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Updated 20 Dec 2016 (What’s New?)

Trig without Tears Part 8:

Double Angle and Half Angle Formulas

Copyright © 1997–2024 by Stan Brown,

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 = cos² A − sin² A

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 sin² x + cos² x = 1 to eliminate either the sine or the cosine from that formula:

cos 2A = cos² A − sin² A = cos² A − (1 − cos² A) = 2 cos² A − 1

cos 2A = cos² A − sin² A = (1 − sin² A) − sin² A = 1 − 2 sin² A

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 sin² x + cos² x = 1.

(59) sin 2A  =  2 sin A cos A

cos 2A  =  cos² A − sin² A  =  2 cos² A − 1  =  1 − 2 sin² A

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 top and bottom by cos² A. Either way, you get

(60) tan 2A = 2 tan A / (1 − tan² A)

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) = ± √(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 cos² A − 1

As before, put  = 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) = ± √(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) = ± √(1 − cos B)/2

cos(B/2) = ± √(1 + cos B)/2

Example: Find sin 75°, which is sin 5π/12.

Solution: 75° is half of 150°, and you know the functions of 150° exactly because they are the same as the functions of 30°, give or take a minus sign.

sin 75° = sin(150°/2) = ±√(1 − cos 150°)/2

Here, cos 150° is negative because 150° is to the left of the origin, in Quadrant II, and 180° − 150° = 30°, so

cos 150° = −cos 30° = −(√3)/2

sin 75° must be positive, because 75° is in Quadrant I. Therefore,

sin 75° = √(1 − (−√3/2))/2

sin 75° = √(2 + √3)/4

sin 75° = √2 + √3 / 2

The expression √2 + √3 is called a nested radical. Some of these can be decomposed to a simpler √a + √b form, but some cannot. Denesting Radicals (or Unnesting Radicals) explains how you can tell whether a particular one can be unnested, and gives an easy method to unnest it. This one can be unnested, leading to

sin 75° = (√6 + √2)/4

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) = ± √(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 √1+cos B:

tan(B/2) = √(1 − cos B) / (1 + cos B)

tan(B/2) = √(1 − cos B) (1 + cos B) / (1 + cos B

tan(B/2) = √1−cos² B / (1+cos B)

Then use equation 38, your old friend: sin² x + cos² x = 1;

tan(B/2) = √sin² B / (1+cos B) = sin B / (1+cos B)

If you multiply top and bottom by √1−cos B instead of √1+cos B, you get another form of the half-angle tangent formula:

tan(B/2) = sin(B/2)/cos(B/2) = ± √(1−cos B) / (1+cos B)

tan(B/2) = √(1−cos B)² / ((1+cos B)(1−cos B))

tan(B/2) = (1−cos B) / √1−cos² B

tan(B/2) = (1−cos B) / √sin² B = (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). Luckily for us, it drops out. Since cos B is always between −1 and +1, (1 − cos B) and (1 + cos B) are never negative 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 + +

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 equation 62.

Practice Problems

To get the most benefit from these problems, work them without first looking at the solutions. Refer back to the chapter text if you need to refresh your memory.

Recommendation: Work them on paper — it’s harder to fool yourself about whether you really understand a problem completely.

You’ll find full solutions for all problems. Don’t just check your answers, but check your method too.

1 Use the half-angle formulas to find sin 90° and cos 90°. Of course you already know those; this problem is just for practice in working with the formulas and easy numbers.
2Use the double-angle formulas to find sin 120°, cos 120°, and tan 120° exactly. Again, you already know these; you’re just getting comfortable with the formulas.
3 3A = 2A + A. Use the double-angle formulas along with the formulas for sine or cosine of a sum to find formulas for sin 3A in terms of sin A only, and cos 3A in terms of cos A only.

(This is actually done, in a later section, by using a different method.)

4 Given sin 3A = (3 − 4 sin² A) sin A and cos 3A = (4 cos² A − 3) cos A, find tan 3A in terms of tan A only. Check yourself by computing tan(2A+A).
5 Find the sine, cosine, and tangent of π/8, exactly.

BTW: Cool Proof of Double-Angle Formulas

I can’t resist pointing out another cool thing about Sawyer’s marvelous idea: you can also use it to prove the double-angle formulas directly. From Euler’s formula for eix you can immediately obtain the formulas for cos 2A and sin 2A without going through the formulas for sums of angles. Here’s how.

Remember the laws of exponents: xab = (xa)b. One important special case is that x2b = (xb)². Use that with Euler’s formula:

cos 2A + i sin 2A = ei(2A)

cos 2A + i sin 2A = (eiA

cos 2A + i sin 2A = (cos A + i sin A

cos 2A + i sin 2A = cos² A + 2i sin A cos A + i²sin² A

cos 2A + i sin 2A = cos² A + 2i sin A cos A − sin² A

cos 2A + i sin 2A = (cos² A − sin² A) + i (2 sin A cos A)

Since the real parts on left and right must be equal, you have the formula for cos 2A. Since the imaginary parts must be equal, you have the formula for sin 2A. That’s all there is to it.

BTW: Multiple-Angle Formulas

The above technique is even more powerful for deriving formulas for functions of 3A, 4A, or any multiple of angle A. To derive the formulas for nA, expand the nth power of (cos A + i sin A), then collect real and imaginary terms.

