Απλοποιήστε (1 - cos theta + sin theta) / (1 + cos theta + sin theta);

Απλοποιήστε (1 - cos theta + sin theta) / (1 + cos theta + sin theta);
Anonim

Απάντηση:

# = αμαρτία (theta) / (1 + cos (theta)) #

Εξήγηση:

# (1-cos (theta) + αμαρτία (theta)) / (1 + cos (theta) + αμαρτία (theta)) #

= (1-cos (theta) + αμαρτία (theta)) * (1 + cos (theta) + sin (theta)

(Θήτα) + 2 sin (θήτα) + 2 cos (θήτα) + 2 cos (2) + 2 sin (theta) cos (theta)) #

= ((1 + sin (theta)) ^ 2-cos ^ 2 (theta)) / 2 + 2 sin (theta) + cos (theta)

= ((1 + cos (theta)) ^ 2-cos ^ 2 (theta)) /

= (1 + sin (theta)) ^ 2-cos ^ 2 (theta)) / ((1 + cos (theta)

= (1/2) (1 + sin (theta)) / (1 + cos (theta)) - + αμαρτία (theta))) #

= (1/2) (1 + sin (theta)) / (1 + cos (theta)) - (1/2) (1 + sin (theta))) #

= (1/2 sin (theta)) / (1 + cos (theta)) - (1/2) (1 + cos (theta)) (1 + sin (theta))) #

(1 + cos (theta)) / (1 + cos (theta)) - (1/2)

# = αμαρτία (theta) / (1 + cos (theta)) #