How to prove $A\cos(\omega t-\phi)$ = $a\cos(\omega t)$ + $b\sin(\omega t)$ using $e^{i\theta}$?

Question

I want to show that $A\cos\left(\omega t-\phi\right)$ = $a\cos\left(\omega t\right)$ + $b\sin\left(\omega t\right)$

First I verified for myself through the angle addition proof that:

$$ \cos\left(\omega t+ \phi\right) = \cos\left(\omega t\right)\cos\left(\phi\right) - \sin\left(\omega t\right)\sin\left(\phi\right) $$

and that

$$ \cos\left(\omega t - \phi\right) = \cos\left(\omega t\right)\cos\left(-\phi\right) - \sin\left(\omega t\right)\sin\left(-\phi\right) $$

therefore I am able to show that:

$$ \cos\left(\omega t - \phi\right) = \cos\left(\omega t\right)\cos\left(\phi\right) + \sin\left(\omega t\right)\sin\left(\phi\right) $$

as cos is an even function and sine is an odd function.

I then multiply through by A:

$$ A\cos\left(\omega t - \phi\right) = A\cos\left(\omega t\right)\cos\left(\phi\right) + A\sin\left(\omega t\right)\sin\left(\phi\right) $$

so that there are constants: $ A\cos\left(\phi\right) = a $ and $ A\sin\left(\phi\right) = b $

which gives the function:

$A\cos\left(\omega t-\phi\right)$ = $a\cos\left(\omega t\right)$ + $b\sin\left(\omega t\right)$

used to model sinusoidal solutions of differential equations.

If we take the case that $\phi = 0$ then there is no phase lag and:

$A\cos\left(\omega t\right)$ = $A\cos\left(\omega t\right)$ with no $\sin\left(\phi\right)$ term, which is bizarre to me because it seems that if $\phi \neq 0 $ the equation $A\cos\left(\omega t-\phi\right)$ can be broken into $a\cos\left(\omega t\right)$ + $b\sin\left(\omega t\right)$ whereas if $\phi = 0 $ the input of $\sin $ is "lost".

Why can't $\cos\left( \omega t\right) $ always be decomposed to $\cos\left( \omega t\right) $ and $\sin\left( \omega t\right) $, I know that for $t = 0$ it is simply because $\sin\left( \omega t\right) = 0$, but what about when $\omega \neq 0$, why can we not describe $\cos\left( \omega t\right) $ in terms of $\cos\left( \omega t\right) $ and $\sin\left( \omega t\right) $?

What are some better proofs or correct/more rigorous proofs of the existence of this relationship? i.e. there any proof using the $e^{i\theta} $ and complex numbers?

Is there a way to relate amplitude $A$ to $ \phi, \omega $, and/or $t$, or is amplitude truly independent of the parameters and variables of any function of this type?

Answer 1

For all $\omega, \cos(\omega t) $ can be decomposed into $ \cos(\alpha t - \beta) $ and the solution can be given as $A\cos(\alpha t)=a\cos(\alpha t)+b\sin(\alpha t)$.

Answer 2

HINT: Use Euler's formula $$ e^{i\theta} = \cos \theta + i \sin \theta $$

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Applying the same formula to the opposite sign argument gives expression $\,e^{-i\theta} = \cos \theta - i \sin \theta,\,$ which when aded to the original one yields expression for $\cos \theta$ in terms of exponents:

$$ \cos \theta = \dfrac12 \left(e^{i\theta} + e^{-i\theta}\right) $$

Then, rewriting expression $\,A\cos\left(\omega t - \phi\right)\,$ in the form of exponent and simplifying should give you the answer. Hope you can pick it from here.

Answer 3

Starting with $$2\cos(x) = e^{ix} + e^{-ix}$$ you have $$F=2\cos(\omega t-\phi)=e^{i(\omega t-\phi )} + e^{-i(\omega t-\phi )}=e^{i\omega t }e^{-i\phi }+e^{-i\omega t }e^{i\phi }$$ $$F=(\cos(\omega t)+i\sin(\omega t))(\cos(\phi)-i\sin(\phi))+(\cos(\omega t)-i\sin(\omega t))(\cos(\phi)+i\sin(\phi))$$ Expand, simplify and you will find again the classical formula for the development of $F$.

Answer 4

A solution I'm fond of is to try to make $x(t)=a\cos \omega t+b\sin \omega t$ the real part of some complex function $z(t)$. But this is accomplished as $z(t)=(a-i b)(\cos \omega +i\sin\omega)$. If we now write the terms in polar form, then we're talking about the real part of $(Ae^{-i \phi})(e^{i\omega t})=Ae^{i(\omega t-\phi)}$ which is just $A\cos(\omega t-\phi)$ as desired. (This additionally makes it simple to swap between polar and Cartesian form via $a+i b=e^{i\phi}$.)

Answer 5

If you want a solution that doesn't use Euler's formula directly, then you might consider that $f(t) = A\cos(\omega t-\phi)$ satisfies the linear homogeneous differential equation $f''+\omega^2f = 0$. We can verify that $\cos(\omega t)$ and $\sin(\omega t)$ constitute linearly independent solutions by calculating their Wronskian: $$\cos(\omega t)\frac{d}{dt}[\sin(\omega t)]-\sin(\omega t)\frac{d}{dt}[\cos(\omega t)] = \cos^2(\omega t)+\sin^2(\omega t) = 1\neq 0$$

Therefore, $\{\cos(\omega t), \sin(\omega t)\}$ spans the solution space of our differential equation, so we must have some $a$ and $b$ such that $A\cos(\omega t-\phi) = a\cos(\omega t)+b\sin(\omega t)$, and we can determine that $a = A\cos(\phi)$ and $b = A\sin(\phi)$ by plugging in $t = 0$ and $t = \frac{\pi}{2\omega}$.

Answer 6

$$Ae^{i(\omega t-\phi)}=Ae^{-i\phi}e^{i\omega t}=(a-ib)e^{i\omega t}.$$

Take the real parts.

Answer 7

To calculate $A$ and $\phi$ from $a$ and $b$:

$$ A = \sqrt{a^2+b^2} $$ $$ \phi = \arccos(\frac{a}{A})=\arcsin(\frac{b}{A}) $$