cos x sin x
68. Sen. If it is desired to obtain the natural functions of angles esti-
mated to seconds, it is necessary that the values in the' tables computed as above
be extended to 7 decimals at least From such a table we may make interpo-
lations for seconds with sufficient accuracy for most practical ends, except
for values near the limits, where the disparity between the variation of the arc
and that of the function changes very rapidly. For example, let it be required
to find sin 34 24' 12" from the data sin 34 24' = .5649670, and sin 34 25' =
.5652070. We observe that an increase of V upon the angle of 34 24' makes
an increase of .5652070 .5649670 = .0002400 in the sine. Hence an increase
of 12", or i of 1', makes an increase of i of .0002400, or .0000480, approxi-
mately Adding, we have sin 34 24' 12" = .5650150. The student must be careful
to notice whether an increase of the angle makes a numerical increase or a de
crease of the function, and add or subtract as the case may require.
69. Prob. To construct a table of logarithmic trigonometrical
SOLUTION. Compute the natural sines and cosines as in the preceding prob-
lem. Take the logarithms of the values thus obtained, and add 10 to each
34 PLANE TRIGONOMETRY.
characteristic. The results are the ordinary tabular logarithmic sines and co
sines. For example, we find from the table ofuatural functions that sin 04 24'
~ .5649670. The logarithm of this number is 1.752023. Adding 10 to the char-
acteristic, we have log sin 34 J 25' = 9.752023, as usually given in the tables. In
like manner the cosines are obtained.
To obtain the tabular logarithmic tangents, we have from tan a; = -^5-^
.eg tan x log sin x log cos x. If we now take the log sin x from the table as
computed by the preceding part of this solution, and from it subtract the cor-
responding log cos x, the result is the trite log tan x, since the extra 10 in the
tabular log sin and log cos is destroyed by the subtraction. Therefore, to this
difference we must add 10 to get the tabular log tan, as above explained. For
example, the tabular log sin 34 24' 9.752023, and log cos 34 24' = 9.916514.
Hence, the tabular log tan 34 24' = 9.752023 - 9.916514 + 10 = 9.835509. In
like mapner the tabular log cot a; = log cos x log sin a; + 10.
If the logarithmic secants are required they can be obtained from the relation
sec x = , which gives log sec x = log cos x. In applying this by means
of the tabular functions, it must be observed that the log cos x, as we get, it
from the table, is 10 too great ; hence, the true log sec a; log cos x + 10.
In tabulating log secants and cosecants, it is not necessary to add 10, since, as
these functions are never less than 1, their logarithms are never negative.
7O. Sen. The interpolations for seconds are usually made in the same way
when using the logarithmic functions, as explained above for the natural func-
tions. But to facilitate the operation, the approximate change of the logarithm
for a change of V of the angle is commonly written in the table, in a column
called Tabular Differences, and marked D.
1. Find from the tables at the close of the volume the natural
trigonometrical functions of 25 18'.
SOLUTION. To find the sine and cosine w 1 look in Table II., and find 25 at
the top of the page. In the extreme left-hand column we find the minutes, and
passing down to 18, find opposite, in the column headed N. sin (natural sine)
42736 ; also in the column N. cos, we find 90408. Now, as these are the lengths
of the sine and cosine as compared with radius, we know they are fractions.
.-. Sin 25 18' = .42736, and cos 25 18' = .90408.
To find tlie tangent we turn to Table IV., and finding 25 at the top of the page,
cass down the column of minutes, on the left-hand of the page, to 18, opposite
which, and under the column headed 25, we find 2698. To this we prefix the
figures 47, which stand in the same column, opposite 11', and belong to the tun-
gents of all the angles from 25 10' to 25 19', and are omitted in the table sim-
ply 10 relieve the eye and to economize space. Thus we find tan 25 18' = .47209$
the number being known to be a fraction because the angle is less than 45
CONSTRUCTION AND USE OF TKIflONOMETltlCAL TABLES. d&
To find the cotangent we look at the bottom of the page in the same table till
we find 25", and then passing up the minutes column at the right hand, find
W)t 25 18' = 2. 11552.
