Transcending Quantities Arising From The Circle

126. After logarithms and exponential quantities have been considered, circular arcs and the sines and cosines of these must be considered; not only because they constitute another kind of transcending quantity, but also because of the logarithms and exponentials of these that arise when they are involved with imaginary quantities, which will become clearer below.

Therefore we may put the radius of the circle or the whole sine to be and it is clear enough that the periphery of this circle cannot be expressed exactly in rational numbers; but by approximations the semi–circumference of this circle has been found to be

for which number for the sake of brevity I may write as

thus so that there shall be   semi–circumference of the circle, whose radius  1, or  will be the length of the arc of 180 degrees.

127. With z denoting the arc of some circle, the radius of which I assume always , the sines and cosines of this arc z mainly are considered. But the sine of the arc z in the following I will indicate in this manner  1

sin A. .z or only sin.z , truly the cosine in this manner cos A. .z or only cos. z . Thus, since  n shall be the 1800 arc, there will be sin 0 0 cos 0 1 . ,.     and

1 1 2 2 3 3 2 2 sin. 1 cos. 0 sin 0 cos. 1 sin. 1 cos. 0 sin.2 0 cos.2 1       Therefore all the sines and cosines will be contained within the limits 1 and 1  . But again there will be     1 1 2 2 cos sin and sin cos .z . z .z . z      and    2 2 sin cos 1 .z .z   . Besides these denominations, these also are well-known : tang.z , which denotes the tangent of the arc z, cot.z the cotangent of the arc z, and agreed to be sin cos tang .z .z .z  and cos 1 sin tang cot .z .z .z .z   , which all are well-known from trigonometry. 128. Hence truly also it is agreed, if the two arcs y and z may be had, to become sin sin cos cos sin . y z .y .z .y .z     and cos cos cos sin sin . y z .y .z .y .z     and likewise sin sin cos cos sin . y z .y .z .y .z     and cos cos cos sin sin . y z .y .z .y .z     Hence in place of y by substituting the arcs 1 3 2 , , 2    etc. there will be

  1 2 sin. z    cos.z   1 2 cos. z    sin.z   1 2 sin. z    cos.z   1 2 cos. z    sin.z sin sin . z      .z cos cos . z      .z sin sin .z .     z cos cos .z .     z   3 2 sin. z    cos.z   3 2 cos. z    sin.z   3 2 sin. z    cos.z   3 2 cos. z    sin.z sin 2 sin . z      .z cos 2 cos . z      .z sin 2 sin . z      .z cos 2 cos . z      .z Therefore if n may denote some whole number, there will be   4 1 2 sin cos n . z  .z      4 1 2 cos sin n . z  .z      4 1 2 sin cos n . z     .z   4 1 2 cos sin n . z     .z   4 2 2 sin sin n . z  .z      4 2 2 cos cos n . z  .z      4 2 2 sin sin n . z     .z   4 2 2 cos cos n . z     .z   4 3 2 sin cos n . z  .z      4 3 2 cos sin n . z  .z      4 3 2 sin cos n . z     .z   4 3 2 cos sin n . z     .z   4 4 2 sin sin n . z  .z      4 4 2 cos cos n . z  .z      4 4 2 sin sin n . z     .z   4 4 2 cos cos n . z     .z Which formulas are true, whether n shall be a positive or negative integer. 129. Let sin and cos ; .z p .z q   there will be pp qq   1; and let sin cos .y m, .y n   , so that also there shall be mm nn   1; the sine and cosine of the arcs of these thus may themselves be considered:

sin.z p  cos.z q  sin. y z mq np     cos. y z nq mp     sin 2 2 . y z mnq nn mm. p       cos 2 . y z nn mm. q mnp        2        23 3 2 sin 3 3 . y z mn m q n m n p     3    3 2 23 cos 3 . y z n m n q mn m p     3 3 etc. etc. These arcs z, y z, y z, y z, .  2 3   etc proceed in an arithmetic progression, truly both the sine as well as the cosine constitute a recurring progression, such as arises from the denominator 1 2   nx mm nn xx   ; for there is sin 2 2 sin . y z n . y z mm nn sin.z          or sin 2 2cos sin sin . y z .y . y z .z       and in a like manner cos 2 2cos cos cos . y z .y . y z .z       . In the same manner again there will be sin 3 2 cos sin 2 sin . y z .y . y z . y z          and cos 3 2cos cos 2 cos . y z .y . y z . y z          and thus sin 4 2cos sin 3 sin 2 . y z .y . y z . y z          and cos 4 2cos cos 3 cos 2 . y z .y . y z . y z          etc. The laws of which, with the benefit of the arcs proceeding in an arithmetical progression, both of the sine as well as of the cosine, can be formed in an expedient manner, as far as it pleases.

