4. log cos a = 6.9218.

2. log cot a = 2.8816; When the small angle is given in seconds. Although the computer may take out his angles to tenths of minutes, cases often arise in which a small angle is given in seconds, or degrees, minutes, and seconds, and in which the trigonometric function is required to five decimals. In this case the preceding method may not always give accurate results, because the arc and its sine or tangent may differ by a greater amount than the error we can admit in the computation. Table II. is framed to meet this case. The following are the quantities given:

In the second column: The argument, in degrees and minutes, as already explained for Table III.

In the first column: This argument reduced to seconds. From this column the number of seconds in an arc of less than 2°, given in degrees, minutes, and seconds, may be found at sight.

Example. How many seconds in 1° 28′ 39′′? In the table, before 1° 28', we find 5280", which being increased by 39" gives 5319", the number required.

Col. 3. The logarithm of the sine of the angle. This is the same as in Table III.

Col. 4. The value of log sine minus log arc; that is, the difference between the logarithm of the sine and the logarithm of the number of seconds in the angle.

Col. 5. The same quantity for the tangent.

Cols. 6 and 7. The complements of the preceding logarithms, distinguished by accents.

The use of the tables is as follows.

To find the sine or tangent of an angle less than 2o:

Express the angle in seconds by the first two columns of the table. Write down the logarithm in column S or column T, according as the sine or a tangent is required.

Find from Table I. the logarithm of the number of seconds. Adding this logarithm to S or T, the sum will be the log sine or log tangent.

Example. Find log sin 1° 2′ 47′′.9.

1° 2′ 47′′.9

S, 4.685 55 3767".9; log, 3.576 10

log sin 1° 2′ 47′′.9, 8.261 65

To find the arc corresponding to a given sine or tangent:

Find in the column L. sin. the quantity next greater or next smaller than the given logarithm.

Take the corresponding value of S' or T' according as the given function is a sine or tangent, and add it to the given function. The sum is the logarithm of the number of seconds in the required angle.

When the cosine or cotangent of an angle near 90° or 270° is required, we take its difference from 90° or 270°, and find the complementary function by the above rules.

Remark. The use of the logarithms of the trigonometric functions is so much more extensive than that of the functions themselves that the prefix "log" is generally omitted before the designation of the logarithmic function, where no ambiguity will result from the omission.

TABLE V.

NATURAL SINES AND COSINES.

19. This table gives the actual numerical values of the sine and cosine for each minute of the quadrant.

To find the sine or cosine corresponding to a given angle less than 45°, we find the degrees at the top of a pair of columns and the minutes on the left.

In the two columns under the degrees and in the line of minutes we find first the sine and then the cosine, as shown at the head of the column.

A decimal point precedes the first printed figure in all cases, except where the printed value of the function is unity.

If the given angle is between 45° and 90°, find the degrees at the bottom and the minutes at the right.

Of the two numbers above the degrees, the right-hand one is the sine and the left-hand one the cosine.

For angles greater than 90° the functions are to be found according to the precepts given in the case of the logarithms of the sines and tangents.

TABLE VI.

ADDITION AND SUBTRACTION LOGARITHMS.

20. Addition and subtraction logarithms are used to solve the problem: Having given the logarithms of two numbers, to find the logarithm of the sum or difference of the numbers.

The problem can of course be solved by finding the numbers corresponding to the logarithms, adding or subtracting them, and taking out the logarithm of their sum or difference. The table under consideration enables the result to be obtained by an abbreviated process.

I. Use in addition. The principle on which the table is constructed may be seen by the following reasonings. Let us put

S = a + b,

a and b being two numbers of which the logarithms are given. We shall have

Since log a and log b are both given, we can find log x from the equation

log x = log blog a,

which is therefore a known quantity.

Now, for every value of log x there will be one definite value of each of the quantities x, 1+x, and log (1 + x). Therefore a table may be constructed showing, for every value of log x, the corresponding value of log (1 + x).

Such a table is Table VI.

The argument, in column A, being log x, the quantity B in the table is log (1 + x).

Therefore, entering the table with log x as the argument, we take out log (1+x), which added to log a will give log S.

We have therefore the following precept for using the table in addition:

Take the difference of the two given logarithms.

Enter the table with this difference as the argument A, and take out the quantity B.

Adding B to the subtracted logarithm, the sum will be the required logarithm of the sum.

It is indifferent which logarithm is subtracted, but convenience: in interpolating will be gained by subtracting the greater logarithm from the lesser increased by 10. The number B will then be added to the greater logarithm.

Example.

Given log m = 1.62974, log n = 2.203 86 ;

; find The required logarithm is found in either of the following two

log (m + n).

The figures in parentheses show the order in which the numbers are written.

EXERCISES,

Log a and log b having the following values, find log (a + b).

II. Use in subtraction. The problem is, having given log a and