Trig without Tears Part 8:

# Double Angle and Half Angle Formulas

revised 1 Jan 2016

Copyright © 1997–2016 Stan Brown, BrownMath.com

Trig without Tears Part 8:

revised 1 Jan 2016

Copyright © 1997–2016 Stan Brown, BrownMath.com

**Summary:**
Very often you can **simplify your work** by expanding
something like *sin*(2A) or *cos*(½A) into functions
of plain A. Sometimes it works the other way and
**a complicated expression becomes simpler** if you see it as a
function of half an angle or twice an angle. The formulas seem
intimidating, but they’re really just variations on equation 48
and equation 50.

With equation 48, you can find *sin*(A+B). What happens
if you set B=A?

*sin*(A+A) = *sin* A *cos* A + *cos* A *sin* A

But A+A is just 2A, and the two terms on the right-hand side are equal. Therefore:

*sin*(2A) = 2 *sin* A *cos* A

The cosine formula is just as easy:

*cos*(A+A) = *cos* A *cos* A −
*sin* A *sin* A

*cos*(2A) = *cos*²A − *sin*²A

Though this is valid, it’s not completely satisfying. It would be nice to
have a formula for *cos*(2A) in terms of just a sine or just a cosine.
Fortunately, we can use
*sin*²*x* + *cos*²*x* = 1
to eliminate either the sine or
the cosine from that formula:

*cos*(2A) = *cos*²A − *sin*²A =
*cos*²A − (1 − *cos*²A) =
2 *cos*²A − 1

*cos*(2A) = *cos*²A − *sin*²A =
(1 − *sin*²A) − *sin*²A =
1 − 2 *sin*²A

On different occasions you’ll have occasion to use all three forms of
the formula for *cos*(2A). Don’t worry too much about where the
minus signs and 1s go; just remember that you can always transform any
of them into the others by using good old
*sin*²*x* + *cos*²*x* = 1.

(59) *sin*(2A) = 2 *sin* A *cos* A

*cos*(2A) = *cos*²A − *sin*²A = 2 *cos*²A − 1 = 1 − 2 *sin*²A

There’s a very cool second proof of these formulas, using Sawyer’s marvelous idea. Also, there’s an easy way to find functions of higher multiples: 3A, 4A, and so on.

To get the formula for *tan*(2A), you can either start with
equation 50 and put B = A to get *tan*(A+A), or use
equation 59 for *sin*(2A) / *cos*(2A) and
divide both parts of the fraction by *cos*²A. Either way, you get

(60) *tan*(2A) = 2 *tan* A / (1 − *tan*²A)

What about the formulas for sine, cosine, and tangent of half an angle? Since A = (2A)/2, you might expect the double-angle formulas equation 59 and equation 60 to be some use. And indeed they are, though you have to pick carefully.

For instance, *sin*(2A) isn’t much help. Put A = B/2
and you have

*sin* B = 2 *sin*(B/2) *cos*(B/2)

That’s true enough, but there’s no easy way to solve for
*sin*(B/2) or *cos*(B/2).

There’s much more help in equation 59 for *cos*(2A). Put
2A = B or A = B/2 and you get

*cos* B =
*cos*²(B/2) − *sin*²(B/2) =
2 *cos*²(B/2) − 1 =
1 − 2 *sin*²(B/2)

Use just the first and last parts of that:

*cos* B = 1 − 2 *sin*²(B/2)

Rearrange a bit:

*sin*²(B/2) = (1 − *cos* B) / 2

and take the square root

*sin*(B/2) =
± *sqrt*([1 − *cos* B] / 2)

You need the plus or minus sign because *sin*(B/2) may be
positive or negative, depending on B.
For any given B (or B/2) there will be only one correct sign, which
you already know from the diagram
that we explored back in Functions of Any Angle.

**Example:**
If B = 280°, then B/2 = 140°, and you know
that *sin* 140° is positive because the angle is in Quadrant
II (above the axis).