If tlie secant were required we should be obliged to obtain it by dividing 1 Iry
the cosine, as our tables do not include this function. Thus sec 25 18' =
cos 25 18' .90408
L NoTE. Tables of secants and cosecants are sometimes given, but they are
not of sufficient importance tc justify their introduction into an elementary
2. Show that sin 37 43' = .61176 ; cos 37 43' = .79105 ; tan 37 43'
==.773353; cot 37 43' = 1.29307; sec 37 43' = 1.264142; cosec37
43' = 1.634628 ; vers 37 43' = .20895 ; covers37 43' = .38824.
3. Find that sin 64 36' = .90334; cos 64 36' = .42894 ; tan 64
36' = 2.10600 ; cot 64 36' = .474835 ; sec 64 36' = 2.331328 ; coseo
64 36' = 1.107003 ; vers 64 36' = .57106; covers 64 36' = .09666.
SUG. In looking for sines and cosines of angles above 45, seek the degrees
at the bottom of the page, and be careful to observe that the columns of sines and
cosines, as named at the top, change names when read from the bottom. The
foundation of this arrangement will be readily perceived. Thus, turning in
Table II. to 24 32', we find sin 24 32-'= .41522. But sin 24 32' = cos (90 - 24 32')
= cos 65 28' = .41522. Thus the degrees and minutes read from the bottom of
the page are the complements of those read from the top.
4. Find that sin 42 27' 12" = .67499; cos 42 27' 12" = .73783;
tan 42 27' 12" = .914834; cot 42 27' 12" = 1.09309.
Sue. Sin 42 27' = .67495, and sin 42 28' = .67516. .'. An increase of 1' in
the angle makes an increase of 21 (hundred-thousandths) in the sine, and 12"
will make $ or i as great an increase, approximately. Observe that in the case
of cosine an increase of the arc makes a decrease of the function.
5. Find that sin 143 24' = 0.596225; cos 151 23' = .877844;
tan 132 36' = 1.08749 ; and cot 116 7' = .490256.
BUG. Sin 143 24' = sin (180 143 24') = sin 36 36'. Also the trigono-
metrical function of any angle is numerically equal to the same function of its
6. Find the logarithmic trigonometrical functions of 32 15'
from the tables at the end of the volume.
36 PLANE TRIGONOMETRY.
SOLUTION. Turning ;o Table II. we find 32 at the top of the page, and
jpposite 15', and in the column L. sin (logarithmic sine), we get 9.727228 ; i. e.
log sin 32 15' = 9.727228. Now from the column of differences, D. 1", we leaTii
that an increase of 1" of the arc at this point makes, approximately, an increase
ol 3.34 ( million ths) in the logarithm of its sine. Hence, we assume thai an in-
crease of 22" makes 22 x 3.34 = 73 (millionths). /. log sin 32 15' 22" = 9.727226
+ .000073 = 9.727301. In a similar manner we have log cos 32 15' = 9.927231.
An increase of l"in the arc makes a decrease of 1.33 (millionths) in the log cos.
/. an increase of 22" makes 29 (millionths) decrease in the log r,os, and log co?
82 15' 22" = 9.927202. Log tan 32 15' 22" = 9.800100 ; and log cot 32' 15 22" =
7. Find that log sin 24 27' 34" = 9.617051 ; log cos 26 12' 20" =
9.952897; log tan 26 12' 20" = 9.692125 ; log cot 126 23' 50" =
SUG. Observe cot (126 23' 50") = cot (180 - 126 23' 50") = cot (53 36' 10").
Also that angles above 45 are found at the bottom of the table ; and remember
to subtract the correction for co-functions, if an increase of arc is assumed.
8. Given the natural sine .45621, to find the angle from the tables.
SOLUTION. Looking for this sine in the table of natural sines, we find the
next less sine to be .45606, and the angle corresponding, 27 8'. Now, at this
point, an increase of 1' in the arc makes an increase of 26 (hundred thousandths)
in the natural sine. But the given sine .45621 is only 15 (hundred thou-
sandths) greater than .45606, the sine of 27 8'. Hence the required angle is
but i of 1' or 60" = 35", greater than 27" 8'. /. sin- 1 .45621 = 27 8' 35",
and its supplement 152 51' 25", which has the same sign, and these arcs in-
creased by every multiple of 2x.
