Copyright © 2007 jsd
Here are some hints on how to do basic math calculations.
These suggestions are intended to make your life easier. Some of them may seem like extra work, but really they cause you less work in the long run.
This is a preliminary document, i.e. a work in progress.
If that’s too much bother, start with a plain piece of paper and sketch in faint guidelines when necessary.
As mentioned in item 7, the name cannot encode every interesting detail. However, it can at least be unique, so you can look up the details.
The classic scenario is this: You start out having an expression for y as a function of x. You solve for the inverse function, so now x is known as a function of y. You want to plot the new function, but for some reason you feel obliged to plot things in the form y versus x, so you are tempted to reverse the meaning of the variables.
This is one of the 77 habits of highly crazy people. Don’t do it! Do not succumb to this temptation.
Instead, you should set the whole calculation aside and start over using variables that don’t need to be redefined. Using meaningful (or at least mnemonic) variable names will help. For instance if you started with current as a function of voltage, use the traditional symbols I and V. You can plot I versus V just as easily as V versus I, with no temptation to change the meaning. If you really have no idea what the symbols mean, use noncommittal names such as p versus q. You can plot q versus p just as easily as p versus q.
More generally, introductory textbooks tend to fall into the following habits:
In mathematics, none of those things is actually required. You should break those habits. (Some spreadsheet apps force on you the idea that the horizontal direction is called x and the vertical direction is called y, but it is still a bad idea.)
Sometimes the penalty for getting this wrong is on the order of three hundred million dollars, as in the case of the Mars Climate Orbiter (reference 3).
Most computer languages do not automatically keep track of the units, so you will have to do it by hand, in the comments. If the calculation is nicely structured, it may suffice to have a legend somewhere, spelling out the units for each of the variables. If variables are reused and/or converted from one set of units to another, you need more than just a legend; you will need comments (possibly quite a lot of comments) to indicate what units are being used at each point in the code.
One policy that is sometimes helpful (but sometimes risky) is to convert everything to SI units as soon as it is read in, even in fields where SI units are not customary. Then you can do the calculation in SI units and convert back to conventional units (if necessary) immediately before writing out the results. (This is problematic when writing an “intermediate” file. Should it be SI or customary? How do you know the difference between an “intermediate” result and a final result?)
It is certainly possible for computer programs to keep track of the units automatically. A nice example is reference 4. It is a shame that such features are not more widely available
Here’s a classic example: The task is to add 198 plus 215. The easiest way to solve this problem in your head is to rearrange it as (215 + (200 − 2)) which is 415 − 2 which is 413. The small point is that by rearranging it, a lot of carrying can be avoided.
One of the larger points is that it is important to have multiple methods of solution. This and about ten other important points are discussed in reference 5.
The classic "textbook" diagram of an inequality uses shading to distinguish one halfplane from the other. This is nice and attractive, and is particularly powerful when diagramming the relationship between two or more inequalities, as shown in figure 2.
However, when you are working with pencil and paper, shading a region is somewhere between horribly laborious and impossible.
It is much more practical to use hatching instead, as shown in figure 3.
Obviously the hatched depiction is not as beautiful as the shaded depiction, but it is good enough. It is vastly preferable on cost/benefit grounds, for most purposes.
Some refinements:
For most people, it is reasonable to just forget about long multiplication. It’s of very little value to most people. On the other hand, if you’re interested in understanding how it works, here is an explanation.
Short multiplication refers to any multiplication problem that consists of multiplying a onedigit number by another onedigit number. That means everything from 0×0 through 9×9.
Long multiplication refers to multiplying multidigit numbers. We are about to discuss a highly efficient algorithm for doing this.
As a prerequisite for doing long multiplication, you need to be able to do short multiplication reliably.
Speed will improve gradually over time. The short multiplication facts will become consolidated in your memory over time. Indeed, long multiplication gives you plenty of practice doing short multiplication, so we can kill two birds with one stone.
Let’s start by doing a supersimple example. Let’s multiply 432 by 1. You can do that in your head. The answer is written out in figure 5, in the standard form used for long multiplication problems.
