So say I wish to go from $$12\sin (t)+4\cos(t)$$
to the form $$A\cos (t+k)$$ by using the double angle formula I can get that $$\cos(k)=4$$ and $$\sin(k)=-12$$ and so we can find $A=\sqrt{4^2+(-12)^2}=4\sqrt {10}$.
But how can I find $k$ from this since $\arcsin(-12)$ is not defined. You can divide the two expressions and then use $\arctan$ to find $k$ but I'm not sure if this is correct?
Could anyone help me here. By the way the is a question in simple harmonic motion if that helps any.
$$A\cos(t+k)=A\cos(t)\cos(k)-A\sin(t)\sin(k).$$
You lost track of the $A$ when you used the addition formula, which is why you wound up trying to take an inverse sine of a number outside $[-1,1]$.
From the above, you need $A\cos(k)=4$ and $-A\sin(k)=12$. There are a number of ways to solve for $A$ and $k$ now. One way to start is to sum the squares: you find $A^2=4^2+12^2=160$, so $A=\pm \sqrt{160}=\pm 4\sqrt{10}$. For convenience, let's take the positive root (so that $A$ is actually the amplitude). If you now divide the equations instead, you find $\tan(k)=-3$, so $k=\arctan(-3)$ or $k=\arctan(-3)+\pi$. Can you figure out which one is correct?