Write the expression in terms of $\sin$ only $\sin(4x)-\cos(4x)$ I am currently taking a Precalc II (Trig) course in college. There is a question in the book that I can't figure out how to complete it. The question follows:
Write the expression in terms of sine only:
$\sin(4x)-\cos(4x)$
So far I have $A\sin(x)+B\cos(x)=k\cdot\sin(x+\theta)$
I believe I have found k: $k=\sqrt{A^2+B^2}=\sqrt{2}$
So I think it would be $\sqrt{2}\cdot\sin(4x+\theta)$ but I do not know how I would find $\theta$.
Thanks in advance for all of your help. You have no idea how much I appreciate it!

Use the two double angle identities that you know:
\begin{align}
\sin2\theta&=2\sin\theta\cos\theta\\
\cos2\theta&=1-2\sin^2\theta
\end{align}
You will also need to know
$$\cos x=\sqrt{1-\sin^2x}$$
As such, we get
\begin{align}
\sin4x-\cos4x&=\sin2(2x)-\cos2(2x)\\
&=2\sin2x\cos2x-(1-2\sin^22x)\\
&=2(2\sin x\cos x)(1-2\sin^2x)+1+2(2\sin x\cos x)^2\\
&=2\sin x\sqrt{1-\sin^2x}(1-2\sin^2x)+1+8\sin^2x(1-\sin^2x)\\
&=2\sin x\sqrt{1-\sin^2x}-4\sin^3x\sqrt{1-\sin^2x}+1+8\sin^2x-8\sin^4x
\end{align}
