How does $c_1 \cos(\omega t) + c_2 \sin(\omega t)$ become $A\cos(\omega t - \delta)$?

Question

When one solves a 2nd order linear ODE, the solution will look something like this:

$$c_1 \cos(\omega t) + c_2 \sin(\omega t)$$, which is sometimes rewritten as $$A\cos(\omega t - \delta)$$

How is the latter derived from the former?

Answer 1

It's actually slightly easier to work backwards, and start with $f(t) = A\cos(\omega t - \delta)$. You might remember from some trigonometry that there's an identity that looks like this:

$\cos(A + B) = \cos A \cos B - \sin A \sin B$

If we apply that to the function in question, we get:

$\begin{eqnarray} A \cos(\omega t - \delta) & = & A \left( \cos(\omega t) \cos \delta + \sin(\omega t) \sin \delta \right) \\ & = & (A \cos \delta)\cos(\omega t) + (A \sin \delta)\sin(\omega t) \\ & = & C_1 \cos(\omega t) + C_2 \sin(\omega t)\end{eqnarray}$

where $C_1 = A \cos \delta$ and $C_2 = A \sin \delta$. You can then do a bit of fiddling around to show that given a formula in the latter form, you can always solve for the former.

Answer 2

Hint: Divide and multiply by $\sqrt{c_1^2 + c_2^2}$ and then write $\frac{c_1}{\sqrt{c_1^2 + c_2^2}}$ and $\frac{c_2}{\sqrt{c_1^2 + c_2^2}}$ as cosine and sine terms respectively. i.e. You will get the value of the constant $A = \sqrt{c_1^2 + c_2^2}$.