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Encyclopedia Britannica - Main :: CLI-COM |
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COE is 2(CD). The sum of the pro-D jections of OD and DE on OA is equal to that of OE, and the sum of the projections of OC and CE is equal to that of OE; hence the sum of the pro- jections of OC and OD is twice that of OR, or cos C+cos D=2 cos I(C+D) cos 2(CD). The difference of the O A projections of OD and OC an OA FIG. . is equal to twice that of ED, hence we have the formula
As another example of the use of projections, we will find the sum of the series cos a+cos (a+fl)+cos (a+29)+ . +cos (a+nIS).Sum of Suppose an unclosed polygon each angle of whichSeries of is >rp to be inscribed in a circle, and let A, A1, A2,Seri s o in A3, , A. be n+i consecutive angular points; Arthmet/cal let D be the diameter of the circle; and suppose a Progression. straight line drawn
a+$, a+2/3, . . . are the angles it makes with Al A3, A2, A3, . . .; we have by projections AA cos (a+2nI,e) =AA' {cos a+ cos a+R+..+Cosa+(nI /31, also AA1=D sin z,5', AA=D sin In/3; hence the sum of the series of cosines is cos (a+%nI 13) sin 20 cosec Z,8. By a double
sin (A1+A2+A3) =sin Al cos As cos A3 +cos Al sin As cos A3+cos Al cos As sin A3 sin Al sin As sin A1; cos (A1+A2+A3) =cos Al cos As cos A3 cos Al sin As sin A3sin Al cos As sin A3 sin Al sin As cos A3. We can by induction extend these formulae to the case of n angles. Assume sin (A1+A2+ ... +An)=S1S3+S3 . . cos (A1+A2+ ... +A) =SoS2+Ss .. . where Sr denotes the sum of the products of the sines of r of the angles and the cosines of the remaining nr angles; then we have sin (A1+A2+ . . . +A+An+1)=cos A+1(S1S3+S5 ...) +sin A,,+1(SoS2+Ss ...). The right-hand side of this equation may be written (S1cosAsia +So sinA+1)(53ensA+1 +S2 sin A,,+1)+ - . or S'1S'3+-. where S', denotes the quantity which cqrresponds for n+i angles to Sr for n angles; similarly we may proceed with the cosine formula
Formulae cos 2A =COs"- A sin2A =2 cos' A I =12 sin' A, for Multiple 2 tan A and Sub- sin 2A =2 sin A cos A, tan 2A = I tang A Multiple Angles. sin 3A =3 sin A -4 sin3 A, cos 3A = 4 cos3 A -3 cosA, sin nA = n cos" A sin A n(n I I(n -2)ensn`3 A sin3 A+ .. . 3 +(I)rtt(nI) . . (n2r) (2r + )( cos" A sine'+1 A cos nA=cos^An(nI) cos"'g A sin' A 2! + . . } ()rn(nI) ...n2r+i) cos""r A sin'* A+ .... 2r! may all be deduced from the addition formulae by making the angles all equal. From the last two formulae we obtain by division n tan A - n(n - 1 '(n - 2) tan, A +... +( - 1)rn(n - 1) ... (n - 2r) tan''-r+1 A+.. . tan nA 3' (2r+1)! In the particular case of n=3 we have tan 3A = -3 tan 3 tan A ant A A The values of sin IA, cos IA, tan IA are given in terms of cos A by the formulae sin a A=(I)n(1 2s A) I cos 2A =(1)Q(I 2s A) l f tan 2 A=( i) (i I cos cos A 1 \ A) A ' where p is the integral part of A/27r, q the integral part of A/2r+1, and r the integral part of A/7r. Sin IA, cos IA are given'{in terms of sin A by the formulae 2 sin ZA=(I)P'(I+sin A)F+(I)4'(Isin A)+, 2 cos IA =(I)i1'(I+sin A)Ir(sin A)t, where p' is the integral part of A/27r+4 and q' the integral part of A/ear 6. In any plane triangle ABC we will denote the lengths of the sides BC, CA, AB by a, b, c respectively, and the angles BAC, ABC, A CB by A, B, C respectively. The fact that the projec- tions Properties of b and c on a straight line perpendicular to the of Tpreriangles. side a. are equal to one another is expressed by the equa- tion b sin C=c sin B; this equation and the one obtained by projecting c and a on a straight line perpendicular to a may be written a/sin A=b/sin B=c(sin C. The equation a=b cos C+c cos B expresses the fact that the side a is equal to the sum of the projections of the sides b and c on itself ; thus we obtain the equations a = b cos C+c cos B b =c cos A+a cos C c=acosB+b cosA If we multiply the first of these equations by a, the second by b, and the third by c, and add the resulting equations, we obtain the formula b'+c2a2=2bc cos A or cos A=(b'-l-c2a')/2bc, which gives the cosine of an angle in terms of the sides. From this expression for cos A the formulae (s-b)(sc) l cos IA s(s-a) 1 I sin 2A = be be tan IA = (s s(s)(a) c) j sin A=be{s(sa)(sb)(sc)}, where