130. Since there shall be sin sin cos cos sin . y z .y .z .y .z     and sin sin cos cos sin . y z .y .z .y .z     , from these expressions either added or subtracted there will be       sin sin 2 sin sin 2 sin cos , cos sin .y z .y z .y z .y z .y .z .y .z .     Because again there will be cos cos cos sin sin . y z .y .z .y .z     and cos cos cos sin sin . y z .y .z .y .z     , in a equal manner,       cos cos 2 cos cos 2 cos cos sin sin . .y z .y z .y z .y z .y .z , .y .z       Let 1 2 y  z v ; from these last formulas there will be     2 1 1 1 cos 1 cos 2 2 22 2 1 1 1 cos 1 cos 2 2 22 cos and cos sin and sin .v .v .v .v . v .v , . v .v ,         from which from a given cosine the sine and cosine of each half angle are found.
131. The arc may be put in place y  z a, y z b   and  ; there will be 2 2 et ab ab y z ,     with which substituted into the above equations, these equations will be obtained, as just as many theorems :

2 2 2 2 2 2 2 2 sin sin 2sin cos sin sin 2 cos sin cos cos 2cos cos cos cos 2sin sin ab ab ab ab ab ab ab ab .a .b . . , .a .b . . , .a .b . . , .a .b . . .      From these again with the help of division these theorems arise : 2 2 tang sin sin sin sin 2 2 tang. sin sin cos cos 2 sin sin cos cos 2 sin sin cos cos 2 sin sin cos cos 2 cos cos tang cot. tang cot tang cot a b a b · .a .b a b a b .a .b .a .b a b .a .b .a .b a b .a .b .a .b a b .a .b .a .b a b .a .b .a . , . , . ,