9. Find sin- 1 .62583; cos- 1 .34268; tan~ 1 .468531 ; 00^.876434.
Results. Sin- '.62583 = 38 44' 35", and 141 15' 25"; and these
arcs increased by every multiple of 2.
cos~ 1 .34268 = 69 57' 36", and 360 - 69 57' 36" =
290 2' 24'', and these arcs increased by every
multiple of Zx.
tan- -'.468531 = 25 6' 16", and 180 + 25 6' 16" = 205
06' 16", and these arcs increased by every multiple
cot-'.876434 = 48 46' 3", and 180 + 48 46' 3" =
228 46' 03", and these arcs increased by everj
multiple of 2ir,
CONSTRUCTION AND USE OF TRIGONOMETRICAL TABLES. 37
SUG. Observe that an increase of the arc makes a decrease of its co-functk-us.
In the table of tangents as given, Table IV., the proportional parts given al the
bottom of each column are the approximate changes which the functions
undergo for a change of 1" in the function. Thus, in finding cotr~ 1 .876434, we
find cot- 1 .876462 = 48 46' ; and at the bottom we find that a change of 8.64
(millionths) in the function makes a change of 1" in the angle. Hence, as the
given cotangent is 28 (millionths) less than the cotangent of 48 46'. the angle
required is 28 -r- 8.64 = 3 (seconds), greater than 48 46'.
71. Sen. It is usually best to take from the table that function which is nearest
in value to the given function, and then increase or diminish the corresponding
arc as the case may require. If we always take from the table the next less
function than that given, in the example for sine, tangent, and secant, and the
next greater for the cosine, cotangent, and cosecant, corrections for seconds
will require always to be added. If we always take from the tables the func-
tions next less than the one given, the corrections for seconds must be added for
sine, tangent, and secant, and subtracted for the co-functions. If we were always
to take from the tables the next greater function than the one given, the
seconds corrections would be added for the co-functions, and subtracted for the
[NOTE. It is very important that the pupil become so familiar with the
nature of these tables as to use them intelligently, and not mechanically. For
this reason we refrain from giving the usual specific, mechanical directions for
their use, and substitute illustrations showing how they are used in accordance
with the principles upon which they are constructed.]
10. Find sin-^-. 34256); cos^- .62584) ; tan-^- 3.41621) ;
Results. sin- 1 (-.34256) = 200 1' 58", and 339 58' 02", and theso
arcs increased by every multiple of 2vr,
cos- 1 (-.62584) = 128 44' 38", and 231 15' 22", and these
arcs increased by every multiple of 2tf.
tan- 1 !- 3.41621) = 106 18' 57", and 286 18' 57", and
these arcs increased by every multiple of 2*.
cot-'(- 1.21648) = 140 34' 42", and 320 34' 42", and
these arcs increased by every multiple of 2*.
BUG. To obtain these results the pupil will need to recall the principles in
the corollaries to (4855). Thus, to find cot- \ 1.21648), we find from
the table that cot- \l. 2 1648) = 39 25' 18" ; and from (55) COR., we learn that
cot (180- x) - - cot x. .: Cot-'t- 1.21648) = 180 - 39 25' 18" = 140 34' 42"
Again, from the same corollary, we learn that cot (360 x) = cot a
. Cot- J (- 1.21468) = 360 - 39 25' 18" = 320 34' 42".
11. Given the logarithmic sine 9.451234, to find the corresponding
00 PLANE TRIGONOMETRY.
SOLUTION. The next nearest log sin found in Table II., is 9.451204 = log
sin 16 25'. Now we learn from the table that an increase of 1" in the angle at
this point, makes an increase in its log sin of 7.14 (milliouths). But the given
log sin, 9.451234, is 30 (millionths) greater than log sin 16 25'. .-. The required
angle is 30 -r- 7.14 = 4 (seconds) greater than 16 25' ; and we have sin 16 25' 4"
= 9.451234. Again, as sin 16 25' 4" - sin (180 - 16 25' 4") = sin 163 34' 50
the latter angle has for its log sin 9.451234. Finally, either of these angles in.
creased by any multiple of 2?r has the same logarithmic sine.
