단순화 (1- cos theta + sin theta) / (1+ cos theta + sin theta)?

대답:

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

설명:

(1-cos (세타) + sin (세타)) / (1 + cos (세타) + sin (세타)) #

(1 + cos (세타) + sin (세타)) ^ 2 # (1-cos (세타) + sin (세타)) *

(θ) + sin2 (θ) +2 sin (θ) +2 cos (θ) = (sinθ) + 2 sin (θ) cos (θ)) #

(θ) +2 sin (θ) +2 cos (θ) +2 sin (θ) cos (θ)) = (1 + sin (θ)

(1 + cos (theta)) / (2 + sin (θ)) 2 sin (θ)

(1 + sin (θ)) # = (1/2) (1 + sin (θ)) ^ 2-cos ^ 2

(1 + cos (세타)) - (1/2) (cos ^ 2 (세타)) / ((1 + sin (세타) + sin (theta))) #

(1 + sin (θ)) - (1/2) (1-sin ^ 2 (세타)) / ((1 + sin (세타) (1 + sin (θ))) #

(1 + sin (세타))) / (1/2) (1-sin (세타)) * (1 + sin (세타) ((1 + cos (θ)) (1 + sin (θ)))

(1 + sin (세타)) / (1 / sin (세타)) - (1/2) (1-sin (세타)) /

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