三角問題一問...(thx)萬分感激!!!

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圖片參考：http://img682.imageshack.us/img682/2987/45614250.png


2010-03-13 13:34:15 補充：
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#1:
cos^2(5pi/8) = [ cos(pi - 3pi/8)]^2 = cos^2(3pi/8)
cos^2(7pi/8) = [ cos(pi - pi/8) ]^2 = cos^2(pi/8)
so,
.cos^2 pi/8 + cos^2 3pi/8 + cos^2 5pi/8 + cos^2 7pi/8
=2 [ cos^2(pi/8) + cos^2(3pi/8) ]
= 2 [ cos^2(pi/8) + sin^2(pi/2 - 3pi/8) ]
=2 [ cos^2(pi/8) + sin^2(pi/8) ]
= 2 (1)
= 2 //

#2:
sinx-siny=5/4
(sinX - sinY)^2 = 25/16
sin^2x + sin^2y - 2sinxsiny = 25 /16 ...(1)
cosx-cosy=3/4
(cosX-cosY)^2 = 9/16
cos^2x + cos^2y - 2cosxcosy = 9 /16 ..(2)
(1)+(2) :
(sin^2x + cos^2x) + ( sin^2y + cos^2y) - 2(cosxcosy + sinxsiny) = 17/8
2 - 2 cos(x-y) = 17/8
cos(x-y) = -1 / 16 //

#3:
cos(B-A)=12/13 即 sin(B-A) =5/13 (由範圍知 0<B-A<pi/4)
sin(B+A)=-3/5 即 cos(B+A)= -4 / 5 (由範圍知 pi < B+A < 3pi / 2)
考慮
cos(B-A)=12/13
cos(2B -(A+B) ) = 12/13
cos2Bcos(A+B) + sin2Bsin(A+B) = 12/13
(-4/5 )cos2B + (-3/5) sin2B = 12/13
故
cos2B + (3/4) sin2B = -15/13 ... (1)
考慮
sin(B+A) = -3/5
sin(2B-(B-A)) = -3/5
sin2Bcos(B-A) - cos2Bsin(B-A) = -3/5
(12/13)sin2B - (5/13)cos2B = -3/5
故
(12/5) sin2B - cos2B = -39/25 ... (2)
(1)+(2) :
(63/20)sin2B = -882 / 325
sin2B = -56 / 65