How do I combine two trigonometric waveforms that have the same $\omega t$ but different $\phi$

Question

I am working on an applied math assignment of mine and there's this problem on waveform combination that is kinda tricky. My lecturer did a problem similar to this as an example. However, the example he did was simple as it only had the angular frequency and no phase shift.

Here is the problem:

The current flowing in an AC circuit is defined by the following trigonometric waveform:
$$i=\cos(3t-1)-2\sin(3t+4)$$

He wants us to combine those two waveforms into a simpler trigonometric waveform of the form $$i=\sin(\omega t+\phi) \quad\text{with}\quad \phi\geq 0$$

Here's my working:

using the identities $cos(A-B)=cos(A)cos(B)+sin(A)sin(B)$ and $sin(A+B)=sin(A)cos(B)+cos(A)sin(B)$

$$cos(3t)cos(-1)+sin(3t)sin(-1)-2[sin(3t)cos(-1)+cos(3t)sin(-1)]$$ $$0.5403023059cos(3t)-0.8414709848sin(3t)-2[0.5403023059sin(3t)-0.8414709848cos(3t)]$$ $$0.5403023059cos(3t)+1.68294197cos(3t)-0.8414709848sin(3t)-1.080604612sin(3t)$$ $$2.223244275cos(3t)-1.922075597sin(3t)$$ Using $Asin(\omega t + \phi)$: $$A[sin(\omega t)cos(\phi)+cos(\omega t)sin(\phi)]$$ $$Asin(\omega t)cos(\phi)+Acos(\omega t)sin(\phi)$$ Equate coefficients:
$$2.223244275=Acos(\phi) eq1$$ $$(2.223244275)^2 = A^2sin^2(\phi) eq2$$ $$-1.922075597=Acos(\phi) eq3$$ $$(-1.922075597)^2=A^2cos^2(\phi) eq4$$ Now add eq2 and eq4: $$(2.223244275)^2 + (-1.922075597)^2 = A^2cos^2(\phi)+A^2sin^2(\phi)$$ $$(2.223244275)^2 + (-1.922075597)^2 = A^2(cos^2(\phi)+sin^2(\phi))$$ $$(2.223244275)^2 + (-1.922075597)^2 = A^2$$ $$A=\sqrt{(2.223244275)^2+(-1.922075597)^2}$$ $$A=2.938909612$$ Now divide eq3 and eq1: $$\frac{2.223244275}{-1.922075597}=\frac{Acos(\phi)}{Asin(\phi)}$$ $$\frac{2.223244275}{-1.922075597}=tan(\phi)$$ $$\phi = tan^-1 (\frac{2.223244275}{-1.922075597})$$ $$\phi = -0.8579234358 rads$$ Note that it is not in the domain so: $$-0.8579234358 rads + 2\pi rads = 5.425261871 rads$$ $$\therefore \phi = 5.425261871 rads$$ So:
The simplified combined waveform is $2.938909612sin(3t+5.425261871)$.
That is the simplified form I came up with. However, the simplified form does not match with the original equation when graphed:
You can see clearly that it is out of phase with the original equation. What am I doing wrong? Is my approach correct? Please help!

Answer 1

Using the identities

$$\begin{align} \cos(A-B) &= \cos A \cos B + \sin A \sin B \\ \sin(A+B) &= \sin A \cos B + \cos A \sin B \end{align}$$

and defining $$s_1 := \sin 1 \qquad c_1 := \cos 1 \qquad s_4 := \sin 4 \qquad c_4 := \cos 4$$

The initial equation becomes $$\begin{align} &\;c_1\cos 3t +s_1\sin 3t-2\left(\;c_4\sin 3t+s_4\cos 3t\;\right) \\ =&\;(s_1-2c_4)\sin 3t + (c_1-2s_4)\cos 3t \tag{1} \end{align}$$ while the target equation (with an appended coefficient of $A$) is $$A \sin \omega t \cos\phi + A\cos \omega t \sin\phi \tag{2}$$

Equating coefficients: $$\begin{align} A\cos\phi = s_1-2c_4 \tag{3} \\ A\sin\phi = c_1-2s_4 \tag{4} \end{align}$$

