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Let be the circular measure of n seconds. Then, when is small, we have en sin 1", approximately.

=

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Hence, if the angle is known, the table gives the value of the expression in parenthesis, and log n can be found from the ordinary table of the logs of numbers; thus log sin n" can be found.

If log sin n" is given, we can find approximately the value of n, and then from the table we have the value of the expression in parenthesis; thus we can find log n, and then n from an ordinary table of logs of numbers.

Rem. When is small (less than 5°),

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Hence a small error in @ will not produce a sensible error in the result, since log will vary much less rapidly than 0.

sin 0
0

(3) Maskelyne's Method. The principle of this method is the same as that of Delambre's. If is a small angle, we have

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.. log sin @= log + log cos 0, approximately.

When is a small angle, the differences of log cos are insensible (Art. 149); hence, if 0 be given, we can find log accurately from the table of natural logarithms, and also an approximate value of log cos 0; the formula then gives log sin at once.

If log sin be given, we must first find an approximate value of from the table, and use that for finding log cos 0, approximately; 0. is then obtained from the formula.

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6. Prove log, 11 = 2.39789527 ....., by (4) of Art. 130.

7. Prove log, 13 = 2.56494935...,

66

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10. Find, by means of the table of common logarithms and the modulus, the Napierian logarithms of 1325.07, 52.9381, and .085623. Ans. 7.18923, 3.96913, -2.4578.

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22. If is the circular measure of an acute angle, prove

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+ and (2) tan 0>0+

24'

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02

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2

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25. Given sin = n sin 0, tan = 2 tan 6: find the limiting values of n that these equations may coexist. Ans. n must lie between 1 and 2, or between 1 and

26. Find the limit of (cos ax) cosec2bx, when x = 0.

2.

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27. From a table of natural tangents of seven decimal places, show that when an angle is near 60° it may be determined within about of a second.

28. When an angle is very near 64° 36', show that the angle can be determined from its log sine within about of a second; having given (log, 10) tan 64° 36' = 4.8492, and the tables reading to seven decimal places.

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(1)

(cos +√1 sin 0)" = cos no + √- 1 sin no.

I. When n is a positive integer.

We have the product

(cos a += 1 sin a) (cos ẞ + √1 sin ẞ)

=

ß)

(cos a cos ß-sin a sin ẞ) + √−1 (cos a sin ẞ+sin a cos ß)

= cos (a + B)+V-1 sin (a + ẞ).

Similarly, the product

[cos (a + B)+√−1 sin (a + ẞ)][cos y +√-1 sin y]

= cos (α + B + y) + √ − 1 sin (a + B + y).

Proceeding in this way we find that the product of any number n of factors, each of the form

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(cos +V-1 sin 0)" = cos no + √−1 sin no,

which proves the theorem when n is a positive integer.

* From the name of the French geometer who discovered it.

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