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### Subtraction of Trig Functions with a Phase Shift

```Date: 09/14/2004 at 14:53:23
From: Jeremy
Subject: Subtraction of Trig Functions

I am a senior Electrical Engineering student and am currently writing
a training manual for my job at an engineering firm.  I am now
covering 3-phase voltages and have a trig question pertaining to
finding the voltage between two phases.

My question is this:  How do you find the difference between two
trigonometric functions with the same frequency, but one is phase
shifted by 120 degrees?

Here's the explicit problem:

cos (wt) - cos (wt - 120), where w = 120*pi.

I've tried using Euler's equations, but keep getting stuck.

```

```
Date: 09/14/2004 at 15:05:05
From: Doctor Schwa
Subject: Re: Subtraction of Trig Functions

Hi Jeremy,

The technique I learned in trig class for this type of problem,

cos (a) - cos(b)

is to subtract the identities

cos(x+y) = cos x cos y - sin x sin y
and
cos(x-y) = cos x cos y + sin x sin y
to get
cos(x+y) - cos(x-y) = -2 sin x sin y.

Then from there, you set x + y = a and x - y = b, because you want
this formula to match the cos(a) - cos(b) pattern, and after a bit of
algebra to solve for x and y, you should get

cos a + cos b = -2 sin ((a+b)/2) sin ((a-b)/2).

In your case, I guess that turns out to be

-2 sin(wt - 60) * sin 60

and it looks to me like that checks out just fine!

I'm sure there's a similar pattern using cos + i sin instead of my
approach--there always is--but at least on my first attempt, it turned
out to be at least as awkward as this one.

I hope that helps clear things up!

- Doctor Schwa, The Math Forum
http://mathforum.org/dr.math/

```

```
Date: 09/14/2004 at 18:11:30
From: Jeremy
Subject: Thank you (Subtraction of Trig Functions)

Dr. Schwa,

Thank you so much for your speedy response.  It seems a bit late in my
college career to become a math guru, but I'm aspiring to become one,
on my own, if possible.

You're a life saver!

Jeremy
```
Associated Topics:
High School Trigonometry

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