This was inspired by a May 2009 reading of Paul Nahin’s An Imaginary Tale: The Story of √−1 (Princeton, 1998).

Just to show the method, I’ll derive the functions of 3A and 4A. You can try the brute-force approach of cos(A + 2A), sin(2A + 2A), and so forth, and see how much effort my method saves.

Functions of 3A

De Moivre’s theorem tells us that

cos 3A + i sin 3A = (cos A + i sin A

Expand the right-hand side via the binomial theorem, remembering that i² = −1 and i³ = −i:

cos 3A + i sin 3A = cos³ A + 3 i cos² A sin A − 3 cos A sin² A − i sin³ A

Collect real and imaginary terms:

cos 3A + i sin 3A = (cos³ A − 3 cos A sin² A) + i (3 cos² A sin A − sin³ A)

Set the real part on the left equal to the real part on the right to find the formula for cos 3A:

cos 3A = cos³ A − 3 cos A sin² A

It would be nice to have a formula that involves only cos A, not sin A. To get that, factor and then use sin² A = 1 − cos² A:

cos 3A = cos A (cos² A − 3 sin² A)

cos 3A = cos A (cos² A − 3(1 − cos² A))

cos 3A = cos A (cos² A − 3 + 3 cos² A)

cos 3A = cos A (4 cos² A − 3)

Going back to the formula with collected terms, set the imaginary part on the left equal to the real part on the right to find the formula for sin 3A:

sin 3A = 3 cos² A sin A − sin³ A

sin 3A = sin A (3 cos² A − sin² A)

sin 3A = sin A (3(1 − sin² A) − sin² A)

sin 3A = sin A (3 − 4 sin² A)

The tangent formula is easy to get: just divide. But it turns out to be easier if you don’t divide the final forms, but rather the “raw” collected terms from above:

sin 3A = 3 cos² A sin A − sin³ A

cos 3A = cos³ A − 3 cos A sin² A

tan 3A = sin 3A / cos 3A

tan 3A = (3 cos² A sin A − sin³ A) / (cos³ A − 3 cos A sin² A)

Factor top and bottom, then divide both by cos² A:

tan 3A = (sin A (3 cos² A − sin² A)) / (cos A (cos² A − 3 sin² A))

tan 3A = (sin A (3 − tan² A)) / (cos A (1 − 3 tan² A))

tan 3A = (sin A / cos A) (3 − tan² A) / (1 − 3 tan² A)

tan 3A = tan A (3 − tan² A) / (1 − 3 tan² A)

(63) sin 3A = sin A (3 − 4 sin² A)

cos 3A = cos A (4 cos² A − 3)

tan 3A = tan A (3 − tan² A) / (1 − 3 tan² A)

Functions of 4A

It’s no more work to find the functions of 4A. Since the technique is similar, I’ll just run through the steps without commentary. First, De Moivre’s theorem:

cos 4A + i sin 4A = (cos A + i sin A)4

cos 4A + i sin 4A = cos4 A + 4 i cos³ A sin A − 6 cos² A sin² A − 4 i cos A sin³ A + sin4 A

cos 4A + i sin 4A = (cos4 A −  6 cos² A sin² A + sin4 A) + i (4 cos³ A sin A − 4 cos A sin³ A)

Set the real parts equal to get cos 4A:

cos 4A = cos4 A − 6 cos² A sin² A + sin4 A

cos 4A = cos4 A − 6 cos² A sin² A + (sin² A

cos 4A = cos4 A − 6 cos² A (1 − cos² A) + (1 − cos² A

cos 4A = cos4 A − 6 cos² A + 6 cos4 A + 1 − 2 cos² A + cos4 A

cos 4A = 8 cos4 A − 8 cos² A + 1

Set the imaginary parts equal to get sin 4A:

sin 4A = 4 cos³ A sin A − 4 cos A sin³ A

sin 4A = 4 sin A cos A (cos² A − sin² A)

sin 4A = 4 sin A cos A (1 − sin² A − sin² A)

sin 4A = 4 sin A cos A (1 −  2 sin² A)

Unfortunately, the first power of cos A can’t be changed to something involving only sin A. Or rather, it can, but at a cost. cos A would be replaced with ±√1 − sin² A, and we’d have to figure out the proper sign every time.

Now the tangent formula:

tan 4A = sin 4A / cos 4A

tan 4A = (4 cos³ A sin A − 4 cos A sin³ A) / (cos4 A − 6 cos² A sin² A + sin4 A)

Divide top and bottom by cos4 A:

tan 4A = (4 tan A − 4 tan³ A) / (1 − 6 tan² A + tan4 A)

There are several possibilities for simplifying this formula, but no one that’s clearly superior to the others, so I’ll just factor tan A from the top of the fraction:

tan 4A = 4 tan A (1 − tan² A) / (1 − 6 tan² A + tan4 A)

(64) sin 4A = 4 sin A cos A (1 −  2 sin² A)

cos 4A = 8 cos4 A − 8 cos² A + 1

tan 4A = 4 tan A (1 − tan² A) / (1 − 6 tan² A + tan4 A)

What’s New

next:  9/Inverse Functions

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