三角函數兩題證明和一題計算

1. 證明 (sec@ - tan@)2 = (1 - sin@) / (1 + sin@)

左方 = (sec@ - tan@)2

= (1/cos@ - sin@/cos@)2

= (1 - sin@)2 / cos2@

= (1 - sin@)2 / (1 - sin2@)

= (1 - sin@)2 / (1 - sin@)(1 + sin@) 注意a2 - b2 = (a + b)(a - b)

= (1 - sin@) / (1 + sin@)

= 右方


2. 證明 cot4@ + cot2@ = csc4@ - csc2@

左方 = cot4@ + cot2@

= cos4@ / sin4@ + cos2@ / sin2@

= (cos4@ + cos2@sin2@) / sin4@

= cos2@(cos2@ + sin2@) / sin4@

= (1 - sin2@) / sin4@

= 1/sin4@ - 1/sin2@

= csc4@ - csc2@

= 右方



3. 注意 sin2@ = 1 - cos2@

所以，

sin2A = 1 - cos2A

= 1 - (cosDsinC)2

= 1 - cos2Dsin2C

sin2B = 1 - sin2Dsin2C


sin2A + sin2B + sin2C

= (1 - cos2Dsin2C) + (1 - sin2Dsin2C) + sin2C

= 2 + sin2C(1 - sin2D - cos2D)

= 2 + sin2C[1 - (sin2D + cos2D)]

= 2 + (1 - 1)sin2C

= 2