Multiplication by 2 is almost as easy, as shown in figure 6:
Multiplication by 20 is also easy; all we need to do is shift the previous result over one place, as shown in figure 7:
Here’s a more interesting example. Let’s multiply 432 by 21. We can reduce the workload if we rewrite 21 as 20+1. We can express that as an equation:
 (1) 
So now we are multiplying 4321×(20+1). The equation is:
 (2) 
There are two ways of understanding where equation 2 comes from. One way is to start with equation 1 and multiply both sides by 432. Another way is to start with the lefthand side of equation 2 and replace the 21 by something that is equal to it; in other words, to do a substitution.
We can apply the distributive law to the righthand side of equation 2. That gives us:

Both of the multiplications in equation 3b are easy, and indeed we have already done them. Plugging in the results of these multiplications gives us:

So at this point, there’s no more multiplying to be done. All that remains is a simple addition problem, as shown in figure 8. The multiplication problem shown in gray is exactly equivalent to the addition problem shown in black.
Note that we can get by without writing the zero at the end of line d in the figure. The bottomline result is the same, whether we write this zero or not, provided we keep things properly lined up in columns, as shown in figure 9. In the number system we are using, the Hindu number system, we normally think of zeros like this as being necessary to indicate place value. However, that’s not strictly true. You don’t need to write the zeros, provided you have some other way of keeping track of place value.
You are allowed to write the zero (as in figure 8) if that’s convenient, but you are also allowed to skip it (as in figure 9) if that’s more convenient.
Long multiplication depends on lining stuff up in columns, as a way of keeping track of place value. Therefore, when learning to do long multiplication, use graph paper. This will help you keep things properly lined up. Tidyness counts for a lot when doing long multiplication.
After you’ve done hundreds of these things, and have learned to be supertidy, you can do without the graph paper, but until then you’ll save yourself a lot of grief if you use graph paper. The best thing is to get a lab book, which is a bound book containing blank graph paper on every page. Such things are cheap and readily available from officesupply stores. The secondbest thing is to use graph paper in pad form, or perhaps looseleaf form. If you have a printer attached to your computer, there are apps that will print graph paper for you, in any imaginable style.
Note that numbers tend to be taller than they are wide. For this reason, rather than writing every row of numbers on the lines of the graph paper, it looks nicer if each row of numbers is 1½ lines below the previous one. You can see how this works in the figures in this section.
The principle here is that the graph paper is supposed to help you, not hobble you. If using the vertical lines is helpful, use the vertical lines. If using the horizontal lines in the obvious way is not optimal, use them in some lessobvious way.There are always lots of people trying to make you color inside the lines, but any kind of originality, artistry, or creativity requires coloring outside the lines. Math, if it’s done right, involves lots of creativity. Don’t be afraid to color outside the lines.
Let’s do one more example. Let’s multiply 432 by 211. We use the same trick as before: We pick 211 apart, digit by digit. That is, we write 211 = 200 + 10 + 1. Multiplying by 200 is easy; just multiply by 2, and then shift the result two decimal places. The whole procedure can be carried out very efficiently, if you organize it properly, as shown in figure 10.
The recipe for doing long multiplication is based on simple ideas:
We will mostly use the same approach as we used in section 4.2. The only difference is that this time we will need to pick apart both of the factors into their individual digits. (Previously we only picked part the second factor.)
As mentioned in section 4.2, the rationale for doing things this way is based on the distributive law and the axioms of the number system. This is discussed in more detail in section 4.7.
There is an easy and reliable method for arranging the calculations. We illustrate the method using the question shown in figure 11.
The first step is to write down the question. It is important to line up the numbers as shown, so that the ones’ place of the top number lines up with the ones’ place of the bottom number, and so on for all the other digits. Keeping things aligned in columns is crucial, since the colums represent place value. As mentioned in item 1, tidiness pays off.
If one of the factors is longer than the other, it will usually be more convenient to place the longer one on top. That’s not mandatory, but it makes the calculation slightly more compact.
You may omit the multiplication sign (×) if it is obvious from context that this is a multiplication problem (as opposed to an addition problem).