 cos cos 2 2 cot cot .b a b a b .b .a . ..     From these finally these theorems are deduced :     2 sin sin cos cos sin sin cos cos 2 2 sin sin cos cos sin sin cos cos 2 cot. tang. .a .b .a .b a b .a .b .b .a .a .b .b .a a b .a .b .a .b , .     132. Since there shall be    2 2 sin cos 1 .z .z   , with the factors taken there will be   cos 1 sin cos 1 sin 1 .z .z .z .z        , which factors, even if imaginary, still perform a huge task in the combinations and in the multiplications of arcs. For the product of these factors may be sought :   cos 1 sin cos 1 sin . z .z . y .y       and there will be found : cos cos sin sin 1 cos sin sin cos .y .z .y .z .y .z .y .z      . EULER’S INTRODUCTIO IN ANALYSIN INFINITORUM VOL. 1 Chapter 8. Translated and annotated by Ian Bruce. page 208 But since there shall be cos cos sin sin cos .y .z .y .z . y z      and cos sin sin cos sin .y .z .y .z . y z     , there will be this product   cos l sin cos l sin cos .y .y .z .z . y z . y z              1 sin   and in a similar manner   cos l sin cos l sin cos .y .y .z .z . y z . y z              1 sin  , likewise   cos l sin cos l sin cos l sin cos .x .x .y .y .z .z . x y z . x y z .                   1 sin   133. Hence therefore it follows :     2 3 cos 1 sin cos 2 1 sin 2 cos 1 sin cos 3 1 sin 3 .z .z . z . z .z .z . z . z           and thus generally there will be   1 sin cos 1 sin n cos.z .z .nz .nz      . From which on account of the ambiguity of the signs there will be  1 sin   1 sin  2 cos n n cos.z .z cos.z .z .nz       and  1 sin   1 sin  2 1 sin n n cos.z .z cos.z .z .nz .        Therefore with these binomials developed in series, there will be        1 1 2 2 2 3 4 4 1 2 1234 12345 6 6 123456 cos cos cos sin cos sin cos sin + etc n n n n nn n n n nn n n n n n .nz .z .z .z .z . .z .z .          and                1 3 1 2 3 1234 1 1 2 3 12 3 4 5 sin cos sin cos sin cos sin etc n n nn n nn n n n n .nz .z .z .z .z .z .z .               n5 5 134. Let the arc z be infinitely small ; there will be sin.z z .z  and cos 1  ; moreover n shall be an infinitely great number, so that the arc nz shall be of finite magnitude, for example nz v  ; on account of sin v n .z z   there becomes 24 6 12 1234 123456 cos 1 + etc vv v .v .       and 35 7 123 12345 1234567 sin + etc vv v .v v .       Therefore from the given arc v with the help of which series the sine and the cosine will be able to be found ; so that the use of which formulas may become more apparent, we may put the arc v to be to the quadrant or 90° as m to n or for there to be 2 m n v    . Because now the value of  is agreed upon, if this is substituted everywhere, the equation will be produced EULER’S INTRODUCTIO IN ANALYSIN INFINITORUM VOL. 1 Chapter 8. Translated and annotated by Ian Bruce. page 210 3 3 5 5 7 7 9 9 11 11 0 sin A 90 1 57079 63267 94896 6192313216 916 0 64596 40975 06246 25365 57565 639 0 07969 26262 46167 04512 05055 495 0 0046817541 35318 68810 06854 639 0 00016 0441184787 3598218726 609 m n m n m n m n m n m n m n . . , , , , ,             13 13 15 15 17 17 19 19 21 21 0 00000 35988 43235 21208 53404 585 0 00000 00569 21729 21967 92681178 0 00000 00006 68803 5109811467 232 0 00000 00000 06066 9357311061957 0 00000 00000 00043 77065 46731374 0 00000 00 m n m n m n m n m n , , , , , ,            23 23 25 25 27 27 29 29 000 00000 25714 22892 860 0 00000 00000 00000 00125 38995 405 0 00000 00000 00000 00000 51564 552 0 00000 00000 00000 00000 00181 240 0 00000 00000 00000 00000 00000 551 m n m n m n m n , , , ,         [The last three places shown here in each power have been corrected from an error in the original work.] and

2 2 4 4 6 6 8 8 10 10 0 cos A 90 1 00000 00000 00000 00000 00000 000 1 23370 0550136169 82735 43113 750 0 25366 95079 01048 01363 65633 664 0 02086 34807 63352 96087 30516 372 0 0009192602 74839 42658 02417162 m n m n m n m n m n m n . . , , , , ,            12 12 14 14 16 16 18 18 20 20 0 00002 52020 42373 06060 54810 530 0 00000 04710 87477 8818171503 670 0 00000 00063 86603 0837918522 411 0 00000 00000 65659 63114 97947 236 0 00000 00000 00529 44002 00734 624 0 00000 00 m n m n m n m n m n     22 22 24 24 26 26 28 28 30 30 000 00003 43773 91790 986 0 00000 00000 00000 01835 99165 216 0 00000 00000 00000 00008 20675 330 0 00000 00000 00000 00000 03115 285 0 00000 00000 00000 00000 00010168 0 00000 00000 00000 00 m n m n m n m n m n , , , , ,           000 00000 029. [The last three places in each power are in error in the original .] Therefore since it is sufficient to know the sines and cosines of angles to 450 , the fraction m n always will be less than 1 2 and hence also from the powers of the fraction m n the series shown will be especially convergent, thus so that only some number of figures more may suffice, particularly if the sine and cosine may not require too many figures. 135. With the sines and cosines found the tangents and cotangents indeed are able to be found by the customary analogies ; but because multiplication and division is such an inconvenience in the generation of numbers of this kind, it is convenient to express these in a particular way. Therefore there will be

