Expert: Josh - 5/2/2009

Question
QUESTION: Dear Josh, would you explain me how to get this:
8sin(50px+p/4)+7sin(50px-p/3)

ANSWER: Dear Patrick,

Since you did not mention your level of studies, I cannot be sure what type of response would be appropriate. I guess we should treat this as a trigonometry exercise, first and foremost. If you are studying harmonics for physics at university, you would probably have other tools at your disposal, such as Fourier analysis.

The main result we need is sin(a+b)=sin(a)cos(b)+cos(a)sin(b).  [#1]

I assume that "p" is meant to be the Greek symbol "pi".

Looking at 8*sin(50*pi*x+pi/4)+7*sin(50*pi*x-pi/3) ...[#2], we can let M=8, N=7, a=50*pi*x, b=pi/4, c=a, d=-pi/3. The idea is to get rid of the phase component, the terms which involve "pi/4" and "pi/3". We can use formula [#1] to expand the expression from [#2] as follows:

M*sin(a+b)+N*sin(c+d) ...this is identical to the question you have given me, all I've done is let M=8, N=7, a=50*pi*x, b=pi/4 etc.
= M*[sin(a)cos(b)+cos(a)sin(b)]+N*[sin(a)cos(d)+cos(a)sin(d)] ...using result from [#1]
=(M/sqrt(2))*[sin(a)+cos(a)] +N*[sin(w*t)/2-(sqrt(3)/2)*cos(w*t)] ...here, we have substituted for the values of "b" and "d", noting that cos(pi/4)=sin(pi/4)=1/sqrt(2), sin(-pi/3)=-sqrt(3)/2 and cos(-pi/3)=cos(pi/3)=1/2.
=[(8/sqrt(2))+7/2]*sin(50*pi*x)+[(8/sqrt(2))-7sqrt(3)/2]*cos(50*pi*x)

Although this may look complicated at first sight, we need not worry. This actually has a simple form A*sin(w*x)+B*cos(w*x), ...[#3]
where the constants A=(8/sqrt(2))+7/2, B=(8/sqrt(2))-7sqrt(3)/2 and w=50*pi.

Can we simplify this further? Reducing the sine and cosine components to a single trigonometric entity? Yes, absolutely! Here is how.

Again, we make use of the result in [#1]. First, we rewrite [#3] as A[sin(w*x)+(B/A)*cos(w*x)] using simple factorisation.

Observe that the content inside the square bracket is similar to sin(w*x+Q) = sin(w*x)cos(Q)+cos(w*x)sin(Q). If we figure out the auxiliary angle Q, we are done. We would succeed in converting the original expression to a single trig. entity.

Comparing sin(w*x)+(B/A)*cos(w*x) to sin(w*x)cos(Q)+cos(w*x)sin(Q), we find that cos(Q)=1 and sin(Q)=B/A, as A is non-zero. Thus, tan(Q)=sin(Q)/cos(Q)=B/A. Taking the inverse tan function, we get Q=tan^-1(B/A) or tan^-1(B/A)+pi as possible solutions, with a "pi" radian ambiguity.

Plugging in numbers, tan(Q)=B/A=[8/sqrt(2)-7sqrt(3)/2]/[8/sqrt(2)+7/2], Q=tan^-1(-0.0443)=-0.0442 (4th quadrant) or -0.0442+pi=3.0974 (2nd quadrant). A quick check would reveal that Q=-0.0442 is the right solution. The reason being: B is a negative number, and we know that sin(x) is always positive (the opposite of what we want here) for any angle pi/2 <= x < pi in the 2rd quadrant.

Retracing the steps:
8*sin(50*pi*x+pi/4)+7*sin(50*pi*x-pi/3)
=A*sin(w*x)+B*cos(w*x),
where A=8/sqrt(2)+7/2, B=8/sqrt(2)-7sqrt(3)/2 and w=50*pi...from [#2]
=A[sin(w*x+Q)] where Q=-0.0442....from [#3]

See attached diagram for better understanding.

