Trig Problem: Sec A = -2 ...

Date: 10/20/97 at 22:47:43
From: Jennifer Rausch
Subject: math 30

I am having problems with the folowing problem:

   sec A  = -2   where 0 < A < 2(pi)

Where would I start? Help, please!

Jennifer

Date: 10/24/97 at 11:02:52
From: Doctor Bruce
Subject: Re: math 30

Hi Jennifer,

I'm going to guess that you want to know how to find all the 
angle(s) in the range between 0 and 2*pi radians that have a
secant equal to -2.

I've always found it a good thing to change trig problems so that 
everything is expressed in terms of sines and cosines. 

We know that

     secant(A) = 1/cosine(A).

So, your problem is the same as finding all the angles A in the 
range between 0 and 2*pi radians that have a cosine equal to -1/2.

When I took trigonometry (shortly after the last Ice Age ended), 
I was made to memorize the sines and cosines of a few "special" 
angles, namely 0,30,45,60, and 90 degrees. So, I can still reach 
into my memory and recall that

     cosine (60 degrees) = 1/2,

or, the same thing in radians,

     cosine (pi/3 radians) = 1/2.

You can certainly check that this is true with a pocket calculator.

We got the number 1/2 all right, but we need an angle with 
cosine -1/2, not +1/2.

Now comes the matter of looking at all the reflections of an angle 
in the four quadrants. For each trig function, two reflections give 
positive values and two give negative. I recall that angles in 
quadrants I and IV have positive cosines, and angles in quadrants II 
and III have negative cosines.  

So, the reflection of (pi/3) into both quadrants II and III gives two 
angles, each with cosine equal to -1/2. Since you gave the restriction 
that 0 < A < 2*pi, we don't have to go any further in solving this 
problem. (Maybe you have seen other problems where you have to add on 
extra multiples of 2*pi to find ALL the angles with a given cosine 
value. If the range for A were larger, you'd have to do that in this 
problem. But 0 < A < 2*pi means to look at just the angles in the four 
quadrants.)

I hope that the counter-clockwise numbering I,II,III,IV of the 
quadrants is familiar to you. I doubt that any other numbering system 
is used. 

What I mean by reflecting an angle into another quadrant is this. 
Say the angle B is in the first quadrant. Then the reflection of B 
into the other quadrants is given by this little table.

quadrant          I     II           III          IV
reflected angle   B     pi - B       pi + B       2*pi - B

Now, I didn't finish working out the problem for you - we Math Doctors 
are supposed to leave you a little work to do on your own! But if you 
don't manage to come up with the answer, write back and we'll try to 
give some more hints.

-Doctor Bruce,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/