정답
안녕하세요.
다음 두 가지 기본 사항 만 기억하세요.
sin (A + B) = sinAcosB + cosAsinB ( 기억 )
그러면 sin (AB)을 쉽게 찾을 수 있습니다.
sin (AB) = sin (A + (-B)) = sinAcos (-B) + cosAsin (-B) = sinAcosB + cosA (-sinB) {since;
cos (-X) = cosX
sin (-X) = sin (X)}
sin (AB) = sinAcosB- cosAsinB
cos (A + B) = cosAcosB-sinAsinB ( 기억 )
cos ( AB) = cos (A + (-B)) = cosAcos (-B) -sinAsin (-B) = cosAcosB-sinA (-sinB)
cos (AB) = cosAcosB + sinAsinB
감사합니다.
답변
\ text {두 변의 차이가 0이면 동일합니다. 즉}
\ left (\ dfrac {\ cos \, A} {\ sin \, A + \ cos \ , B} + \ dfrac {\ cos \, B} {\ sin \, B-\ cos \, A} \ right)-\ left (\ dfrac {\ cos \, A} {\ sin \, A-\ cos \, B} + \ dfrac {\ cos \, B} {\ sin \, B + \ cos \, A} \ right) = 0
\ text {L eft hand side}
\ left (\ dfrac {\ cos \, A} {\ sin \, A + \ cos \, B} + \ dfrac {\ cos \, B} {\ sin \ , B-\ cos \, A} \ right)-\ left (\ dfrac {\ cos \, A} {\ sin \, A-\ cos \, B} + \ dfrac {\ cos \, B} {\ sin \, B + \ cos \, A} \ right)
= \ left (\ dfrac {\ cos \, A} {\ sin \, A + \ cos \, B}-\ dfrac {\ cos \, A} {\ sin \, A-\ cos \, B} \ right) + \ left (\ dfrac {\ cos \, B} {\ sin \, B-\ cos \, A}- \ dfrac {\ cos \, B} {\ sin \, B + \ cos \, A} \ right)
= \ cos \, A \ left (\ dfrac {1} {\ sin \ , A + \ cos \, B}-\ dfrac {1} {\ sin \, A-\ cos \, B} \ right) + \ cos \, B \ left (\ dfrac {1} {\ sin \, B-\ cos \, A}-\ dfrac {1} {\ sin \, B + \ cos \, A} \ right)
= \ cos \, A \ left (\ dfrac {- 2 \ cos \, B} {(\ sin \, A + \ cos \, B) (\ sin \, A-\ cos \, B} \ right) + \ cos \, B \ left (\ dfrac {2 \ cos \, A} {(\ sin \, B + \ cos \, A) (\ sin \, B-\ cos \, A} \ right)
= \ dfrac {-2 \ cos \, A \ cos \, B} {\ sin ^ 2 \, A-\ cos ^ 2 \, B} + \ dfrac {2 \ cos \, A \ cos \, B} {\ sin ^ 2 \, B-\ cos ^ 2 \, A}
=-2 \ cos \, A \ cos \, B \ left (\ dfrac {1} {\ sin ^ 2 \, A-\ cos ^ 2 \, B}-\ dfrac {1} {\ sin ^ 2 \, B-\ cos ^ 2 \, A} \ right)
= -2 \ cos \, A \ cos \, B \ left (\ dfrac {\ sin ^ 2 \, B-\ cos ^ 2 \, A-\ sin ^ 2 \, A + \ cos ^ 2 \, B} {(\ sin ^ 2 \, A-\ cos ^ 2 \, B) (\ sin ^ 2 \, B-\ cos ^ 2 \, A)} \ 오른쪽)
= -2 \ cos \, A \ cos \, B \ left (\ dfrac {\ sin ^ 2 \, B + \ cos ^ 2 \, B-(\ cos ^ 2 \ , A + \ sin ^ 2 \, A)} {(\ sin ^ 2 \, A-\ cos ^ 2 \, B) (\ sin ^ 2 \, B-\ cos ^ 2 \, A)} \ right )
= -2 \ cos \, A \ cos \, B \ left (\ dfrac {1-1} {(\ sin ^ 2 \, A-\ cos ^ 2 \, B) ( \ sin ^ 2 \, B-\ cos ^ 2 \, A)} \ right)
= 0
\ implies \ left (\ dfrac {\ cos \, A} {\ sin \, A + \ cos \, B} + \ dfrac {\ cos \, B} {\ sin \, B-\ cos \, A} \ right)-\ left (\ dfrac {\ cos \, A} {\ sin \, A-\ cos \, B} + \ dfrac {\ cos \, B} {\ sin \, B + \ cos \, A} \ right) = 0
\ implies \ left (\ dfrac {\ cos \, A} {\ sin \, A + \ cos \, B} + \ dfrac {\ cos \, B} {\ sin \, B-\ cos \, A} \ right) = \ left (\ dfrac {\ cos \, A} {\ sin \, A-\ cos \, B} + \ dfrac {\ cos \, B} {\ sin \, B + \ cos \, A} \ 오른쪽)
\ text {QED}