s denotes z(a+b+c), can be deduced by means of the dimidiary formula. From any general relation between the sides and angles of a triangle other relations may be deduced by various methods of transformation, of which we give two examples. a. In any general relation between the sines and cosines of the angles A, B, C of a triangle we may substitute pA+gB+rC, rA+pB+qC, qA+rB+pC for A, B, C respectively, where p, q, r are any quantities such that p+q+r+1 is a positive or negative multiple of 6, provided that we change the signs of all the sines. Suppose p+q+r+i =6n, then the sum of the three angles 2n7r (pA +qB +rC), 2n7r (rA +pB +qC), 2nir (gArB +pC) is 7r ;and, since the given relation follows from the condition A+B+C =7r, we may substitute for A, B, C respectively any angles of which the sum is ar; thus the transformation is admissible.0. It may easily be shown that the sides and angles of the triangle formed by joining the feet of the perpendiculars from the angular points A, B, C on the opposite sides of the triangle ABC are respectively a cos A, b cos B, c cos C, 7r2A, 7r2B, 7r2C; we may therefore substitute these expressions for a, b, c, A, B, C respectively in any general formula. By drawing the perpendiculars of this second triangle and joining their feet as before, we obtain a triangle of which the sides area cos A cos 2A, b cos B cos 2B, c cos C cos 2C and the angles are 4A7r, 4B7r, 4C7r; we may therefore substitute these expressions for the sides and angles of the original
cos 4A =a' cos' A cos' 2A b' cos'- B cos' 2Bc2 cos' C cos' 2C 2bc cos B cos C cos 2B cos 2C This transformation obviously admits of further exten- sion. Solution of (I) The three sides of a triangle ABC being given, Triangles. the angles can be determined by the formula L tan zA=1o+2 log (sb) +z log (s-c)2 log s; log (sa) and two corresponding formulae for the other angles. Formulae for Sine and Cosine of Sum of Angles. 1 n(n2 1) tan, A+...+(- 1)rn(, 1)* "((n 2r-1 1) tang A+... Spherical Trigonometry. 7. We shall throughout assume such elementary propositions in spherical geometry as are required for the purpose of the investigation of formulae given below. A spherical triangle is the portion of the surface of a sphere bounded by three arcs of great
AB denote these arcs, the circular measure of the Definition angles subtended by these arcs respectively at the of Spherical centre of the sphere are the sides a, b, c of the spherical Triangle. triangle ABC; and, if the portions of planes passing through these arcs and the centre of the sphere be drawn
great
which the sides are arcs of the same three great circles. If we consider one of these triangles ABC as the fundamental one, then one of the others is equal in all respects to ABC, and the remaining six have each one side equal to, or common with, a side of the triangle ABC, the opposite angle equal to the corresponding angle of ABC, and the other sides and angles supplementary to the corresponding sides and angles of ABC. These triangles may be called the associated triangles of the fundamental one .4BC. It follows that from any general formula containing the sides and angles of a spherical triangle we may obtain other formulae by replacing two sides and the two angles opposite to them 1,y their supplements, the remaining side and the remaining angle being unaltered, for such formulae are obtained by applying the gig en formulae to the associated triangles. If .4', B', C' are those poles of the arcs BC, CA, AB respectively a hich lie upon the same sides of them as the opposite angles A, B, C, then the triangle A'B'C' is called the A A polar triangle of the triangle ABC. The sides of the polar triangle are ,rA, ,rB, ,rC, and the angles ,ra, ,rb, ,rc. Hence from any general formula connecting the sides and angles of a spherical triangle we may obtain another formula by changing each side into the supplement of the opposite Bangle and each angle into the supplement of the opposite side. 8. Let 0 be the centre of the sphere on which is the spherical triangle ABC. Draw AL perpendicular to OC and AM perpendicular to the plane End of Article: COE If you wish, you can link directly to this article.
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