COS B + — COS A = COS B + -n — r cos A 



a bill A 



182 Demonstrations of Trigonofiietrical Formulce. 



Theorem 2. If a perpendicular be let fall from B to the 

 opposite side b, we have 



b = a cos C + c cos A 

 = c cos A — « cos (A + B) 



■r-r I > T.\ c k ^ c n sinB 



Hence cos (A + B) = — cos A = — cos A : — - 



^ ' a a a sin A 



Now c = a cos B + b cos A 

 ,'. ^ ~- cos B + 



a 



Hence by substitution 



/ » -ox / -r» , sin B . \ . sin B 



COS (A -f B) = (cos B + -. — - cos A) cos A — -. — ^.- 



^ ' \ sm A ' sin A 



T-, . , sin B 3 . sin B 

 = .*. cos 15 cos A + -; — — cos '^A — ^ r- 



sin A sin A 



T-, A sin B •■!-»•* sin B 



= COS B cos t\. + T— I sui B sni A : — - 



sin A sm A 



= .*, cos B . COS A — sin B sin A 



Theorem 3. Let A be an exterior angle of the triangle. 

 Then by Euclid c = A' - B. 

 Also cf sin C = c sin A = c sin A' 



or a sin (A' — B) = c sin A' .'. sin (A' — B) = — sin A' 



Now c — a cos B + i cos A 

 = a cos B — 6 cos A' 



.*. - = COS B cos A = cos D — -. — - cos A 



a a sin A 



Bsin B ., 

 : — r, cos A 



sin A' 



Hence by substitution 



sin (A'— B) = (cos B — "" ,, cos A') sin A' 



^ ' ^ sin A' / 



= .•. sin A' cos B — cos A' sin B 

 Theorem 4. Since b = a cos C + c cos A 



or = a cos (A'— B) — c cos A' 



b c 



.'. cos (A — B) = 1 cos A' 



a a 



sin B c . , sin B c ,, 



— ^hTT + T ^"^ ^ = -^t; + COS A 



sin A a sm A' a 



Now c = a cos B + i cos A = « cos B — Zi cos A' 



.',{ =cosB-.-AcosA'= cosB- -^cosA' 



" a sm A 



Bsin B . , 

 r-jT cos A' 



sin A 



Hence 



