is there an easy arithmetic way to calculate the sin 13 or cos 26. Without any technical tools (calculator)?

Question

i want to be able to calculate a good estimate, like 3 digits behind comma. without tools., just private memory calculations.

these values must have been constructed once. I hope there is an easier way than MacLaruin's or Taylor's formula to estimate it.

Answer 1

Not really. 13 is an odd-ball number, a prime in fact, which makes some of they typical patterns and relationships somewhat difficult to adapt.

Recall the trigonometric relationships:
sin 2x = 2 sin x cos x, among all the others

Its possible to convert one expression into another of a different angle value. But you would be left with the same problem there. Some angles, if you can manage to get there, like 0, 30, 45, 60, 90 are easy to memorize and simply substitute.

Then, if need be, and this is hardly mathematically sound... but if you want to draw out a geometric figure of a right triangle... making sure you get the angles right... you could simply measure out and divide the lengths of the sides. The values would only be as accurate as the figure you draw.

There are other alternative routes you could take. In calculus and beyond, we learn of writing trig functions in terms of Eulers constant.
sin x = [ e^ix - e^-ix ] / 2i
for example. But of course, you would have to know the values of e raised to a power which includes imaginary values. Not necessarily a good alternative, as you would still have to rely on a calculator, a table, or yet another relationship.

Also, in calculus, we learn of Taylor Expansions. We could express the trig functions in terms of an infinite series, where each term follows a simple pattern. Of course you cannot evaluate an infinite series in your head... but you could limit yourself to a sufficient number of terms to guarantee some degree of accuracy. This is actually how calculators solve for trig values... as quickly as electricity.

Whether you use the Eulers expressions or strictly the trig functions, simply having a pre-written list of values would be helpful... thats what they used to do a long time ago (along with logarithm tables, they had trig tables). But those tables are only accurate to so many digits... and only have specific values listed as functional inputs.

But of course, the trigonometric relationships would fill in many of the gaps. These tables only need to include sine, from 0 to 45. All other integer angles from 46 to 360, for each of the other trig functions, have relatively simple relationships to sine from 0 to 45.

Then, Im sure, you could construct values using an iteration method. Given a relationship, you could reiterate it to solve for increasing accuracy. I cant think of a practical method/example off the top of my head at the moment.

Not that this would help, or would be any more practical, but you could rewrite the trig functions in terms of the hyperbolic trig functions. As necessary.

Regardless of the technique you use, I highly doubt you could get to three significant figures in your head. You would at least rely on the use of pen-paper math.

I would also like to point out that the most practical method, Taylor Expansions, which would probably be the most accurate... and gives you the leeway to get as accurate as you wish to on your own, if youre willing to do the work. This method requires the use of angles in radian measure. Radian measures use the value pi. In your example, 13 and 26 degrees, you would have to convert to radians... and therefore you would be applying technique to irrational values. Most difficult to do in the head.

Answer 2

Sorry, but the difficulty in calculating trig functions (and logarithms, for that matter) is such that mathematicians long ago compiled the results in tables so they could look them up rather than calculate them every time. Once calculated to a given accuracy, the calculations will always yield the same result. Consider that we don't even calculate the "easy" ones, like 30, 45, and 60, every time we need them. We memorize these if we use them often, since √3 and √2 aren't all that much fun to calculate, either, to say nothing of nπ/180.

13 = 15 - 2 = (60 - 45) - (1 + 1)
sin (15 - 2) = sin(15)cos(2) - cos(15)sin(2)
sin(15) = sin(60)cos(45) - cos(60)sin(45)
cos(15) = cos(60)cos(45) + sin(60)sin(45)
sin(2) = π/90 to 5 digits
cos(2) = √(1 - (π/90)^2)
Putting all these together,
(√(1 - (π/90)^2)) [(1/2)(1/2)(√3)(√2) - (1/2)(1/2)(√2))] - (π/90)[(1/2)(1/2)(√2) + (1/2)(1/2)(√3)(√2)] =

(1/4)(√2)[(√(1 - (π/90)^2))(√3 - 1) - (π/90)(1 + √3)]

is about as easy as it gets.

Answer 3

No. Sorry.

Before we had calculators we had books filled with sin and cos values for you to look up. There's no easy head-maths you can use.