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### Solving Trigonometric Equations in Radians

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Date: 12/14/97 at 00:26:48
From: alex r
Subject: Solving trigonometric equations in radians

Solve for theta in radians to 2 decimals if 0 < theta < 2 pi tan theta
= -0.318.

I tried inverse tan -0.318 (in radians mode) and it gave me -0.307;
however, it said between 0 and 2pi so I added 2pi and got one of the
answers, 5.9. However, I can only get the other 2.8 if I add 1 pi.

Why am I adding one pi? Whenever I do something in degrees, if I am
given an angle and want to find another angle with the same tan or
whatever, I add 360. Here pi = 180 - why am I adding pi (180) and not
2pi (360)?

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Date: 12/14/97 at 04:28:51
From: Doctor Luis
Subject: Re: Solving trigonometric equations in radians

This is a good question. To understand why you only add one pi, you
need to understand something about a special property of trigonometric
functions called periodicity.

As you may know, trigonometric functions are periodic. A periodic
function f(x) has the following special property,

f(x+P) = f(x)

This means that the function repeats itself (it gives you the same
values over again) when you add a special number. Since there's an
infinite number of such numbers, we usually take the smallest positive
number such that the equation above is still true, and we call that
number the period of the function f(x).

let p = period of f(x).

You can see that if you add p to x a lot of times you'll still get the
same value

f(x+np) = f(x)  , where n is an integer (negative, positive or zero)

This gives you a problem if you want to solve some equation like the
following for x (there's an infinite number of values of x that will
give you the same value of y!)

f(x) = y

---->   x + np = inversef(y)

or          x = inversef(y) + mp  ,   where m = -n

Since you know that you have an infinite values of x, and you also
know the period of f(x), we usually take the solution x with m = 0, so
that
x = inversef(y)

Notice that, in this case, x will have to be between 0 and p, because
if it were greater than p you could always subtract p until you got x
in between 0 and p, or if x were less than 0 you could always add p
until you got x in between 0 and p. Then,

0 < x < p

The solution with this restriction on x is called the principal
solution of the periodic function f(x) = y

You are already familiar with the trigonometric functions sine and
cosine: their period is the same and is equal to 2pi radians or
360 degrees. In fact, the cosine is the same curve as the sine, except
that it is shifted pi/2 radians to the right (that's why they have the
same period)

sin(x+2pi) = sin(x)   and   cos(x+2pi) = cos(x)

Now, the tangent doesn't have a period of 2pi like the sine and
cosine. It has a period of 1 pi. This is because of the way the
tangent is defined,

sin(x)
tan(x) = --------
cos(x)

So,

sin(x+pi)
tan(x+pi) = -----------
cos(x+pi)

Since the cosine is just a shifted sine curve,

cos((pi/2)-x) = sin(x)

This is the same as saying that

sin((pi/2)-x) = cos(x)

Now, you can rewrite sin(x+pi) and cos(x+pi) like this,

sin(x+pi) = sin(pi+x)
= sin((pi/2)-(-x-(pi/2)))
= cos(-x-(pi/2))
= cos(-((pi/2)+x))
= cos((pi/2)+x)         (b/c cos(-x) = cos(x) )
= cos((pi/2)-(-x))
= sin(-x)
= -sin(x)               (b/c sin(-x) = sin(x) )
and,

cos(x+pi) = cos(pi+x)
= cos((pi/2)-(x-(pi/2)))
= sin(x-(pi/2))
= -sin((pi/2)-x)         (b/c sin(-x) = sin(x) )
= -cos(x)

Then,

sin(x+pi)
tan(x+pi) = -----------
cos(x+pi)

- sin(x)
= -----------
- cos(x)

sin(x)
= --------
cos(x)

= tan(x)             (that's just the definition!)

And so, you see

tan(x+pi) = tan(x)

that's just a periodic function with period pi !!

Now, your problem said: solve for x in

tan(x) = -0.318   with   0 < x < 2pi

so,

but remember, our solution for a periodic function has to be between
0 and the period p. In this case, we just proved that the period p of
the tan(x) function is pi  (radians). So we add pi to x, and get

x + pi = 2.83         (radians)

This brings our solution to be between 0 and pi (3.14...) and gives us
the principal solution of the equation tan(x) = -0.318

Remember that a periodic function has an infinite number of solutions?
Since the problem asks for solutions between 0 and 2 pi there are
going to be two solutions. One between 0 and pi, and the other one
between pi and pi+pi=2pi. So, to get the other solution we just add pi
(pi is the period, remember?) to our principal solution, and get

(x+pi)+pi = 2.83 + pi

If the problem had asked for a solution with 0 < x < 3pi, there would
have been three solutions, and we would have had to add pi (the period
of tan(x)) again,

(x+2pi)+pi = 5.97 + pi

If the problem had asked for a solution between -pi and 0, we would
have kept the solution your calculator gave you in the first place.

x = invtan(-0.318)
= -0.307

So, you see, knowing the principal solution and the period of a
periodic function gives you all of the solutions. All you have to do
is find the solutions that you want (for example, all the solutions
between -10 and 345).

solve for x,

tan(2x) = -0.318        with   0 < x < 2pi

what would the solutions be? (hint: there are four solutions)

Something to think about: If a periodic function f(x) has period p,
what is the period of the periodic function f(rx+s), where r is any
positive real number and s any real number?

I hope this helped :)

-Doctor Luis,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
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Associated Topics:
High School Trigonometry

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