12. Show from the table that the angle whose log cos is 9.778151,
is 53 C r 49", and also 306 52' 11", and each of these angles increased
by any multiple of 2*.
13. What angles correspond to the logarithmic cosines 9.246831
14. Find from the table what angles have for their logarithmic
tangents 9.895760, 10.531054, and 11.216313.
Results. The first two are the log tans of 38 11' 20", and 73
35' 43", and also of 180 + either of these angles, and each increased
by any multiple of 2*.
15. Find the angles corresponding to the logarithmic cotangents
10.008688, 9.638336, and 9.436811.
Results. The first two are the log cots of 44 25' 37", and 66 29'
54", and also of 180 + either of these angles, and each increased by
any multiple of 2*.
72. SCH. Strictly speaking, negative numoers have no logarithms; since
no base can be assumed, such that all negative numbers can be represented by
said base affected with exponents. It is therefore customary to say that nega-
tive numbers have no logarithms. Nevertheless, we do apply logarithm* to nega-
tive trigonometrical functions. Thus, if we have cos x, the sign is inter-
preted as simply telling hi what quadrants x may end ; while, in other respects
the function is treated exactly like + cos x.
16. Given log (- cosz) = 9.346251, to find x.
SOLUTION. The logarithmic cosine 9.346261, considered independently of its
sign, corresponds to 77 10' 35". But the sign requires that the arc shall end
in the 2d or 3d quadrant, for such angles, and such only, have negative cosines.
. The angles required are 180 =F 77 10' 35" = 102 49' 25", and 257 10' 35",
and these increased by entire circumferences, as all these angles have Joga-
rithmic cosines, which are numerically equal to 9.346261, and the cosines them-
selves are negative.
17. What angle less than 180 has a negative cosine whose tabu
lar logarithmic value is 9.653825 ? Ans. 116 47' 4"
CONSTRUCTION AND tJSE OF TRIGONOMETRICAL TABLES. 39
J8. What angle lefej than 180 has a negative tangent whose tabu
m- logarithmic value is 9.884130? Ans. 142 33' 15"
19. What angle less than 180 has a negative sine whose tabulai
logarithmic value is 9.341627 ?
20. What angle less than 180 has a negative cotangent whose
tabular logarithmic value is 9.564299 ? Ans. 110 8' 15".
21. Find values of x < 180 which fulfil the following conditions :
log (- cos x] = 9.562468 ; log (- tan x) = 10.764215 ;
log (- sin x) = 8.886432; log (- cot x) = 11.152161.
Results. 111 25'; 99 45' 54"; none; 175 58' 14".
22. Having at hand only the common logarithmic tables of trig-
onometrical functions, and the table of logarithms of numbers, I
wish to find the number of degrees, minutes, and seconds corre-
sponding to the natural tangent 2.16145. How is it done, and what
is the result?
Answer: Find the logarithm of 2.16145, to this add 10, and find
the angle corresponding to this tabular logarithmic tangent. The
angle is 65 10' 20".
23. From the same tables as above find the natural cosine of
35 23'. Also what angle corresponds to natural tangent 2.
24. From the same tables as above find the angle corresponding to
natural tangent 1.82645. Also to natural cosine .42536.
25. Why is it in the table of logarithmic functions that the sine
of an angle minus its cosine 4- 10 gives the tangent ? Why that cosine
the sine + 10 gives the cotangent ? Why that the sum of the tan-
gent and cotangent of any angle = 20 ? Why is but one column of
tabular differences needed for tangents and cotangents, while the
sines and cosines require each a separate column ?
FUNCTIONS OF ANGLES NEAR THE LIMITS OF THE QUADRANT.
[NOTE. This subject may be omitted in an elementary course, the first time
going over, if thought best.]