We proceed digit by digit, starting with the rightmost digit in the bottom factor, and proceeding systematically righttoleft. We do things in order of increasing place value, right to left. That makes sense, especially when doing additions, because a carry can affect the next column (to the left), but never the previous column. This is opposite to the direction of reading English text, which is unfortunate, but there’ s nothing we can do about it.
So, let’s look at the rightmost digit in the bottom factor in our example. It’s a 1. We multiply this digit into each digit of the other factor, digit by digit, working systematically righttoleft across the top factor.
We place these results in order in row c, with due regard for place value. The 1×7 result goes in the ones’ place, the 1×6 result goes in the tens’ place, and the 1×5 result goes in the hundreds’ place, and so forth, as shown in row c in figure 12. Actually, multiplication by 1 is so easy that you can just copy the whole number 4567 into row c without thinking about it very hard. So far, this is just like what we did in section 4.2.
Note: You may wish to leave a little bit of space above the numbers in row c, for reasons that will become apparent later.Also note: In all the figures in this section, anything shown in black is something you actually write down, whereas anything shown in color is just commentary, put there to help us get through the explanation the first time.
That is all we need to do with the loworder digit of the bottom factor. We now move on to the next digit of the bottom factor. In this case it is a 2.
At this point in this example, things get interesting, because some of the shortmultiplication results are two digits long. There is no way to write all of them in a single row. Fortunately, we are allowed to use multiple rows. One possibility is to spread things out as shown in figure 13. We write 2×7=14 in row d, and then write 2×6=12 on row e, and so on.
It is theoretically possible to write the product 4567×2 on a single row, but this would require doing a bunch of additions, but we choose not to do that. We are going to do all the short multiplications first, and then do all the additions later. This requires slightly more paper, but it is vastly easier to do, easier to understand, and easier to check.
As previously mentioned, the procedure is to pick a digit from the lower factor, and multiply this digit into each digit of the upper factor. The result of the first such multiplication is 2×7=14, which we place in row d. Alignment is crucial here. The 14 must be aligned under the 2 as shown. That’s because it “inherits” the place value of the 2.
Next we multiply 2×6=12. You might be tempted to write this in row d, but there is no room for it there, so it must go on row e, as shown in the figure. Again alignment is critical. This product is shifted one place to the left of the previous product, because it inherits additional place value. It inherits some from the 2 (in the bottom factor) and some from the 6 (in the top factor). Since we are working systematically righttoleft, you don’t need to think about this too hard; just remember that each of these shortdivision products must be shifted one place to the left of the previous one, as suggested by the diagonal dotted blue line in figure 13.
Let me explain the colorcoded backgrounds. In row d, the 14 has a yellow background, because it is derived from the 7 in row a, which has a yellow background. Similarly the 12 in row e yas a pink background, because it is derived from the 6 in row a, which has a pink background.
Although figure 13 is good for illustrating the principles involved, it is unnecessarily spread out. It would be better to write things more compactly, as shown in figure 14, because this makes it easier to keep things properly aligned. (The objective here is not to save paper. Remember, paper is cheap, and trying to save paper is almost always a bad idea. The advantage of the compact representation is that it makes it easier to keep things lined up in columns.)
Note the pattern of placing successive shortdivision results on alternating rows. There is guaranteed to be enough room to do this, because the product of two onedigit numbers can never have more than two digits.
We are now more than halfway finished.
At this point (or perhaps earlier), if you are not using gridruled paper, you should lightly sketch in some vertical guide lines. The tableau has become large enough that there is some risk of messing up the alignment, i.e. putting things into the wrong columns, if you don’t put in guide lines.
That’s all for the “2” digit in the bottom factor. We now proceed to the “3” digit. We multiply this by each of the digits in the top factor, in the nowfamiliar fashion. All the shortmultiplication results from this step are put in rows f and g.
At this point, all of the short multiplications are complete. In figure 15, you can see where everything comes from. The colorcoded background indicates which digit of the upper factor was involved, and the row indicates which digit of the bottom factor was involved.
All that remains is a big addition problem, adding up rows c through g inclusive. Draw a line under row g as shown in figure 16, and start adding.
