Here N is the tabulated difference in the logarithms corresponding to the difference 1 in the numbers at any part n of the table. Hence we can find N different numbers between n + 1 and n whose logarithms differ in the first 7 places of decimals. Hence, at this part of the table, we can find accurately to an Nth of unity any intermediate number whose logarithm is given correctly to 7 decimal places.

Now N varies inversely as n.

Hence, as we proceed in the table, we can interpolate, between numbers differing by the constant amount unity, a continually smaller number of values corresponding to given logarithms calculated to a fixed number 7 of decimal places.

566. To find the difference of the sines of two angles approximately in terms of the difference of the angles.

Let 81 be the circular measure of an angle. Then

sin (a + 8) — sin a = cos a sin 8 — sin a (1 — cos d)

= cos a (8 – 1-83 + ...) − sin a (1♪2 – 21484 + ...).

In this series after 8 cos a the terms, taken in pairs, are continually decreasing and alternate in sign.

Hence the series differs from 8 cos a by a quantity less than 182 sin a + 183 cos a.

Let & be the circular measure of an angle less than 1'. Then

Also sin a and cos a are < 1.

.. 182 sin a + 18 cos a < 1. 10−7.

Hence, if 8 is not greater than the circular measure of l′, the equation

sin (a + 8) - sin a ==

8 cos a

26

holds as far as 7 decimal places.

J. T.

567. To interpolate the sine of any given angle between a and a + 1', we may use the above formula which will give the required sine correctly to 7 decimal places.

568. But conversely, to interpolate an angle corresponding to a given sine, in the sine-difference formula put for 8 the circular measure of 1'. Then

sin (a + 1') - sin a =

π

10800

cos a = N. 10-7 say.

Here N is the tabulated difference in the sines corresponding to the difference l' in the angles at any part a of the table. Hence we can find N different angles between a + l' and a whose sines differ in the first 7 decimal places.

Now as the angle a increases up to 90°, cos a and therefore N decreases. Hence we can find intermediate angles from their sines with continually decreasing accuracy as we approach 90°.

For example, sin 78° 5′ and sin 78° 6' differ by 600. 10-7. Hence up to 78° 6' we can, from the sines of angles given correctly to 7 places, interpolate angles correctly to within a 600th of a minute; i.e. a 10th of a second. Again sin 84° 3′ and sin 84° 4' differ by 300. 10-7. Hence up to this point we can similarly interpolate correctly to within a 5th of a second. And so on.

569. To find the difference of the tangents of two angles approximately in terms of the difference of the angles.

Thus tan (a +8) – tan a differs from 8 sec2 a by a quantity less

than

=

This quantity increases as a and as 8 increase.

Now it will be found by examining the tables that approximately tan a sec2 a = = 1, when a is 34°; = 10, when a is 64°; = 100, when a is 78°; = 1000, when a is 84°: and so increases more and more rapidly up to 90°, when it is infinite. Hence the tangent-difference formula is less and less accurate as the angle approaches 90°.

570. Hence, to interpolate a tangent corresponding to a given angle between a and a + 1', the equation

holds for 7 places, up to 34°; for 6 places, up to 64°; for 5 places, up to 78°; for 4 places, up to 84°; and so on.

571. Conversely, to interpolate an angle corresponding to a given tangent, in the tangent-difference formula put for 8 the circular measure of 1'. Thus

П

tan (a + 1') - tan a =

sec2 a =

= N. 10-7 say.

10800

Its

Here N, which varies as sec2 a, increases as a increases. smallest value, viz. the difference between tan 0 and tan l', is about 3000. Hence even here we could find an angle correctly to within nearly of a second corresponding to a tangent given correctly to 7 decimal places.

Now the difference in the tangents for l' is at 34° about 4000 × 10-7; at 64°, 1500 × 10-6; at 78°, 600 × 10−5; at 84°, 270 × 10-4; and so on.

1

Hence we can find angles, when the tangents are given correctly to 7 decimal places from 0 to 34°, with an accuracy increasing from to of a second: when the tangents are given to 6 places from 34° to 64°, with an accuracy increasing fromto of a second: when the tangents are given to 5 places from 64° to 78°, with an accuracy increasing from to of a second: when the tangents are given to 4 places from 78° to 84°, with an accuracy increasing from 1 to 1 of a second and so on.

5

572. To find the difference of the secants of two angles approximately in terms of the difference of the angles.

Here we have

sec (a + 8) - sec a =

sec a sec & 1- tan a tan S

sec a

= sec a sec 8 (1 + tan a tan 8 + tan2 a tan2 8+ ...) - sec a.

Now sec &>1 and tan &> 8.

secants > 8 sec a tan a.

Hence the difference of the

But cos 8>1-182 and sin d<8. Hence the difference of the secants

Hence sec (a + d) - sec a differs from 8 sec a tan a by a quantity

82 sec a (1⁄2 + tan2 a + 18 tan a) – nearly d2 sec a (sec2 a — §).

1 Stan a -182

-

This quantity increases as a and as 8 increase.

Now it will be found by examining the tables that approximately sec a (sec2 a-1)= 1 when a is 33°; = 10 when a is 63°; = 100 when a is 77°; = 1000 when a is 84°; and so increases more and more rapidly up to 90° where it is infinite.

Hence the secant-difference formula is less and less accurate as we approach 90°.

573. Hence, to interpolate a secant corresponding to a given angle between a and a + 1', the equation

sec (a + 8) - sec a = 8 sec a tan a

holds for 7 places up to 33°; for 6 places up to 63°; for 5 places up to 77°; for 4 places up to 84°; and so on.

574. Conversely, to interpolate an angle corresponding to a given secant, in the secant-difference formula put for d the circular measure of 1'. Thus

sec (a+1') - sec a =

π

sec a tan a = N. 10-7 say.

10800

Here N increases as a increases. Hence, as we approach zero, we can interpolate an angle with less and less accuracy corresponding to a given secant. For instance sec 11° 28′ and sec 11° 29' differ by 600 × 10-7. Hence here we can find angles to within of a second corresponding to secants given correctly to 7 places. Again sec 3° 13′ and sec 3° 14′ differ by 160 × 10-7. Hence here we can interpolate an angle correctly to within only second. And so on.

of a

On the other hand as we approach 90° we find the following results. The difference in the secants for 1' is at first about 2 × 10-7, and at 33° about 2000 × 10-7; at 63°, 1200 x 10-6; at 77°, 600 × 10−5; at 84°, 250 × 10-4; and so on.

Hence we can find angles, when the secants are given correctly to 7 decimal places from 1' to 33°, with an accuracy increasing from 30" to of a second; when the secants are given correctly to 6 decimal places from 33° to 63°, with an accuracy increasing fromto of a second; when the secants are given correctly to 5 decimal places from 63° to 77°, with an accuracy increasing fromto of a second; when the secants are given correctly to 4 decimal places from 77° to 84°, with an accuracy increasing from 1 to of a second; and so on.

575. The secondary ratios cosine, cotangent, and cosecant of an angle being simply the sine, tangent, and secant respectively