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 that
$$\cos(4x)=1-2\sin^2(2x)=1-2(2\sin x \cos x)^2=1-8\sin^2 x(1-\sin^2x)$$