In any case, the result of the big addition is the result of the overall multiplication problem, as shown on row h of figure 16.
Let’s be superexplicit about the rules for keeping things lined up in columns.
It pays to think of the shortmultiplication products in groups. The groups are labeled in figure 16. Row c is the “by one” group. Rows d and e form the “by twenty” group. Rows f and g form the “by three hundred” group.
The first rule is that each group is shifted over one place relative to the group above it. For example, the rightmost digit in row d is shifted one place relative to the rightmost digit on row c. Similarly, the rightmost digit in row f is shifted one place relative to the rightmost digit in row d.
This rule has a simple explanation: The groups inherit place value from the digit in the lower factor that gave rise to the group.
As another way of expressing the same rule: Each group is lined up under the digit in the lower factor that gave rise to the group. For example, in figure 16, the group on lines f and g is lined up under the 3 in the lower factor. Similarly, the 8 on line e is shifted over one place relative to the 10 on line d.
The second rule is that within each group, each shortmultiplication product is shifted over one place relative to the previous factor in the group. For example, in figure 16, the 12 on line e is shifted over one place relative to the 14 on line d.
This rule also has a simple explanation: Each of the products inherits some place value from the digit in the upper factor that gave rise to the product.
Let’s do another example, as shown in figure 17, which illustrates one more wrinkle.
This example calls attention to the situation where some of the shortmultiplication products have one digit, while others have two. You can see this on row c of the figure, where we have 3×3=9 and 3×8=24. In most cases it is safer to pretend they all have two digits, which is what we have done in the figure, writing 9 as 09. Similarly on line d we write 6 as 06. This makes the work fall into a nice consistent pattern. It helps you keep things lined up, and makes the work easier to check.
In some cases, such as 345×1 or 432×2, all the shortmultiplication products have one digit, so you can write them all on a single line, if you wish. This is less work, and more compact. On the other hand, remember the advice in item 3: paper is cheap. You may find it helpful to write the shortmultiplication products as two digits even if you don’t have to.If even one of the products (other then the leftmost one) requires two digits, you will almost certainly need two rows anyway, so you might as well get in the habit of using two rows.
Mathematically speaking, writing onedigit products as two digits is unconventional, but it is entirely correct. It has the advantage of making the algorithm more systematic, and therefore easier to check.
In any case, the result of the addition is the result of the overall multiplication problem, as shown on row g of figure 17. That’s all there is to it.
This algorithm ordinarily uses two rows of intermediate results for each digit in the bottom factor. In special cases, we can get by with a single row.
Another hallmark of this algorithm is that we first do all the short multiplying, and then do all of the adding. Every time we do a short multiplication, we just write down the answer. This has several major advantages:
Conversely, if any of the shortmultiplication results looks suspicious, you can easily figure out where it came from, and recheck the multiplication.
This differs from the oldfashioned “textbook” approach, which uses only one row per digit, as shown below. The oldfashioned approach supposedly uses less paper – but the advantage is slight at best, and if you allow room keeping track of “carries” throughout the tableau, the advantage becomes even more dubious. Besides, paper is cheap, so it is pennywise and poundfoolish to save paper at the cost of making the algorithm more complex, more laborious, and less reliable.
The oldfashioned approach may look more compact, but it involves more work. You have to do the same number of short multiplications, and a greater number of additions. It requires you to do additions (including carries) as you go along.
Last but not least, it makes it much harder to check your work.
4  5  6  7  
×  3  2  1  
4  5  6  7  
9  1  3  4  
1  3  7  0  1  
1  4  6  6  0  0  7 
The oldfashioned approach.
Not recommended.
Remember, paper is cheap (item 3) and it is important to be able to check your work (item 9).
The longmultiplication algorithm as presented here is guaranteed to work. That’s because it is quite directly based on the fundamental axioms of arithmetic, including the axioms of the number system.
Consider the numbers 4567 and 321 which appeared in section 4.3. We can expand 4567 as 4000+500+60+7. That’s what 4567 means, according to the rules of the Hindu number system. Similarly, we can expand 321 as 300+20+1.
The reason for expanding things this way is that in the sum 4000+500+60+7, each term consists of a single nonzero digit, with some place value.
