✔ 最佳答案
1. cos0/1+sin0 - 1-sin0/cos0
= cos^2 0 - (1-sin0)(1+sin 0)/(1+sin0)cos0
= (cos^2 0 - 1 + sin^2 0)/(1+sin 0)cos0
= 0
證明下列各式是恆等式
1(a) cos0/tan0 +sin0 = 1/sin0
L.H.S. = cos0/tan0 +sin0
= cos0/(sin0/cos0) + sin0
= cos^2 0/sin0 + sin0
= (cos^2 0 + sin^2 0)/sin0
= 1/sin0
= R.H.S.
1(b) cos^20-sin^20 = 1-2sin^20
L.H.S. = cos^2 0 - sin^2 0
= (1-sin^2 0) - sin^2 0
= 1-2sin^2 0
= R.H.S.
1(c) 1-cos^20/cos0+cos0 = 1/cos0
L.H.S. = (1-cos^20)/cos0+cos0
= (sin^2 0 + cos^2 0)/cos 0
= 1/cos0
=R.H.S.
1(d) 1/cos0 - cos0 =tan0sin0
L.H.S. = 1/cos0 - cos0
= (1-cos^2 0)/cos0
= sin^2 0 / cos0
= sin0/cos0 * sin0
= tan 0 sin0
= R.H.S.
1(e) 3-2cos^20 = 1+2cos^2(90度-0)
L.H.S. = 3 - 2 cos^2 0
= 1+2(1-cos^2 0)
= 1+2sin^2 0
R.H.S. = 1+2cos^2 (90deg-0)
= 1+2sin^2 0
=L.H.S.
2010-02-19 17:06:25 補充：
好似1a咁,我遇到要+sin0就唔知點計好
我自己計到咁樣不知所措
1a)
cos0/(sin0/cos0)+sin0 = 1/sin0
cos^20/sin0 + sin0 = 1/sin0
????????????????????
如果可以教埋我點算
15點
[Notice:通分母~^^"]
2010-02-19 17:22:41 補充：
0代表Θ
