trigonometric...thz

solve the following:

1 - cosB - (sin^2)B = 0

where 0 < B < 360

solve B



1 - cosB - sin²B = 0

1 - cosB - (1-cos²B) = 0 【Applying identity cos²x+sin²x = 1】

1 - cosB - 1 + cos²B = 0

cos²B - cosB = 0

cosB(cosB - 1) = 0

cosB = 0 or cosB-1 = 0

cosB = 0 or cosB = 1

For cosB = 0

B = 90° or B = 270° ... (1)

For cosB = 1

There's no solution for (0° < B < 360° as cos0° = cos360° = 1) ... (2)

Combining (1), (2) the solution is B = 90° or B = 270°


2007-01-20 21:17:58 補充：
小小補充：由於問題裡 B 的範圍是「0° ＜ B ＜ 360°」，所以 0° 和 360° 不包括在答案值裡。^^

1 - cosB - (sin^2)B = 0
1 - cosB - (1 - (cos^2)B) = 0
(cos^2)B - cosB = 0
cosB = 0 or cosB = 1

B = 90 , 270 or 0 , 360