phk69202
幼苗
共回答了26个问题采纳率:92.3% 举报
令f(A) = I / J + cos^2(A) ,
其中I = [{[cos(π-2A)-1]sin(π+A/2)sin(π/2-A/2)} ,
J = sin^2(π/2-A/2)-sin^2(π-A/2)].
分别化简两式:
由于cos(π-2A)-1 = -cos2A - 1 = -[cos2A + 1] = -2(cosA)^2 ,
sin(π+A/2)sin(π/2-A/2) = -sin(A/2)·cos(A/2) = -(sinA)/2
故 I = (sinA)·(cosA)^2 =
sin^2(π/2-A/2) = [cos(A/2)]^2 ,sin^2(π-A/2) = sin^2(A/2)
所以sin^2(π/2-A/2)-sin^2(π-A/2)] = [cos(A/2)]^2 - [sin(A/2)]^2 = cosA,
所以f(A) = I / J + cos^2(A)
= 【sin(2A)】/2 + 【1 + cos(2A)】/2
= (1/2)【sin(2A) + cos(2A)】 + (1/2)
= (1/2)【√2·sin(2A + π/4)】 + (1/2)
因为0
1年前
5