I've constructed the proof for the cos difference formula
cos(A-B) = cosAcosB + sinAsinB
using the distance formula, but I'm not sure how to derive the sine sum/difference formula from this. I've thought about using a co-function identity, but that seems to have taken me nowhere.
Can someone help? Greatly appreciated.
Geronimo
Favorite Answer
cos(A – B) = cos(A) • cos(B) + sin(A) • sin(B)
cos(x) = sin [ (π ⁄ 2) – x ] ... identity 1 (used later)
sin(x) = cos [ (π ⁄ 2) – x ] ... identity 2 set: x = A – B
sin(A – B) = cos [(π ⁄ 2) – (A – B) ]
sin(A – B) = cos [(π ⁄ 2) – A + B) ]
sin(A – B) = cos [ { (π ⁄ 2) – A } + B ]
sin(A – B) = cos [ { (π ⁄ 2) – A } – (- B) ] ... equation 1
cos [ { (π ⁄ 2) – A } – (- B) ] = cos[ (π ⁄ 2) – A ] • cos(- B) + sin[ (π ⁄ 2) – A ] • sin(- B)
since:
sin(- B) = - sin(B) and cos(- B) = cos(B)
cos [ { (π ⁄ 2) – A } – (- B) ] = cos[ (π ⁄ 2) – A ] • cos(B) – sin[ (π ⁄ 2) – A ] • sin(B)
... and using identities 1 and 2 (earlier) ...
cos [ { (π ⁄ 2) – A } – (- B) ] = sin(A) • cos(B) – cos(A) • sin(B)
... and because of equation 1 ...
sin(A – B) = sin(A) • cos(B) – cos(A) • sin(B)