Therefore, $$\begin{align} A^2 &= A^2\cos^2\phi + A^2\sin^2\phi \\ &= (s_1-2c_4)^2+(c_1-2s_4)^2 \\ &= \left(s_1^2+c_1^2\right) - 4\left(s_1 c_4+c_1 s_4\right) + 4\left(c_4^2+s_4^2\right) \\ &= 1 - 4\sin(1+4) + 4 \\ &= 5 - 4\sin 5 \\ \to A &= \sqrt{5-4\sin 5} = 2.972\ldots \tag{5} \end{align}$$ and $$ \tan\phi = \frac{A\sin\phi}{A\cos\phi}= \frac{c_1-2s_4}{s_1-2c_4} \quad\to\quad \phi = \arctan\frac{\cos 1-2\sin 4}{\sin 1 - 2\cos 4} = 0.7628\ldots \tag{6}$$

Thus, we have

$$2.972\ldots\cdot\sin(3t+0.7628\ldots) \tag{$\star$}$$

Answer 2

Find the amplitude and phase shift of \begin{equation} y=a\cos(\omega t+\phi_1)+b\sin(\omega t+\phi_2)\tag{1} \end{equation}

Notice that \begin{eqnarray} \omega t+\phi_1&=&\omega\left(t+\frac{\phi_1+\phi_2}{2\omega}\right)-\frac{\phi_2-\phi_1}{2}\\ \omega t+\phi_2&=&\omega\left(t+\frac{\phi_1+\phi_2}{2\omega}\right)+\frac{\phi_2-\phi_1}{2} \end{eqnarray}

So if we make the substitutions \begin{eqnarray} T&=&t+\frac{\phi_1+\phi_2}{2\omega}\\ \phi&=&\phi_2-\phi_1 \end{eqnarray} we can rewrite equation $(1)$ in a symmetric form as \begin{equation} y=a\cos\left(\omega T-\frac{\phi}{2}\right)+b\sin\left(\omega T+\frac{\phi}{2}\right)\tag{2} \end{equation}

Note that

\begin{eqnarray} \frac{dy}{dT}&=&b\omega\cos\left(\omega T+\frac{\phi}{2}\right)-a\omega\sin\left(\omega T-\frac{\phi}{2}\right)\\ \frac{d^2y}{dT^2}&=&-a\omega^2\cos\left(\omega T-\frac{\phi}{2}\right)-b\omega^2\sin\left(\omega T+\frac{\phi}{2}\right)\\ \end{eqnarray} Therefore, we get the differential equation \begin{equation} \frac{d^2y}{dT^2}+\omega^2y=0\\ \end{equation} which has solution \begin{eqnarray*} y&=&y(0)\cos(\omega T)+\frac{y^\prime(0)}{\omega}\sin(\omega T)\\ &=&\left[a\cos\left(\frac{\phi}{2}\right)+b\sin\left(\frac{\phi}{2}\right)\right]\cos(\omega T)+ \left[b\cos\left(\frac{\phi}{2}\right)+a\sin\left(\frac{\phi}{2}\right)\right]\sin(\omega T) \end{eqnarray*}

Now let \begin{equation} \psi=\arctan\left(\frac{a\cos\left(\frac{\phi_2-\phi_1}{2}\right)+b\sin\left(\frac{\phi_2-\phi_1}{2}\right)}{b\cos\left(\frac{\phi_2-\phi_1}{2}\right)+a\sin\left(\frac{\phi_2-\phi_1}{2}\right)}\right) \end{equation}

Noting that

\begin{equation} \left[a\cos\left(\frac{\phi}{2}\right)+b\sin\left(\frac{\phi}{2}\right)\right]^2+\left[b\cos\left(\frac{\phi}{2}\right)+a\sin\left(\frac{\phi}{2}\right)\right]^2=a^2+b^2+2ab\sin(\phi) \end{equation} we have \begin{eqnarray} y&=&\sqrt{a^2+b^2+2ab\sin(\phi_2-\phi_1)}\,\sin(\omega T+\psi)\\ y&=&\sqrt{a^2+b^2+2ab\sin(\phi_2-\phi_1)}\,\sin\left(\omega t+\frac{\phi_1+\phi_2}{2}+\psi\right) \end{eqnarray} giving the resulting amplitude and phase shift.

For the particular variables in your exercise the result is

$$ y=2.97249005022\sin(3t+0.762832763) $$