35 7 123 12345 123 7 24 6 12 1234 123 6 etc sin cos 1 e tang vv v vv v v . .v .v . .v                         tc and 24 6 12 1234 123 6 35 7 123 12345 123 7 1 e cos sin etc cot vv v vv v . .v .v v . .v .                         tc Now if the arc shall be 0 90 m n v  , in the same manner as before there will be 0 tang A 90 m n . . 0 cot A 90 m n . . 2 0 6366197723 676 mn nn mm ,    0 6366197723 676 n m   , 0 29755 67820 597 m n   , 4 4 0 31830 98861838 mn mn mm ,    3 3 0 01868 86502 773 m n   , 0 20528 88894145 m n   , 5 5 0 00184 24752 034 m n   , 3 3 0 0065510747 882 m n   , 7 7 0 0001975800 715 m n   , 5 5 0 00034 50292 554 m n   , 9 9 0 0000216977 373 m n   , 7 7 0 00002 02791061 m n   , 11 11 0 00000 24011370 m n   , 9 9 0 0000012366 527 m n   , 13 13 0 00000 02664133 m n   , 11 11 0 00000 00764 959 m n   , 15 15 0 00000 00295 865 m n   , 13 13 0 00000 00047 597 n m   , 17 17 0 00000 00032 868 m n   , 15 15 0 00000 00002 969 m n   , 19 19 0 00000 00003 652 m n ,   17 17 0 00000 00000185 m n   , 21 21 0 00000 00000 406 m n   , 19 19 0 00000 00000 012 m n   , 23 23 0 00000 00000 045 m n   , 25 25 0 00000 00000 005 m n   , of which the account of the series will be set out further below.

136. Certainly from the above it is agreed, if the sines and cosines of all the angles less than half a right angle were known, thence likewise the sines and cosines of all the greater angles are to be had. Truly if only the sines and cosines of the angles less than 30° were had, from these by addition and subtraction alone the sines and cosines of all the greater angles are possible to be found. For since there shall be 0 1 2 sin 30 .  on putting from §130 0 y  30 cos sin 30 sin 30 .z . z . z        and sin cos 30 cos 30 .z . z . z        and thus from the sines and cosines of the angles z and30  z there are found sin 30 cos sin 30 . z .z . z        and cos 30 cos 30 sin .z .z       .z , from which the sine and the cosine of the angles from 300 to 600 and hence all the greater angles are defined.

137. With tangents and cotangents a similar aid in use comes along. For since there shall be   tang tang 1 tang tang tan .a .b .a .b .a b ,     there will be 2tang cot tang l tang tang 2 tang 2 and cot 2 .a .a .a .a .a . a .a ,     from which the tangents and cotangents of the arcs less than 300 are found and the cotangents as far as to 600 . Now let ; there becomes ; therefore there will be a   30 b 2 60 2 and cot 2 tang 30 2 a b .a .      b   cot 30 tang 30     2 tang 30 2 .b .b . b     , from which also the tangents of arcs greater than 30° are obtained. But secants and cosecants are found from tangents by subtraction alone ; indeed there is EULER’S INTRODUCTIO IN ANALYSIN INFINITORUM VOL. 1 Chapter 8. Translated and annotated by Ian Bruce. page 214 1 2 cosec cot cot .z . z .z   and hence   0 1 2 sec cot 45 tang .z . z .z.   Therefore it is seen clearly enough from these, how a canon of sines will be constructed.

138. An infinitely small arc z may be put anew in the formulas §133 and n shall be an infinitely large number i, so that iz may maintain a finite value v. Therefore there will be

and from which sin and cos 1 v v i i nz v z ,   .z .z  ; with these put in place there becomes     1 1 1 1 2 cos i i v v i i .v      and     1 1 1 1 2 1 sin i i v v i i .v       But in the preceding chapter we have seen that 1 1 z i z e i          with e denoting the base of hyperbolic logarithms ; therefore for z in one part I write    v, v 1 and for the other part 1 , and there becomes 1 1 2 cos v v e e .v      and 1 1 2 1 sin v v e e .v       From which it is understood, how imaginary exponential quantities may be reduced to the sine and cosine of real arcs. Truly there will be

