Double Angle and Half Angle Formulas

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.

Sine or Cosine of a Double Angle

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) = cos²A − sin²A

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 sin²x + cos²x = 1 to eliminate either the sine or the cosine from that formula:

cos(2A) = cos²A − sin²A = cos²A − (1 − cos²A) = 2 cos²A − 1

cos(2A) = cos²A − sin²A = (1 − sin²A) − sin²A = 1 − 2 sin²A

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 sin²x + cos²x = 1.

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.

Tangent of a Double Angle

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 cos²A. Either way, you get

Sine or Cosine of a Half Angle

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 = cos²(B/2) − sin²(B/2) = 2 cos²(B/2) − 1 = 1 − 2 sin²(B/2)

Use just the first and last parts of that:

cos B = 1 − 2 sin²(B/2)

Rearrange a bit:

sin²(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 cos²A − 1

As before, put A = B/2:

cos B = 2 cos²(B/2) − 1

Rearrange and solve for cos(B/2):

cos²(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.

Tangent of a Half Angle

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−cos²B)] / [1+cos B]

Then use equation 38, your old friend: sin²x + cos²x = 1;

tan(B/2) = sqrt(sin²B) / (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−cos²B)

tan(B/2) = (1−cos B) / sqrt(sin²B) = (1−cos B) / sin B

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

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.)

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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