as

$y=-2\sqrt{10}\sin\left(4x+\tan^{-1}\left(\dfrac13\right)\right)$