The next step is to expand the product 4567×321 as (4000+500+60+7)×(300+20+1). We can then carry out the product in the expanded form, term by term, by repeated application of the distributive law. In our example, there are four terms in the first factor and three terms in the second factor, which give us 12 terms in the product, as shown in table 1. Each term in the product consists of a singledigit multiplication – a short multiplication – with some place value.
300  20  1  
7:  2100  140  7  
60:  18000  1200  60  
500:  150000  10000  500  
4000:  1200000  80000  4000 
So, the main thing the algorithm needs to do is to help you write down these 12 numbers in a convenient, systematic way.
One convenience is that the algorithm saves you from writing down all those zeros. You pay a small price for this. The price is that you absolutely must keep the numbers properly lines up in columns.
The rules about how to write the numbers in columns are not magic; they are just a way of keeping track of place value. They are a direct consequence of the axioms of arithmetic, including the axioms of the number system.
In particular, the “12” that appears in the lowerleft element of table 1 is the same “12” that appears at the left of row g in figure 16 ... and the latter is lined up in columns so as to give it the correct place value.
The word “multiplicand” is moreorless Latin for “the thing being multiplied”.
In figure 11, some people insist that the top number should be called the “multiplier” and the bottom number should be called the “multiplicand”. Meanwhile, other people insist on naming them the other way around. Quite commonly, they use the words in ways inconsistent with the supposed definition.
All of this is nonsense. Multiplication is commutative. Pretending to see a distinction between the two numbers is the silliest thing since the holy war between the BigEndians and the LittleEndians.
Furthermore, multiplication is associative, so we could easily have three or more things being multiplied, as in 12×32×65×99, in which case it is obviously hopeless to attempt to distinguish “the” multiplier from “the” multiplicand.
The modern practice is to call both of the numbers in figure 11 the factors. You can equally well call both of them multipliers, or call both of them multiplicands. Along the same lines, modern practice is to say that an expression such as 12×32×65×99 is a product of four factors.
There are also holy wars as to whether 4567×321 means 4567 “times” 321, or 4567 “multiplied by” 321. The distinction is meaningless. Don’t worry about it.
The usual “textbook” instructions for how to do long division are both unnecessarily laborious and unnecessarily hard to understand. There’s another way to organize the calculation that is much less mysterious and much less laborious (especially when long multidigit numbers are involved).
Note that as discussed above, keeping things lined up in columns is critical. It may help to use gridruled paper, or at least to sketch in some guidelines.
Suppose the assignment is to divide 7527 by 13. After writing down the statement of the problem, and before doing any actual dividing, it helps to make a crib, as shown in figure 18. This is just a multiplication table, showing all multiples of the assigned divisor, which is 13 in our example.
It is supereasy to construct such a table, since no multiplication is required. Successive addition will do the job. We need all the multiples from ×1 to ×9, but you might as well calculate the ×10 row, by adding 117+13. This serves as a valuable check on all the additions that have gone before.
We now begin the main part of the algorithm. The first step is to consider the leading digit of the dividend (which in this case is a 7). This is less than the divisor (13). There is no way we can subtract any nonzero multiple of 13 from 7. Therefore we leave a blank space above the 7 and proceed to the next digit. We could equally write a zero above the 7, as shown in gray in figure 20, but that’s just extra work and confers no lasting advantage, since leading zeros are meaningless.
Now consider the first two digits together, namely the 7 and the 5. Look at the crib to find the largest entry less than or equal to 75. It is 65, as in 5×13=65. Copy this entry to the appropriate row of the division problem, namely row b. Align it with the 75 on the previous row. Since this is 5 times the divisor, write a 5 on the answer line, aligned with the 75 and the 65, as shown. Check the work, to see that 5 (on the answer line) times 13 (the divisor, on line a), equals 65 (on line b).
Now do the subtraction, namely 75−65=10, and write the result on line c as shown.
When we take place value into account, the 5 in the quotient is short for 500, and the 65 that we just subtracted is short for 6500. Normally we would need to write the zeros explicitly as part of these numbers, to indicate place value. However, in this situation we are using columns to keep track of place value. If we keep everything properly lined up, we don’t need to write the zeros. In fact, the zeros would just get in the way (especially in the quotient) so we are better off not writing them.
