 # Calculations by hand

How does one find the sine, cosine, tangent of an angle without looking at a table or using a calculator/computer? Likewise, how does one find the logarithm of a number under the same conditions?
For the trigonometric functions, let me add one more condition: I don’t want to have to construct a triangle, measure two sides, and find a ratio. I’m looking for a formula of some kind.

“[He] beat his fist down upon the table and hurt his hand and became so
further enraged… that he beat his fist down upon the table even harder and
hurt his hand some more.” – Joseph Heller’s Catch-22

As one who excelled in math as a student, and as one of the rare people who use trigonometric functions on a regular basis at work, I’d have to say you’re SOL. Even if there was an algorithm for each of these functions, which I’m sure there is (especially for logarithms), the algorithms themselves would be enough of a pain in the ass to practically require the use of a calculator or a computer.

There are some infinite series, but I don’t have the book in front of me. That’s probably your best bet for a formula to memorize. There’s a lot of calculation involved, though.

sin x = x - x^3/(32) + x^5/(5432) - x^7/(76543*2) + …

cos x = 1 - x^2/2 + x^4/(432) - x^6/(65432) + …

(Note, x must be in radian measure, not degree…to convert degrees to radians, multiply by pi/180)

To get tangent, calculate sin and cos and divide. There’s a series for natural log too; I’d have to look it up (and if you’re truly desperate to know, I will).

The … means these are infinite series; but just the few few terms generally give a good approximation within 4 or 5 decimal places. They’re fairly easy to remember because of the pattern, but it’s still a pain to work out by hand. Which leaves me wondering: why do you WANT to?

Just curious because they weren’t listed in any of my math books.

“[He] beat his fist down upon the table and hurt his hand and became so
further enraged… that he beat his fist down upon the table even harder and
hurt his hand some more.” – Joseph Heller’s Catch-22

Isn’t it cute to see all those log tables people used to use? In my engineering school library, they had an encyclopedia-looking set of books containing just numbers in tables (like logs, sines, etc.) An inch of dust on them. One little calculator makes 'em all obsolete.

P.S. Most people cant even divide wihtout a calculator. Calculating sines is just masochism, pure and simple.

If the angle is in a right triangle, you can calculate the ratios of the sides.

sin x = o / h
cos x = a / h
tan x = o / a

o = the length of the leg opposite angle x
a = the length of the leg adjacent to angle x
h = the length of the hypoteneuse

quadell, you should riffle through 'em to see if anyone left humorous notes, pictures, or money in there. You never know what you might find in books that have lain undisturbed for decades.

Cave Diem! Carpe Canem!

In school I learned the algorithm for finding square roots by hand. I never used it again. Wish I could remember it, though.

APB9999:

Guess the square root (zero is NOT a good guess!)

Divide the number by your guess.

Use this average as your next guess.

Repeat as long as you like.

APB9999 said:

>In school I learned the algorithm for finding square roots by hand. I never used it again. Wish I could remember it, though.

I think I remember the algorithm you’re thinking of. It’s a little reminiscent of long division. Here goes:

To find the square root of 156.25:

place the number under a radical sign (like a long division sign) and group the
digits by two both left and right of the decimal point.

<code>
___.
V 1 56.25
</code>

over the left-most group (the “1” in my example), place the square root of the
largest integer greater than or equal to that group. E.g., a 1 for the groups 1 2 3, a 2 for the groups
4 5 6 7 8, a 3 for 9 10 11 12 13 14 15, etc. Then square this number, place
it under the first group, and subtract.
<code>
1__.__
V 1 56.25
-1
0
</code>

Bring down the next group (“56”) as in long division. To the left of this lowest line,
write double the “quotient” in progress, followed by a blank space.

<code>
1__.__
V 1 56.25
-1
2_ )0 56
</code>

Now comes the tricky part: the “2_” represents twenty-something. Your task is to find a
single digit that will go both above the “56” and next to the 2. Their product will be placed
under the “0 56” and subtracted; it needs to be as large as possible without going over “0 56”. E.g.,

1 * 21 = 21
2 * 22 = 44
3 * 23 = 69 too large

So “2” is the magic digit this round.

<code>
1__2._
V 1 56.25
-1
22 )0 56
-44
12
</code>

Next, bring down the “25” and write a “24_” next to “12 25”

<code>
1__2._
V 1 56.25
-1
22 )0 56
-44
24_ )12 25

</code>

1 * 241 = 241
2 * 242 = 484
3 * 243 = 729
4 * 244 = 976
5 * 245 = 1225 !

<code>
_1__2._5
V 1 56.25
-1
22 )0 56
-44
245 )12 25
12 25
0
</code>

If the subtraction comes out zero, you’ve found an exact answer. If not,
keep bringing down “00” pairs to calculate more places.

Shoot! My spacing got all messed up! 8-(

Let me know, and I’ll send you the ASCII file