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### 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.

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/
```
Associated Topics:
College Trigonometry
High School Trigonometry

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