73. Failure of Table IT. The method which has been given
in the preceding pages for finding the logarithmic functions of angles
involving seconds, by means of the Tabular Differences, Table II., is
sufficiently accurate in most cases for practical purposes, but is
40 PLANE TRIGONOMETRY.
entirely too rude for the sines, tangents, and cotangents of angles
near the beginning of the quadrant (those less than 2 or 3), and
for cosines, tangents, and cotangents of angles near the close of the
quadrant (those between 87 or 88 and 90). An example will
render this clear. Suppose we wish to find log sin 1' 12". We find
from Table II., log sin 1' = 6.463726 ; and also that the average
increase of the log sin between 1' and 2' is 5017.17 (milliontLs) for
every second increase of the angle. Bat this average rate of increase
f the function during the minute is much less than its real rate of
increase in the first part of the minute, as from 1' to 1'12", and much
greater than the real rate of increase in the latter part, as the angle
approaches 2'. In fact, we see from this table, that we should use
2934.85 as the increase of log sin 2' for 1" increase of the arc. Now,
in our proposed example, we want the increase of the log sin while
the angle is passing from 1' to 1' 12". This, as shown above, is con-
siderably more than 5017.17 (millionths) for every second.
The cosine being the sine of the complement is subject to the same
law of change near the close of the quadrant, that governs the sine
at the beginning.
The case of the tangent of a small angle is similar to that of the
sine ; and since the cotangent is the reciprocal of the tangent, it
has the same law of change, only that the one increases as the other
decreases. Thus, since doubling a small arc, as 1", doubles its tan-
gent (approximately), it divides its cotangent by 2.
Finally, while the law of change in the sine is very different neai
the close of the quadrant from what it is near the beginning, the
sine changing very rapidly at the beginning and very slowly at the
close, and the cosine is just the opposite, the tangent, and cotangent
have the same law of change at both extremities of the quadrant.
Thus, if near the beginning of the quadrant a certain small increase
of the arc increases the tangent at a particular rate, it decreases the
cotangent at the same rate, since these functions are reciprocals of
each other. Moreover, since tan 1 = cot (90 1) = cot 89,
cot 89 changes according to the same law as tan 1 ; and tan 89
changes reciprocally with cot 89.
74. Description of Table III. The first page of the table
enables us to find the sines of angles less than 2 36' 15" (and con-
sequently the cosines of angles between 87 23' 45" and 90) with
very great accuracy. The columns headed Angles contain the degrees,
minutes, and seconds of the proposed angles, and the columns at
their right give the same angles in seconds. The columns headed
CONSTRUCTION AND USE OF TRIGONOMETRICAL TABLES. 41
Diff. contain the corrections to be used according to the> following
problems. The second and third pages answer a similar purpose
with reference to tangents and cotangents of arcs within 2 36' 20"
of the limits of the quadrant.
7. JProp. Letting x represent any number of seconds less than
2 36' 15", ice have,
log sin x' 1 = 4.685575 + logz - Diff.
DEM. The length of 1" of an arc to radius unity is 3.14159265358979 (the
length of the semicircuinferenee) -*- 648000 (the number of seconds in 180), and
= .00000484812. For practical purposes this fraction may also be taken as the
sine of 1". Though, actually, the sine is less than the arc, the expressions for
arc 1" and sine 1" agree to as many places of decimals as we have here. Again,
for these small arcs the sine increases at nearly the same rate as the arc, so that
ain3" = .00000484812 x 3 nearly; sin 102" = .00000484812 x 102 nearly; these
results being slightly in excess of the true values. It is the correction for this
excess that is furnished by Table III. in the columns marked Diff. But this
table is adapted to logarithmic computation ; hence we have log sin x" = log
sin 1" + logo; Diff. In this expression log sin 2" is the logarithm of the natu-
ral sine of x" (not increased by 10,- as each function in Table II. is) ; log sin 1"
+ log#, that is, the logarithmic sine of 1" plus the logarithm o f the number of
seconds, corresponds to multiplying the sin 1" by the number of seconds, and
gives the If garithm of the product, or strictly, the logarithm of the length of the
arc of a?". Now, the sine of x" being less than the arc, its logarithm is less than
the logarithm of the length of the arc. Just how much less the table tells. This
difference, therefore, between the logarithm of the arc and the logarithm of its
Bine, which is given in the table, is to be subtracted. Finally, to make this
result agree with Table II. we must add 10 to the result. Now, log sin 1" =
log .00000484812 = (T.685575, and adding 10, we have 4.685575.