1 cos 1 sin v e .v      .v and 1 cos 1 sin v e .v      .v

139. Now in the same formulas § 133 n shall be an infinitely small number or 1 i n  with i being an infinitely great number ; the equations become cos cos 1 and sin sin ; z z i i .nz .    .nz . z i for with the arc vanishing z i is equal to the sine itself, truly the cosine . With these in place there is had  1     1 1 cos 1 sin cos 1 sin 2 1 i i .z .z .z .z       and     1 1 cos 1 sin cos 1 sin 2 1 i i .z .z .z .z z i .        Moreover with the hyperbolic logarithms obtained above (§125) we have shown that   1 1 1 1 1 1 or 1 i i i l xi x y l      y on putting y in place of1 x . Now therefore on putting y in place of cos 1 sin .z .z   for the one part and cos.z   1 sin.z for the other part, it will produce     1 1 1 11 1 2 1 1 i i       l cos.z sin.z l cos.z sin.z   on account of the vanishing logarithms, thus so that nothing can be concluded. Truly the other equation for the sine may be put in place     1 1 1 1 2 1 i i l cos.z sin.z l cos.z sin.z z i        and thus 1 1 21 1 cos.z sin.z cos.z sin.z z l       , from which it is apparent, in what way imaginary logarithms relate back to circular arcs. EULER’S INTRODUCTIO IN ANALYSIN INFINITORUM VOL. 1 Chapter 8. Translated and annotated by Ian Bruce. page 216
140. Since there shall be sin. cos tang z .z  .z,the arc z may be expressed by its tangent, so that there shall be 1 1 1 2 11 1 tan g.z tan g.z z l       . Truly above (§ 123) we have seen that 257 1 22 2 2 1 13 5 7 etc xxx x x x l .       Therefore on putting z   1 tang.z there becomes

tang tang tang tang 13 5 7 etc .z .z .z .z z .     Therefore if we may put , so that the arc shall be z, of which the tangent is t, that we will indicate thus , and thus there will be tang.z t  A tang . .t z..  A tang t. Therefore with the tangent t known the corresponding arc 3579 13 5 7 9 etc tt t t t z .      Since therefore, if the tangent t is equal to the radius 1, the arc becomes z  to the arc 450 or 4 z   , there will be 111 4 357 1 etc.,      , which is the series first produced by Leibnitz for the value being expressed for the periphery of the circle. [Usually called Gregory’s series, which predates the derivation of Leibnitz.] 141. But so that the length of a circular arc can be defined readily from a series of this kind, it is evident that for the tangent t a small enough fraction must be substituted. Thus with the aid of this series the length of the arc z will be found, the tangent t of which is equal to 1 10 ; for this arc becomes 11 1 10 3000 500000 z .    etc ,

the value of which series may be shown as a decimal fraction approximately without difficulty. But truly from such a known arc nothing will be able to be concluded about the length of the whole periphery, since the ratio shall not be assignable, that the arc of which the tangent is 1 10  , may hold to the whole periphery. Hence on this account towards finding the periphery of this kind an arc must be sought, which shall be at the same time some part of the periphery and the tangent small enough to be able to be expressed conveniently. Therefore according to this it is customary to take the arc of 30°, the tangent of which is 1 3  because the tangents of smaller arcs commensurable with the periphery become exceedingly irrational. Whereby on account of the arc of 6 30    there will be 2 11 1 6 3 33 3 53 3 etc.       and 2 3 23 23 23 23 1 33 53 73  etc.,     with the aid of which series, the value of  would be determined by incredible labour. 142. But here the labour with that is greater, because in the first place the individual terms shall be irrational, then truly any term is only about a third less than the preceding. A remedy to this inconvenience can arise. The arc of 450 or 4  may be taken; the value of which itself can be expressed by the scarcely converging series 111 357 1 e  tc., yet this may be retained and separated into two arcs a and b, so that there shall be 4 a b 45     . Therefore since there shall be   tang tang 1 tang tang tan 1 , .a .b .a .b .a b     the equation becomes 1 tang tang tang tang   .a .b .a .b and 1 tang 1 tang tang .a .a .b .    Now let 1 2 tang.a  ; then 1 3 tang.b  ; hence each of the arcs a and b may be expressed by rational series converging much more quickly than above, and the sum of these will give the value of the arc 4  ; and hence thus it will be EULER’S INTRODUCTIO IN ANALYSIN INFINITORUM VOL. 1 Chapter 8. Translated and annotated by Ian Bruce. page 218 3579 3579 11 1 1 1 12 32 52 72 92 11 1 1 1 13 33 53 73 93 etc 4 + etc . .                     Therefore in this manner the length of the semi–circumference  would be able to be found, as indeed it has been made with the aid of the series mentioned before.