TRIGONOMETRICAL ANALYSIS. 1. By Art. 211, R sin (a+2a)=sin a cos 2a+sin 2a cos a 2. 3. 4. R cos (a+2a)=cos a cos 2a-sin a sin 2a R sin 3a sin a cos 2a+sin 2a cos a R cos 3a=cos a cos 2a-sin a sin 2a 5. By Art. 94, cos2 3a+sin2 3a=R2 6. Transp. in the 4th, cos a cos 2a-sin a sin 2a=R cos 3a 7. Subtracting 6th from 5th, 2sin2 3a=R2-R cos 3a 8. Adding the 5th and 6th, 2cos2 3a=R2+R cos 3a 9. Dividing the 7th by 2, sin2 3a=R2-R cos 3a 10. Dividing the 8th by 2, cos2 3a=R2+R cos 3a 11. Extr. sqr. root in the 9th, sin 3a=√†R2—FR cos 3a 12. Extr. sqr. root in the 10th, cos 3a=√1⁄2R2 +†R cos 3a. We might proceed in the same manner to find expressions for the sines and cosines of other multiple arcs, substituting successively 4a, 5a, 6a, &c. in the equations in Art. 208, as we have 3a. 2. 66 1. By Art. 215, sin (a+b)=sin a cos b+sin b cos a sin (a-6)=sin a cos b-sin b cos a 3. Multiplying together the 1st and 2d, (Alg. 235,) sin (a+ b)sin (a-b)=sin2 a cos2 b-sin2b cos2 a 4. By Art. 94, Rasin2 a+cos2b 5. Considering radius equal to 1, 1-sin2b=cos2b, and 1sina a cos2 a 6. Substituting in the 3d, sin (a+b) sin (a—b)=sin2 a×(1 -sin b)-(sin b×1-sin2a) 7. Expanding the 6th, sin (a+b) sin (a—b)=sin3a—sin3 a sin3b-sin3b+sin2a sin2 b' 8. Cancelling terms, sin (a+b)× sin (a-b)=sin3a-sin3b 9. By Alg. 238, sin2 a-sin3 b=(sin a+sin b)x(sin a sin b) 10. Making the 8th and 9th equal, sin (a+b) sin (a—b)= (sin a+sin b) × (sin a-sin b.) It is the object of the following series of equations, to find different expressions for the tangents, cotangents, &c. of arcs. The principal results which are required to be found, are given in the 7th, 12th, 20th, 27th, 31st, 36th, 38th, 40th, 45th, 48th, 53d, 56th, and 59th equations, which, for facility of finding them, the numbering is designated by a star. Expressions for the tangents, &c. of arcs. Art. 216. 1. By Art. 93, R:tan::cos: sin 2. Multiplying extremes and means, R tan=cos tan 3. By division, R÷tan-cos÷sin, and COS R -; also tan sin tan R 4. By the given equation, (Art. 216,), tan (a+b)= R sin (a+b) 6. Cancelling R from the numerator and denominator of R (sin a cos b+sin b cos a) +sin b cos a÷ R tan a+b=(sin a cos b cos a cos b-sin b sin a R 7.* Uniting terms, (Alg. 162,) tan (a+b)= R (sin a cos b+sin b cos a) cos a cos b-sin b sin a 8. Dividing by cos a cos b, the sin a cos b, sin b cos a' numerator becomes, sin b +: sin a cos a cos b cos a cos b cos a and the denominator becomes, COS cos a cos b × R; tan a tan b merator, R R 10. By substituting these expressions, we have for the nu the donominator, 1 =tan a+tan b; and, for R Hence, 11. The equation becomes, tan (a+b)=. tan a+tan b tan a tan b 1 R2 12.* Multiplying both numerator and denominator of the second member of the equation by R2, tan (a+b)= Rax(tan a+tan b) R2-tan a tan b 13. By Art. 216, tan (a+b)=R×sin (a−b) . cos (a-b) substituting for sin (a-b), and cos (a-b), as given in Art, 208, we have tan (a—b)=(sin a cos b—sin cos a cos b+sin a sin by 16. Dividing the denominator by cos a cos b, it becomes, (s+d) cos 21. By Art. 219, sin 66 sin (s—d)=R(sin s+sin d) (s-d) cos (s+d)=R(sin s-sin d) by the 22d, and rejecting sin (s+d) cos (s-d)_sin s+sin d 23. Dividing the 21st = sin (s-d) cos (s+d) sin s-sin d R2, 24. By Alg. 155, the first member of the equation in the 23d= 'cos (s-d) (sin (s+d)) x 25. By Art. 216, tan= RX sin ; that is, tan (s+d)= COS Rxsing(s+d); therefore, cos (s+d) tan (s+d) ___sin 1(s+d) R cos (s+d) ; that is, tan (s—d)= tan (s—d) __sin (s—d) R cos (s-d) tan (s-d) sin (s-d). Ꭱ cos (s-d) (s—d)=R×sin (s-d); therefore, cos (s-d) R tan (s-d) sin (s-d) 27.* Substituting the expressions as obtained above, sin (s+d) cos(s-d) __tan (s+d) x cos (s+d) sin (s-d) R tan (s-d) tan 1 (s+d). X. = the 1st R tan (s-d) tan (s-d) member of the equation; and, the equation is 30. Cot (a+b)= ( 1 tan (a-b) 29. Converting the equation into a proportion, sin a+sin b sin a-sin b::tan (a+b): tan R2 R2 (a+b)=(R2—. X. (a—b) -cot a cot cot a cot b ) = 31.* Multiplying by cot a cot b, cot (a+b)= R2 (tan a tan b) tan a-tan b 34. By Art. 93, tan: R::R: cot; therefare, tan = 35. Substituting the values, as obtained above, for tan a and 36.* Multiplying by cot a cot b, cot (a—b)= R2 cot a cot b+R__cot a cot b+R2 R2 cot b-R2 cot a cot b-cot a We have obtained in the 12th and 20th equations express |