To find *cos*(B/2), start with a different piece of the
*cos*(2A) formula from equation 59:

*cos*(2A) = 2 *cos*²A − 1

As before, put A = B/2:

*cos* B = 2 *cos*²(B/2) − 1

Rearrange and solve for *cos*(B/2):

*cos*²(B/2) = [1 + *cos* B]/2

*cos*(B/2) =
± *sqrt*([1 + *cos* B] / 2)

You have to pick the correct sign for *cos*(B/2) depending on the
value of B/2, just as you did with *sin*(B/2). But of course the
sign of the sine is not always the sign of the cosine.

(61) *sin*(B/2) = ± *sqrt*([1 − *cos* B] / 2)

*cos*(B/2) = ± *sqrt*([1 + *cos* B] / 2)

You can find *tan*(B/2) in the usual way, dividing sine
by cosine from equation 61:

*tan*(B/2) = *sin*(B/2)/*cos*(B/2) =
± *sqrt*([1 − *cos* B]/[1 + *cos* B])

In the sine and cosine formulas we couldn’t avoid the square
roots, but in this tangent formula we can. Multiply top and
bottom by *sqrt*(1+*cos* B):

*tan*(B/2) =
*sqrt*([1 − *cos* B] / [1 + *cos* B])

*tan*(B/2) =
*sqrt*([1−*cos* B][1+*cos* B] / [1+*cos* B]²)

*tan*(B/2) =
[*sqrt*(1−*cos*²B)] / [1+*cos* B]

Then use equation 38, your old friend:
*sin*²*x* + *cos*²*x* = 1;

*tan*(B/2) =
*sqrt*(*sin*²B) / (1+*cos* B) =
*sin* B / (1+*cos* B)

If you had multiplied top and bottom by
*sqrt*(1−*cos* B) instead of
*sqrt*(1+*cos* B), you would get

*tan*(B/2) = *sin*(B/2)/*cos*(B/2) =
± *sqrt*([1−*cos* B]/[1+*cos* B])

*tan*(B/2) =
*sqrt*([1−*cos* B] / [1+*cos* B])

*tan*(B/2) =
*sqrt*([1−*cos* B]² / [1+*cos* B][1−*cos* B])

*tan*(B/2) =
(1−*cos* B) / *sqrt*(1−*cos*²B)

*tan*(B/2) =
(1−*cos* B) / *sqrt*(*sin*²B) =
(1−*cos* B) / *sin* B

The half-angle tangent formulas can be summarized like this:

(62) *tan*(B/2) = (1 − *cos* B) / *sin* B = *sin* B / (1 + *cos* B)

You may wonder what happened to the plus or minus sign in
*tan*(B/2). Fortuitously, it drops out. Since *cos* B
is always between −1 and +1, (1 − *cos* B) and
(1 + *cos* B) are both positive for any B. And the
sine of any angle always has the same sign as the tangent of the
corresponding half-angle.

Don’t take my word for that last statement, please. There are only four possibilities, and they’re easy enough to work out in a table. (Review interval notation if you need to.)

B/2 | Q I, (0°;90°) | Q II, (90°;180°) | Q III, (180°;270°) | Q IV, (270°;360°) |
---|---|---|---|---|

tan(B/2) |
+ | − | + | − |

B | (0°;180°), Q I or II | (180°;360°), Q III or IV | (360°;540°), Q I or II | (540°;720°), Q III or IV |

sin B |
+ | − | + | − |

1−cos B |
+ | + | + | + |

Of course, you can ignore the whole matter of the sign of the sine and just assign the proper sign when you do the computation.

Another question you may have about
equation 62: what happens if
*cos* B = −1, so that
(1 + *cos* B) = 0?
Don’t we have **division by zero** then? Well, take a little closer look at
those circumstances. The angles B for which
*cos* B = −1
are ±180°, ±540°, and so on. But in this case
the half angles B/2 are ±90°, ±270°, and so on:
angles for which the tangent is not defined anyway. So the problem of
division by zero never arises.

And in the other formula, *sin* B = 0 is not a
problem. Excluding the cases where *cos* B = −1,
this corresponds to B = 0°, ±360°,
±720°, etc. But the half angles B/2 are 0°,
±180°, ±360°, and so on. For all of them,
*tan*(B/2) = 0, as you can verify from the second half
of formula equation 62.

next: 9/Inverse Functions

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