The previous post shows how you could use linear interpolation to fill in gaps in a table of logarithms. You could do the same for a table of sines and cosines, but there’s a better way. As before, we’ll assume you’re working by hand with just pencil, paper, and a handbook of tables.

## Linear interpolation

Suppose you want to find the sine of 12.3456° and you have a table of sines for angles in increments of 0.1°. In Table 4.10 of A&S we find

sin 12.3° = 0.21303 03862 74977

sin 12.4° = 0.21473 53271 67063

If we were to use linear interpolation, we’d estimate

sin 12.3456° = sin 12.3° + 0.456(sin 12.4° − sin 12.3°) = 0.21380 78393 21768

which is accurate to six decimal places.

## Better approach

Another approach would be to use the identity

sin(θ + φ) = sin θ cos φ + cos θ sin φ

rather than linear interpolation, setting θ = 12.3° and φ = 0.0456°. We can look up the sine and cosine of θ in our table, but how do we find the sine and cosine of φ?

The cosine is easy: set it to 1. For a small angle *x* (in radians) the cosine of *x* is approximately 1 with an error of less than *x*²/2. In radians,

φ = 0.0456 π/180 = 0.00079 58701 38909

and so the truncation error in approximating cos φ with 1 is about 3×10^{−7}.

Computing the sine of φ is easy, but it requires converting φ to radians. You could probably find the conversion factor in your handbook, e.g. in Table 1.1 of A&S.

0.0456° = 0.0456 × 0.01745 32925 19943

Once φ is in radians, sin φ = φ with an error of less than φ³/6 (see here).

Putting the pieces together we have

sin(θ + φ) = sin 12.3° × 1 + cos 12.3° × φ

which, using the numbers above, gives us 0.21380785249034476, which is off by about 6×10^{−8}.

## More accuracy

If we want even more accuracy, we need to find the weakest link in our calculation. The error in approximating sin φ as φ is on the order of φ³ while the error in approximating cos φ as 1 is on the order of φ², so the latter is the largest source of error.

If we approximate cos φ as 1 − φ²/2, the error is on the order of φ^{4}, and the weakest link would be the sine approximation which has error on the order of φ³, which is still quite small. The overall error in computing sin 12.3456° would be less than 10^{−10} if we use this higher order approximation for cosine φ.

## Compare and contrast

Let’s go back to approximating the cosine of a small angle by 1 and compare the two approximation approaches above.

Linear interpolation:

sin 12.3456° = sin 12.3° + 0.456(sin 12.4° − sin 12.3°)

Addition formula:

sin 12.3456° = sin 12.3° + 0.0456 (π/180) (cos 12.3°)

The the second terms in the two approaches are

0.0456(sin 12.4° − sin 12.3°)/0.1

and

0.0456 (π/180) (cos 12.3°).

The two are similar because

(sin 12.4° − sin 12.3°)/0.1 ≈ (π/180) (cos 12.3°).

The term on the left is the difference quotient for sine at 12.3° with step *h* = 0.1 and the term on the right is the derivative of sine at 12.3°.

Wait, isn’t the derivative of sine just cosine? It is when you’re working in *radians*, which is why calculus almost always uses radians, but when you work in degrees, the derivative of sine is π/180 times cosine.

What this shows is that if you approximate cosines of small angles as 1, the sum formula reduces to a one-term Taylor approximation.