===========================================
Physical interpretation of sinusoidal waveforms
(This is a long read, but it does give you some intuition if you are interested)

In the expression A*sin(w*t+m), "A" represents the amplitude of the sine wave, "w" represents the angular frequency, and "m" represents the phase offset.

You know that a sine wave, sin(t) normally oscillates between +1 and -1. With vertical scaling by A, its height is bounded between +A and -A. With regard to angular frequency, consider the case where w=1 and m=0. Then, we have sin(wt), a periodic function which repeats itself every 2*pi. We can compress the wave (horizontally along the "t" axis) by doubling the frequency to w=2. Now, we have sin(2t), whereas before, we had sin(t). So, when t=pi/2, we see that sin(2t)=sin(pi) attains the same value as sin(t) when t=pi. Using the same reasoning, we can dilate the sine wave. For example, to expand the sine wave by a factor of 2, we would choose w=0.5. Essentially, the value of "w" controls the horizontal scale of the wave function. The variable "m" basically translates the sine wave by a certain amount. Consider the function sin(t). We know that it attains a value of 1 when t=pi/2. But a function like sin(t+pi/2) already achieves the same amplitude when t=0. The latter is said to "lead" the former by pi/2. i.e., when we draw both functions on a piece of graph paper, sin(t+pi/2) has the same shape as sin(t), except it has been shifted to the left by pi/2 radian. You may recognise that sin(t+pi/2) is in fact the cosine function cos(t). Reversing the sign of "m", sin(t-pi/2) shifts sin(t) to the right by pi/2 radian. In this situation, sin(t-pi/2) is said to "lag" behind sin(t) by pi/2 radian.

Now, the tricky bit. sin(50pi*t+(pi/4)) has a frequency of F=50*pi/(2*pi)=25 Hz. That is, it completes 25 cycles in a second if we treat "t" as the time variable. When "w" is anything but 1, to find the actual phase offset in terms of the number of cycles with respect to the compressed waveform, we do this in 2 steps:
1. Factorise 50pi*t+(pi/4) as (50*pi)*[t+(pi/4)/(50*pi)]
2. phi=(pi/4)/(50*pi)=1/200 is seen as the phase offset. If you like, it is AHEAD of sin(50pi*t) by 0.005 second in time. To interpret this in terms of the number of cycles, we divide "phi" by the period of the sine waveform (T=1/F), (1/200)*(25/1) gives a displacement of (1/8)th of a cycle.

By the same logic, sin(50pi*t-(pi/3)) is BEHIND sin(50pi*t) by 1/150 second. This corresponds to (1/150)*(25/1)=(1/6)th of a cycle at the carrier frequency of 25 Hz. [Refer to the diagram]

---------- FOLLOW-UP ----------

QUESTION: Dear Josh, i can't find the amplitude and the phase shift of the question i asked you before:
8*sin(50*pi*x+pi/4)+7*sin(50*pi*x-pi/3)and would you check if the graph found is the same as for the question. i am still on college.  

Answer
Dear Patrick,

The amplitude and phase shift are given by A and Q, respectively. The approximate values of A and Q are 9.156 and -0.0442, as shown on the graph. Yes, the graph is based entirely on your question; it shows what happens when two sinusoidal waveforms are added together.

I really hope that you get something from this. As you go through my worked solution, particularly the trigonometry part, if you have any difficulty understanding my answer, please raise it with me so that we can clear up any confusion. Please be very specific (e.g., mention which step you have a problem with, what you get and what you don't get) so that I can respond accordingly and address the issues which are bugging you.

Regarding the graph, notice that a function like g(t)=7*sin(wt-pi/3) is shifted to the right with respect to sin(wt) whereas f(t)=8sin(wt+pi/4) is shifted to the left with respect to sin(wt). This is the physical interpretation of the phase shift. When the phase shift (displacement) is expressed relative to the angular frequency "w", the phase shift is measured in cycles. This makes it easier to see how far one waveform leads/lags behind another.