tan x+y = (tan x + tan y) / (1 - tan x tan y)

We already know (see trig equivalences) that

cos x+y = cos x cos y - sin x sin y

and

sin x+y = sin x cos y + cos x sin y

So tan x + y = (sin x+y) / (cos x+y)

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

Dividing both numerator and denominator by (cos x)(cos y), we get:

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

= (tan x + tan y) / (1 - tan x tan y)

That was too easy, wasn't it?  Now let's have some fun with this.

You may run into a problem in your math career that goes something like this:

Evaluate the following sum exactly:

Tan^-1(1/2) + Tan^-1(1/5) + Tan^-1(1/8)

The way you solve such a problem is this:

Let x = Tan^-1(1/2),  so tan x = 1/2

Let y = Tan^-1(1/5), so tan y = 1/5, and

Let z = Tan^-1(1/8), so tan z = 1/8

Now if you can find w = tan x+y+z, the answer will be tan^-1 w.

We start by finding tan x+y = (tan x + tan y) / (1 - tan x tan y)

= (1/2 + 1/5) / (1 - (1/2)(1/5))

= (7/10) / (9/10)

= 7/9

Now we apply the formula again to find tan (x+y)+z = (tan(x+y) + tan z) / (1 - tan(x+y)tan z)

= (7/9 + 1/8) / (1 - (7/9)(1/8))

= (56/72 + 7/72) / (1 - 7/72)

= (63/72) / (63/72)

= 1

So w = tan x+y+z = 1, so tan^-1 w = tan^-1 1 = pi/4

Isn't that cute? Tan^-1(1/2) + Tan^-1(1/5) + Tan^-1(1/8) = pi/4

What's so magical about these numbers, 1/2, 1/5, and 1/8 that make their arctans add up to such a "round" number?  Let's investigate...

In general, the formula for combining three angles, a=Tan^-1(x), b=Tan^-1(y), and c=Tan^-1(z) can be derived by repeating the formula one more time:

Tan^-1(a+b+c)=
=((a+b)/(1-ab)+c)/(1-((a+b)/(1-ab))c)
=(a+b+c-abc)/(1-bc-ab-ac)

Combining another angle, d,

Tan^-1(a+b+c+d)=
=((a+b+c-abc)/(1-bc-ab-ac)+d)/(1-((a+b+c-abc)/(1-bc-ab-ac))d)
=(a+b+c+d-bcd-acd-abd-abc) / (1-ab-ac-ad-bc-bd-cd+abcd)

Coming up with magical formulas -- like combining the arctans of 1/8, 1/5, and 1/2 to get the arctan of 1 -- is really fairly easy.  Just set the four-angle combination, above, equal to 1 and solve for d

1=(a+b+c+d-bcd-acd-abd-abc) / (1-ab-ac-ad-bc-bd-cd+abcd)
d=(1-a-b-c-ab-ac-bc+abc)/(1+a+b+c-ab-ac-bc-abc)

Using this formula, we can see, for example, that for the sums of 2/7, 1/8, 1/5 and d to be equal to the arctan of 1, then d must be equal to 3/16.

You can see that if a, b, and c are rational numbers then d will be rational, too.  It is surprising and fun to see that d is not only rational, but fairly simple in most cases, because a lot of canceling happens when you add calculate this fraction.

Internet References

Mathworld: Tangent -- Equation 20 is the generalized tangent sum formula

Related Pages in this website

Trig Equivalences

Relationship between cos 2x and tan²x -- cos(2x) = (1-tan²(x))/(1+tan²(x))

The webmaster and author of the Math Help site is Graeme McRae.
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