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I don't understand number 5

I don't understand number 5-example-1

1 Answer

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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


(a\sin c)/(a\cos c)=(-2)/(-6)\iff\tan c=\frac13

If we assume
-\frac\pi2<c<\frac\pi2, then we can take the inverse tangent of both sides to get
c=\tan^(-1)\left(\frac13\right). If we do this, then we find


a\sin\left(\tan^(-1)\left(\frac13\right)\right)=\frac a{√(10)}=-2\implies a=-2√(10)

So we can rewrite


y=-6\sin4x-2\cos4x

as


y=-2√(10)\sin\left(4x+\tan^(-1)\left(\frac13\right)\right)
User LukeWaggoner
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