ions for the tangent both of a+b and a−b in terms of the tangent. But the tangent of a+b and a-b may be expressed in terms of the cotangent. 38. By substituting, tan (a+b)=R2=1÷ cot a cot b-R2 cot b cot a 40.* Substituting, tan (a—5)=1÷(Cot a cot b+R2 cot b-cot a cot a cot b+1 = cot b-cot a 41. Dividing the first equation by the second, (Art. 208.) we R sin (a+b)_sin a cos b+sin b cos a have R sin (a-b) sin a cos b—sin b cos a 42. Dividing by R, 46. Dividing the numerator, cos (a+b), by sin a sin b, it becomes, by Art 216, b+cot a R R -=(Art. 220) cot 47. Dividing the denominator, cos (a-b), by sin a sin b, it 49. Dividing 3d (Art. 212,) = equation by the 4th. and dividing by R, cos (a+b) cos a cos b-sin a sin b cos (a-b) cos a cos b+sin a sin b 50. Dividing both the numerator and the denominator by cos b sin sin b cos a, we obtain the fraction, (cos b 'cos b, sin a cos a COS a sin b 54. Dividing both the numerator & the denominator by sin a cos b, we have the fraction, cos a sin cot b-tan cot b-tan a 57. In Art. 218, tan a=tan (a+‡a)=: R2 (2 tan a) tan b cot a-tan + = R R R X R cot a+tan b tan a=. R22 and because R2 = 1, we have a 2 cot a-tana. Having proceeded thus far in these analytical investigations, it will be readily seen that they may be carried on to a very great extent. By combining expressions for the sines, tangents, &c. to a sufficient degree, the formula we have obtained, may be multiplied almost indefinitely. TEN THEOREMS FOR PRACTICE. Theorem 1. Tan la cota-cot a 1. By Art. 220 b. tan a=2÷(cot a-tana) 2. By Art. 93. R2÷cot a=tan a, and R2-cot a=tan ļa 3. Subst. in the 1st, R÷cot a=2÷(cot a-R2÷cota) 4. Mult. by cot a, R2 ×cot ļa—(R1÷cot 1a)=2 cot a 5. Transposing, R4÷cot a=R2 cota-2cot a Theorem 2. Tana=(1-cos a)÷sin a 1. By theorem 1st, tan acota-2cot a 2. By Art. 220, cot a=R2÷tana, & 2cot a=2R2÷tan a 3. By substituting, tan ļa=(R2÷tan a)-(2R2÷tan a) 4. By Art. 216, R2÷tan a=cot a÷sina, and 2Ra ÷ tan a 2cos asin a 5. Substituting, tan a=(cos la÷÷sin a)-(2cos a÷sin a) 6. By Art. 210, R cos a=cos2 a-sin2 a 7. Multiplying by 2, 2R cos a=cos2 a-2sin2 la 8. Substi. tan }a= cos 1a__2cos2 1a-2sin2 a sin a 9. By Alg. 148, tan ja= 2sina cosa 2 sin la cos2 a-cos la sin la+2sin2 a 10. Cancelling 2 a=2sin2 2sin2 la cosa sin la cosa, and dividing by sin la, tan a÷2sin la cos a 11. Art. 210, 2sin2 a=R3-R cos a=numerator in 10th 12. Transp. in equation 1st, Art. 210, 2sin'a = R sin a cosa, which, multiplied by cosa, =R sin a cos, Ja÷cos ja = R sin a 13. Substituting these values for the numerator and denomin the 10th, tan a=(R2 — R cos a)÷R sin a 14. Substituting 1 for R2 and R, tan da =(1—cos a)÷sin a Theorem 3. tan a sin a÷(1+cos a) 1. By the preceding formula, tan a=(1-cos a)÷sin a 2. By Art. 210, equa. 5th, sin2 a=R2-R cos a=1—cos a = num. in 1st den. in 1st a 3. By Art. 210, equa. 1st, sin a=2 sin ja cos ta 9. Substituting in the 7th, tan a=R sin a÷(R2+R cos a) 10. Substi. 1 for R2 and R, tan a sin a÷(1+cos a) Theorem 4. tan2a-(1-cos a)÷(1+cos a) 1. By Art. 216, tan ÷cos a =R sin cos, and tan a=R sin a = 2. Involving to the 2d power, tan2a=R2 sin2±a÷cos2 a sina cos2a 3. Multiplying by 2, tan2 a 2 sin2 a÷2 cos2 a 4. By Art. 210, 2 sin2 a⇒R2 – R cos a 5. 66 661 66 2 cos2a=R2+R cos a 6. Substi. in the 3d, tan2a=(R2-R cos a)÷R2+Rcos a 7. Putting 1 for R2 and R, tan2a=(1—cos a)÷(1+cos a) Theorem 5. sin a=2 tan ja÷(1+tana) 1. By Art 210, equation 1st, sin a=2 sin ja cos a 2. By Art. 216. sin tan cos÷R; that is, sin atan a cos a 3. Subst. in 1st, sin a=2tan a cosa cos a=2tan a cos2 a 4. By Art. 93, cos=R2÷sec.; that is, cos2 a=R2÷sec2 a; also, by Art. 94, sec2=R2+tan2; that is, seca R2+tan3a 5. By substituting, R2-÷sec2a-R2÷(R2+tan2a) 6. Substi. in 4th, sin a=2 tan a × R2 ÷ (R2 +tana) 7, Putting 1 for R2, sin a=2 tan ja÷(1+tan2ža). |