Migliore risposta
Ehi,
Ricorda solo queste due basi:
sin (A + B) = sinAcosB + cosAsinB ( Ricorda )
Quindi, puoi facilmente trovare sin (AB).
sin (AB) = sin (A + (- B)) = sinAcos (-B) + cosAsin (-B) = sinAcosB + cosA (-sinB) {da;
cos (-X) = cosX
sin (-X) = sin (X)}
sin (AB) = sinAcosB- cosAsinB
cos (A + B) = cosAcosB-sinAsinB ( Ricorda )
cos ( AB) = cos (A + (- B)) = cosAcos (-B) -sinAsin (-B) = cosAcosB-sinA (-sinB)
cos (AB) = cosAcosB + sinAsinB
Buon apprendimento!
Risposta
\ text {I due lati saranno uguali se la loro differenza = 0. Ovvero}
\ 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 \ sinistra (\ 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)} \ destra)
= -2 \ cos \, A \ cos \, B \ sinistra (\ 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
\ implica \ 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
\ implica \ 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)
\ text {QED}