.-. log sin a" = 4.685575 + logz - Diff.,
a result which agrees with the logarithmic functions in Table II. Q. E. D.
76. COR. I. To obtain the log cos of an angle between 87 23' 45'
ind 90, /row this table, take the log sin of its complement.
77. COR. 2. To obtain the log tan of an angle less than 2 36' 20",
from this table, use the formula,
log tan x" = 4.685575 + logx + Diff.
DEM. For as small an arc as 1", sine, arc, and tangent are practically
equal; hence, log tan 1" log sin 1'' = 4.685575 (10 being added). Moreover,
for these small arcs the tangents increase (like the sines) in nearly the same
ratio as the arcs; hence, we add log a?. Finally, the tangent is a little in excess
>f the arc, which excess is given in the table, and is to be added. Q. E. D.
12 PLANE TRIGONOMETRY.
78. COR. 3. To obtain the log cot of an angle less than 2 36 15' ,
from this table, use the formula,
log cot a;" = 15.314425 logo; Diff.
DEMONSTRATION. Since cot x" = } S cot " = log 1 - log tan &
s. 20 (4.685575 + log x + Diff.) = 15.314425 log x Diff. The 20 arises
from adding 10 twice to log 1 (= 0). One 10 is added because 4.G85575 is 10 in
excess of the true log tan 1" ; and the other 10 is added in order to make the
log cot x" agree with the ordinary tabulated value, as in Table II. Q. E. D.
7,9. COR. 4. To obtain the log tan of an angle between 87 23' 45"
and 90, take the log cot of its complement ; and to obtain the log cot
of an angle between the same values, take the log tan of its complement.
80. Prob. Having given a log sin less than 8.657397 (the log
*in 2 36' 15"), to find the corresponding angle.
SOLUTION. From log sin x" = 4.685575 + log x Diff., we have, log x =
log sin x" 4.685575 + Diff. Hence, if from the given log sin, we subtract
4.685575, and then add the proper correction as furnished by Table III., we have
the logarithm of the number of seconds sought. But we cannot tell what Diff.
to take till we know the number of seconds. To meet this difficulty, find the
angle corresponding to the given log sin from Table II., and reduce it to seconds.
This will be sufficiently accurate to furnish the required Diff.
81. COR. 1. Having given a log cos less than 8.65?o97 (the log cos
o/87 23' 45"), to find the corresponding angle, treat it as if it were a
log sin, and having found the corresponding angle, take its comple-
82. COR. 2. For log tan and log cot, the formula in (77,
log a; = log tan x" 4.685575 - Diff.,
and, log x = 15.314425 log cot z" - Diff.
These are applied as in (80) ; that is, the Diff. to be sub tract ad is
found by getting from Table II. the required angle in seconds, as
near as may be, and then take from Table III. the corresponding
CONSTRUCTION AND USE OF TRIGONOMETRICAL TABLES. #
1. Find the log sin, tan, and cot of 1 11' 15".
SOLUTION. Log sin 1 11' 15" = 4.685575 + log 4275 - .000031 = 8.316480.
l e 11' 15" = 4275". Since 4275" is between 4230" and 4300, the Diff. is 81
Log tan 1 11' 15" = 4.685575 + log 4275 + .000062= 8.316573.
Log cot 1 11' 15" = 15.314425 - log 4275 - .000062 = 11.683427.
Or, log cot can be found by subtracting log tan from 20.
2. Verify the following by using Table III. :
log sin 56' 26" = 8.215242 ; log tan 56' 26" 8.215301 ;
log cot 56' 26" =11.784699.
3. Verify the following by using Table III. :
log cos 88 17' 44" = 8.473396 ; log tan 88 17' 44" = 11.526412 ;