We now shift attention to the situation shown in figure 21. Bring down the 2 from the dividend, as shown by the red arrow. So now the number we are working on is 102, on line c.
The steps from now on are a repeat of earlier steps.
Look at the crib to find the largest entry less than or equal to 102. It is 91, as in 7×13=91. Write copy this entry to the division problem, on row d, directly under the 102. Since this came from row 7 of the crib, write a 7 on the answer line, aligned with the 102 and the 91, as shown. Check the work, to see that 7 (on the answer line) times 13 (the divisor, on line a), equals 91 (on line d). Do the subtraction.
As a check on the work, when doing this subtraction, the result should never be less than zero, and should never be greater than or equal to the divisor. Otherwise you’ve used the wrong line from the crib, or made an arithmetic error.
We now shift attention to the situation shown in figure 22. Bring down the 7, look in the crib to find the largest entry less than or equal to 117, which is in fact 117, as in 9×13=117. Since this came from line 9 of the crib, write a 9 on the answer line, properly aligned.
The final subtraction yields the remainder on line g. The remainder is zero in this example, because 13 divides 7527 evenly.
The crib plays several important roles.
Perhaps the crib’s most important advantage, especially when people are first learning long division, is that the crib removes the mystery and the guesswork from the long division process. This is in contrast to the “trial” method, where you have to guess a quotient digit, and you might guess wrong. Using the crib means we never need to do a short division or trial division; all we need to do is skim the table to find the desired row.
We have replaced trial division by multiplication and tablelookup. Actually we didn’t even need to do any multiplication, so it would be better to say we have replaced trial division by addition.
Another advantage is efficiency, especially when the dividend has many digits. That’s because you only need to construct the crib once (for any given divisor), but then you get to use it again and again, once for each digit if the dividend. For long dividends, this saves a tremendous amount of work. (This is not a good selling point when kids are just learning long division, because they are afraid of big multidigit dividends.) Setting up the crib is so fast that you’ve got almost nothing to lose, even for small dividends.
Another advantage is that it is easy to check the correctness of the crib. It’s just sitting there begging to be checked.
When bringing down a digit, you may optionally bring down all the digits. For instance, in figure 21, if you bring down all the digits you get 1027 on row c. One advantage is that 1027 is a meaningful number, formed by the expression 7527−13×5×100. This shows how the steps of the algorithm (and the intermediate results) actually have mathematical meaning; we are not not blindly following some mystical mumbojumbo incantation. I recommend that if you are trying to understand the algorithm, you should bring down all the digits a few times, at least until you see how everything works.A small disadvantage is that bringing down all the digits requires more copying. The countervailing small advantage is that it may help keep the digits lined up in their proper columns. Whether the advantages outweigh the cost is open to question, and probably boils down to a question of personal preference.
Another remark: Division is the “inverse function” of multiplication. In a profound sense, for any function that can be tabulated, you can construct the inverse function – if it exists – by switching columns in the table. That is, we interchange abscissa and ordinate: (x,y)↔(y,x). That’s why we are able to perform division using what looks like a multiplication table; we just use the table backwards.
The modern numeral system is based on place value. As we understand it today, each numeral can be considered a polynomial in powers of b, where b is the base of the numeral system. For decimal numerals, b=10. As an example:
 (5) 
This allows us to understand long multiplication (section 4) in terms of the more general rule for multiplying polynomials.
Given two expressions such as (a+b+c) and (x+y), each of which has one or more terms, the systematic way to multiply the expressions is to make a table, where the rows correspond to terms in the first expression, and the rows correspond to terms in the second expression:
 (6) 
In the special case of multiplying a twoterm expression by another twoterm expression, the mnemonic FOIL applies. That stands for First, Outer, Inner, Last. As shown in figure 23, we start with the First contribution, i.e. we multiply the first term from in each of the factors. Then we add in the Outer contribution, i.e. the first term from the first factor times the last term from the last factor. Then we add in the Inner contribution, i.e. the last term from the first factor times the first term from the last factor. Finally we add in the Last contribution, i.e. we multiply the last terms from each of the factors.
