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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.

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$$A\cos(t+k)=A\cos(t)\cos(k)-A\sin(t)\sin(k).$$

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?

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  • $\begingroup$ I don't see how to know which is correct unfortunately? Also you made a mistake when simplifying $A$. $\endgroup$ – Craig May 17 '15 at 17:57
  • $\begingroup$ Any chance how you would could explain how you could possibly know which one is right and how you could know it isn't $\arctan (-3)+k\pi~k\in \mathbb{Z}$ $\endgroup$ – Craig May 17 '15 at 18:05
  • $\begingroup$ @Craig Thanks for pointing out the mistake. Anyway, it actually could be $\arctan(-3)+k \pi$ for $k \in \mathbb{Z}$, but actually there are only two functions here, since $\sin$ and $\cos$ are $2 \pi$ periodic. So the same function with $k=\arctan(-3)$ could be written with $k=\arctan(-3)+2 \pi$ or $\arctan(-3)+200000000000\pi$ if you want. That is, the choice of the phase shift is not unique. As for how to check which one of those two it should be, just plug it back in with a value of $t$, say $t=0$. You should have $\sqrt{160} \cos(k)=4$. Which of the two $k$s has that property? $\endgroup$ – Ian May 17 '15 at 18:06
  • $\begingroup$ $k=\arctan(-3)$ works. Thanks. $\endgroup$ – Craig May 17 '15 at 18:58

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