Half-Angle Formula - Getting the Right Sign

Date: 06/11/2003 at 08:48:49
From: Prizzi
Subject: Half-angle formulae: I can't get the sign right

Find the possible values of tan(1/2)theta when tantheta = 7/24.

I get the correct numerical answers 1/7(tan in the first quadrant) 
and 7(tan in the third quadrant), but for my answer when considering 
tan in the third quadrant, the answer at the back of my book gives -7.

I used the half-angle formulae for cos^2(1/2)theta and sin^2(1/2)
theta, substituting it into the trig identity 
sintheta/costheta = tantheta to find tan^2(1/2)theta, and taking the 
square root I got tan(1/2)theta.

Date: 06/11/2003 at 12:39:36
From: Doctor Peterson
Subject: Re: Half-angle formulae: I can't get the sign right

Hi, Prizzi.

Our Formulas FAQ lists these half-angle identities:

   |sin(x/2)| = sqrt([1-cos(x)]/2),
   
   |cos(x/2)| = sqrt([1+cos(x)]/2),
   
   |tan(x/2)| = sqrt([1-cos(x)]/[1+cos(x)]),
   
   tan(x/2) = [1-cos(x)]/sin(x),
            = sin(x)/[1+cos(x)].

The absolute values on the first three indicate that you have to 
decide what sign to use based on the quadrant in which you expect the 
answer to lie. You apparently used these, dividing to get tan^2 
before taking the square root, which is a good idea. But when you 
took the square root, you were not done, since there are TWO square 
roots of any positive number, and you had to choose one. I hope the 
version of the half-angle identities you used makes this clear.

If theta is in the first quadrant, so is theta/2, and the tangent is 
positive. But if theta is in the third quadrant, then theta/2 is in 
the second quadrant, and you have to choose the negative tangent.

If you used the last identity I showed above, you would not have a 
square root to find at the end; rather, you would make the choice of 
sign when you decided what the sine and cosine of theta itself are, 
before putting them into the formula. Try that method, and see that 
you get the right signs out if you put the right signs in.

If you have any further questions, feel free to write back.

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

Date: 06/11/2003 at 14:36:16
From: Prizzi
Subject: Thank you (Half-angle formulae: I can't get the sign right)

Thank you for your help! I tried your version and finally have the 
correct signs. However I'm finding it difficult to follow how you 
derived tan(x/2) = [1-cos(x)]/sin(x),= sin(x)/[1+cos(x)] from 
|tan(x/2)| = sqrt([1-cos(x)]/[1+cosx)]). If it is not too much 
trouble, could you show me what you did? I'll probably be able to 
remember this formula if I understand how it is derived. Thank you!

Date: 06/11/2003 at 14:58:13
From: Doctor Peterson
Subject: Re: Thank you (Half-angle formulae: I can't get the sign 
right)

Hi, Prizzi.

I didn't derive these formulas myself, but took them from the Dr. Math 
Trigonometry Formulas FAQ:

  http://mathforum.org/dr.math/faq/formulas/faq.trig.html#identities 

But let's see what we can do. We know that

  |tan(x/2)| = sqrt([1-cos(x)]/[1+cos(x)])

which we can write as

                      1-cos(x)
  tan(x/2) = +/- sqrt(--------)
                      1+cos(x)

Multiplying numerator and denominator by 1-cos(x) gives

                      [1-cos(x)][1-cos(x)]            [1-cos(x)]^2
  tan(x/2) = +/- sqrt(-------------------- = +/- sqrt(------------)
                      [1+cos(x)][1-cos(x)]            1 - cos^2(x)

                 1 - cos(x)
           = +/- ----------
                   sin(x)

All that is left is to choose the sign. Since tan(x/2) is positive 
when x/2 is in the first or third quadrant, it is positive when x is 
in the first or second quadrant, that is, when sin(x) > 0. That means 
we can just drop the "+/-", since 1-cos(x) > 0 and we just need the 
sign of the sine.

You can derive the other version the same way.

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