If you can do long division, you can do square roots.
Most square roots are irrational, so they cannot be represented exactly in the decimal system. (Decimal numerals are, after all, rational numbers.) So the name of the game is to find a decimal representation that is a sufficientlyaccurate approximation.
We start with the following idea: For any nonzero x we know that x÷√x is equal to √x. Furthermore, if s_{1} is greater than √x it means x/s_{1} is less than √x. If we take the average of these two things, s_{1} and x/s_{1}, the average is very much closer to √x. So we set
 (7) 
and then iterate. The method is very powerful; the number of digits of precision doubles each time. It suffices to use a rough estimate for the starting point, s_{1}. In particular, if you are seeking the square root of an 8digit number, choose some 4digit number as the s_{1}value.
This is a special case of a more general technique called Newton’s method, but if that doesn’t mean anything to you, don’t worry about it.
Note that the square of 1.01 is very nearly 1.02. Similarly, the square of 1.02 is very nearly 1.04. Turning this around, we find the general rule that if x gets bigger by two percent, then √x gets bigger by one percent ... to a good approximation.
We can illustrate this idea by finding the square root of 50. Since 50 is 2% bigger than 49, the square root of 50 is 1% bigger than 7 ... namely 7.07. This is a reasonably good result, accurate to better than 0.02%.
If we double this result, we get 14.14, which is the square root of 200. That is hardly surprising, since we remember that the square root of 2 is 1.414, accurate to within roundoff error.
Sine and cosine are transcendental functions. Evaluating them will never be supereasy, but it can be done, with reasonably decent accuracy, with relatively little effort, without a calculator.
In particular:
The following facts serve to “anchor” our knowledge of the sine and cosine:

Actually, that hardly counts as “remembering” because if you ever forget any part of equation 8 you should be able to regenerate it from scratch. The 0^{∘} and 90^{∘} values are trivial. The 30^{∘} is a simple geometric construction. Then the 60^{∘} and 45^{∘} values are obtained via the Pythagorean theorem. The value for 45^{∘} should be particularly easy to remember, since √2 = 1.414 and √½ = ½√2.
The rest of this section is devoted to the Taylor series. A loworder expansion works well if the point of interest is not too far from the nearest anchor.
For most purposes, the best option is to use the Taylor[1,3] approximation anchored at zero. This requires a couple more multiplications, but the result is accurate to better than 0.07%.
If you really want to minimize the number of multiplications, we can start by noting that the Taylor[1] extrapolation coming up from zero is better than the Taylor[0,1] extrapolation coming down from 30^{∘}, so rather than using the closest anchor we use the 0^{∘} anchor all the way up to 20^{∘} and use the 30^{∘} anchor above that. This has the advantage of minimizing the number of multiplications. Disadvantages include having to remember an obscure fact, namely the need to put the crossover at 20^{∘} rather than halfway between the two anchors. The accuracy is better than 2.1%, which is not great, but good enough for some applications. The error is shown in figure 25.
There are many other options, but all the options I know of involve either more work or less accuracy.
The spreadsheet that produces these figures is given by reference 6.
Here are some additional facts that are needed in order to carry out the calculations discussed here.

Last but not least, we have the Pythagorean trig identity:
 (10) 
and the sumofangles formula:
 (11) 
If you can maintain even a vague memory of the form of equation 11, you can easily reconstruct the exact details. Use the fact that it has to be symmetric under exchange of a and b (since addition is commutative on the LHS). Also it has to behave correctly when b=0 and when b=π/2.
If we assume b is small and use the smallangle approximations from equation 9, then equation 11 reduces to the secondorder Taylor series approximation to sin(a+b).
 (12) 
If we drop the secondorder term, we are left with the firstorder series, suitable for even smaller values of b:
 (13) 
You can use the Taylor series to interpolate between the values given in equation 8. Since every angle in the first quadrant is at least somewhat near one of these values, you can find the sine of any angle, to a good approximation, as shown in figure 24.
Copyright © 2007 jsd