One More Path To Trigonometric Addition Formulas

There’s something truly beautiful about the internal consistencies of mathematics. Thanks to its nature we get to hike on such a variety of paths to reach the same destination.  Math becomes a mental traveler’s paradise.

Consider the trigonometric addition formulas:

sin(x + y) = sin(y) cos(x) + sin(x) cos(y)  and

cos (x + y) = cos (x) cos(y)  –  sin(x) sin(y) .

These lead to highly useful double angle formulas, which are needed to evaluate integrals. They are also used in Fourier analysis and in communication systems. How do we derive them? I’ll first explore a slightly more difficult route. Secondly we will briefly refer to a conventional geometrical approach and finally we’ll examine my own offbeat path (but I can’t guarantee it’s original) .

(1) A circuitous route using slightly higher mathematics

Knowing the rate of change of sine, cosine and natural base functions allows one to express sin(w), cos(w) and ew as power series. Replace the w with xi, where i is the square root of -1, the so-called imaginary number, and exi turns out to be the sum of the cosine series and i times the sine series, namely Euler’s relationship:

exi = cos (x) + i sin(x).

If we subsequently use the above to express ei(x+y)  we obtain

ei(x+y) = cos (x + y) + i sin(x + y).                        equation(1)

But using exponent laws, ei(x+y) = exi (eyi ) , and then applying Euler’s formula again:

ei(x+y) = exi [eyi ] = [cos(x) + i sin(x) ] [cos(y) + i sin(y) ] and expanding the right hand side:

ei(x+y) = cos(x) cos(y) + i sin(y) cos(x) +  i sin(x) cos(y)  – sin(x) sin(y).

If we regroup the above by pairing and separating the real components from the imaginary ones :

ei(x+y) = cos(x) cos(y)   – sin(x) sin(y) + i [sin(y) cos (x) + sin(x) cos (y)].        equation(2)

we can compare equations (1) and (2) and have to conclude that:

cos(x + y) = cos(x) cos(y)  –  sin(x) sin(y).

sin(x + y) = sin(y) cos(x) + sin(x) cos(y).

(2) A Geometric Approach

Elementary math can be used if we begin with the following diagram. The details of this particular derivation are found at the site indicated in the caption.


(3) A simpler off beat path from working backwards

After deriving the Sine Law from simple trig ratios, I subsequently thought of using the same diagram to make substitutions into the sine addition formula. Surely enough, after some manipulations, I obtained the Sine Law, implying that the latter can be used to derive the other.

A derivation of the sine addition formula

By using the same approach with the cosine addition formula, I obtained another unexpected starting point. Apologies for not typing it out. The convenience of simply photographing my handwritten versions proved to be too tempting!

A derivation of the cosine addition formula

When Math-Shortcuts Involved Little or No Technology

There are some great tools available to anyone who wants to do math, regardless of the level involved. And whether it’s a calculator, spreadsheet or software package, they don’t just cater to time-savers. Many are programmable and therefore thought-provoking. But in the old days of hand-calculations, people did not necessarily surrender to tedium. They often came up with shortcuts, some of which were excellent approximations, while others gave an exact result. Let’s look at an example of each variety.

Whether you’re on a plane, hill or mountain, if you know your altitude, what is the easiest way of calculating the distance to the horizon?

How far is the horizon from the top of the 390 foot (119 m) peninsula(Cape St Mary’s)? About 24.2 miles (almost 39 km), but in this old picture my daughter was looking for birds and was not concerned with distances.

It turns out you don’t have to resort to the tangent to a circle and the Pythagorean theorem. The shortcut consists of multiplying the altitude in feet by 1.5 (its units not shown) and to square root the product. The answer is a close approximation of the distance in miles!

To understand why the shortcut works we need to look at the longer and more accurate way of solving the problem. The distance, d, from an altitude, h, is simply the leg of a right-angled triangle. One angle is 90 degrees because the line of sight is tangent to the radius (R) of the earth, an approximate circle. Finally the hypotenuse is the sum of R and h.

circleSince d² + R² = (h+R)², after expanding and eliminating the common R² term, we obtain:

d² = h² + 2hR

The radius of the earth is about 3959 miles and there are 5280 feet in a mile, so plugging in R and solving for d we obtain

d = √( h² +  41 807 040 h ) in feet.

But the trick claims that d =√(1.5h) in miles. Converting to  feet ,

d =5280√(1.5h).

Equating the two expressions for d:

√( h² +  41 807 040 h) = 5280√(1.5h).

Squaring both sides:

h² +  41 807 040 h = 5280²(1.5)h.

Dividing through by h and simplifying the right hand side:

h +  41 807 040  = 41 817 600.

For any h value between 1 and 40 000 feet, the left hand sum will differ from the right hand side by a maximum of 0.07%. And so the trick works well. But how was it devised?

My guess is that if someone plotted various values of height in feet versus the distance in miles using the Pythagorean expression, she would have noticed a square root function going through the origin of the form y = a√h, where a is constant. To verify it, she likely calculated successive changes in distance per changes in the square roots of height. I’ve tried that and I obtained values in the neighborhood of 1.224 and 1.225. Then all she had to do was recognize that those values are pretty close to the square root of 1.5.

The second shortcut is a simple algorithm for computing averages mentally.  I came up with it independently when I was a high school student, and needless to say this trick has also occurred to many other people!

Instead of adding up values x1, x2, x3 …. xn and dividing by the number of values, n,  if you have no calculator, we begin with a convenient estimate. Then add up the differences from the estimate, divide the sum by n and add the result to the estimated value. For example, for the average of 78, 82, 97, 60 and 77, we could use an estimate of 80. The differences from the values are -2, +2, +17, -20 and -3. Mentally the numbers are smaller so it’s easy to come up with the sum of -6. Finally the average is -6/5 + 80 = -1.2 + 80 = 78.8, also a very feasible mental calculation.

Here’s why it works. Let E = estimate for the average(Xavg) of an n number of values: x1, x2, x3…xn

The first step involves adding up the differences (call the sum Sd) from E:

x1 – E  + x2 – E + x3 – E… + xn – E = Sd =  x1 + x2 + x3 … + xnnE

The we divide through by the sum by n:

Sd/n = (x1 + x2 + x3 … + xn nE)/n = (x1 + x2 + x3 … + xn)/nnE /n.

But on the right hand side , the first term is the definition of average, so:

Sd/n = Xavg – E

Or  Xavg =  Sd/n + E

And how did I and others devise such a trick? Intuitively, at least that’s how I came up with it, from what I recall. That’s is even more fun than trying to figure out how someone else stumbled upon it.

Up ↑