evident that OA2+AE.AG ; OA2 :: EG : AF', and hence R2+tan ABlanBC: R:: tanAB-tan BC: tan(AB-BC). 6. The radius being expressed by unit, the sum of the tangents of the angles of any triangle is equal to the number arising from their continued product. For, let A, B, and C, denote the several angles of the triangle; and since two of these, such as A and B, are supplementary to the remaining one C, the tangent of A+B is the same (schol. def. Trig.) as that of the third angle in an opposite direction. Whence tanA+tanB tanC, and therefore tan A+ tan B = 1–tanAtanB = tan C+tan A tan B tan C, or tan A+ tan B + tan C tan A tan B tan C. = 7. The properties of the tangents are easily derived from those of the sines, 2. Change the sign of B in the last article, and tan A—tanB= sin (A-B) cos AcosB 3. Instead of A and B in art. 1. substitute their complements, 4. Make the same substitution in art. 2, and cot B-cot A = tanA+tanB cot B+cot A sinA sinB, gives tan(A+B)=-tan AtanB cotB cotA-1° 6. Change the sign of B in the last article, and tan(A-B)= tanA–tanB cot B-cot A 1+tanA tanВcotB cotA+1' 7. Divide the expression in the first article by that in the sesin(A+B)_tanA+tanB_cotB+cotA sin(A–B) tan A—tanB cot B—cotA cond, and 8. In the last article, change the sign of B, and instead of A take its complement, and cos(A+B)_cot B - tan A cos(A-B)—cot B + tan B cot A-tanB cot A+tanB = 9. Divide the expression of art. 12. NO. 3. by that of art. 5., 12. Decompose the expression in art. 9., and tanA= 1 sin2A 13. In the last article, change A into its complement, and cot A cosec2A+cot2A. 14. Subtract the last expression from the one preceding it, and tan A—cot A——2cot2A, or tan A—cotA—2cot2A. 15. In art. 9, 10, and 11, for 2A and A, take A and A, and 1-cosA sin A -cos A tan A= sinA1+cos A1+cos A 16. Multiply the expressions of art. 1. and 2., and (tanA+tauB) (tanA–tanB)=tanA*—tanB*= sin(A+B)sin(AB cos A cos B 2 17. Multiply the expressions of art. 3. and 4., and (cotB+cot A) (cot B-cot A) cot B2-cot A2= sin(A—B) sinA+B) 18. Divide art. 28. of NO. 3. by art. 29., and 2sin(A+B) cos( A—B) __tanž (A+B) 2cos (A+B) sin (A-B)-tan (A-B) sin A+ sinB 19. Divide art. 30. of the same NO. by art. 31., and cosB+cosA cos B-cos A Since by art. 14. cot A-2cot2A=tanA, if the arc A and its compound expression be continually bisected, there will arise: cot A cot A=tanА cot A-cotA=tan A cot A-cotA={tan} A Wherefore, collecting these successive terms, and observing the effects of the opposite signs, the general result will come out, 1 —cotA={tan‡Ã+žtan‡A+žtan‡A.....+ — tan A If n be supposed to become indefinitely large, then This neat and very simple investigation is given in the second French edition of Cagnoli's Trigonometry, printed at Paris in 1800, and forming the completest treatise which has yet appeared on the subject. It was also, and nearly about the same time, communicated by my friend Mr Wallace of the Royal Military College at Sandhurst, a geometer of the first order, to the Royal Society of Edinburgh; another instance of that accidental coincidence which has occurred so frequently in the history of mathematical discovery. 8. It is obvious that the terms of the series for the tangent of the multiple arc are formed out of the coefficients of the powers of a binomial. Wherefore, 9. The series for the tangent in terms of the arc, is easily derived, by the theory of functions, from the expression of the 2t tangent of the double arc. Since tan2a = 24 +213 + 25+ &c. Put t = a + Aa3 + Ba5+ &c., and, by substitution, tan2a = 2a + 8Aa3 + 32Ba5 + &c. = 2a + (2A+2) a3+ (2B + 6A+2)a5 +, &c. Equating, therefore, the corresponding terms, we obtain, 8A 2A +2, or A, and 32B = 2B + 6A + 2, or 30B = 4, and B = • Whence, in general, tan a=a+}a3+‚1⁄2a3, &c. Again, revert this series, and a-t-}ť3+3t5——ť2+&c. The last series affords the most expeditious mode for the rectification of the circle. Assume an arc a, whose tangent t 41-413 120 is one-fifth part of the radius, and tan4a=. 1-6t+t 119 1 consequently (Prop. 5. Trig.) tan (4a 45°) = = 239 .004,184,100,418. Wherefore, computing the terms of the series, a=.197,395,559,850, and 4a=.789,582,239,400. In like manner, we find 4a-45°.004,184,076,000, and hence the difference between these values, or .785,398,1634 exhibits the length of the octant; which number, multiplied by 4, gives 3,1415926536 for the circumference of a circle whose diameter is 1. 10. Proposition sixth, with its corollaries, would furnish a simple quadrature of the circle. The sine of a semiarc being equal to half the chord, it follows that the ratio of an arc to its chord is compounded of the successive ratios of the radius to the cosines of the continued bisections of half that arc. Assuming therefore the arc of 60°, whose chord is equal to the radius, the logarithm of the ratio of the circumference of a circle to its diameter will be thus computed: exceeds only by 3 in the last place the logarithm of 3,141592654. As the successive terms come to form very nearly a progression that descends by quotients of 4, the third of the last one is, for the reason stated in page 245, considered as equal to the result of the continued addition. 11. An elegant mode of forming the approximate sines corresponding to any division of the quadrant, may be derived from the principles stated in the calculation of trigonometric lines: For the successive differences of the sines for the arcs AB, A, and A+ B, are sinA sin(A — B,) and sin (A+B) sin A; and consequently the difference between these again, or the second difference of the sines, is sin(A + B) + sin( A — B)—2sin A=(Prop. 3. cor. 3. Trig. ) — 2vers BsinA. The second differences of the progressive sines are hence subtractive, and always proportional to the sines themselves. Wherefore the sines may be deduced from their second differences, by reversing the usual process, and recompounding their separate elements. Thus, the sines of A-E |