cos^2A+cos^2(2π/3+A)+cos^2(4π/3+A)

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邂后松鹰 共回答了26个问题 | 采纳率84.6%
cos^2A+cos^2(2π/3+A)+cos^2(4π/3+A)=COS^2A+(-COS(-π/3+A))(-COS(-π/3+A))+(-COS(π/3+A))(-COS(π/3+A))=COS^2A+COS^2(-π/3+A)+COS^2(π/3+A)=1/2(1+COS2A)+1/2{(1+COS[-2π/3+2A])}+1/2{(1+COS[2π/3+2A])}=3/2+根号3/2sin2A
1年前

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cos^2A+cos^2(A+60)-cosA*cos(A+60)
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5cos^2a+4cos^2β=4cosα则cos^2a+cos^2β
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5cos^2a+4cos^2β=4cosα
令cosα = t
5t^2+4cos^2β=4t
4cos^2β=4t-5t^2
4cos^2β≥0
4t-5t^2≥0
5t^2-4t≤0
t(5t-4)≤0
0≤t≤4/5
5cos^2a+4cos^2β=4cosα
4cos^2a+4cos^2β=4cosα-cos^2α=4-4+4cosα-cos^2α=4-(2-cosα)^2=4-(2-t)^2
0≤t≤4/5
0≥ -t ≥ -4/5
2 ≥ 2-t ≥ 6/5
4 ≥ (2-t)^2 ≥ 36/25
-4≤-(2-t)^2≤-36/25
0≤4-(2-t)^2≤64/25
即0≤4cos^2a+4cos^2β≤64/25
0≤cos^2a+cos^2β≤16/25
设ABC同时满足sinA+cosB+sinC=0 cosA+cosB+cosC=0的任意角,求证cos^2A+cos^2
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cosa + cosb + cosc = sina + sinb + sinc = 0
(cosa)^2 = (cosb + cosc)^2
= (cosb)^2 + (cosc)^2 + 2*cosb*cosc .(1)
(sina)^2 = (sinb + sinc)^2
= (sinb)^2 + (sinc)^2 + 2*sinb*sinc .(2)
(1) + (2),得cos(b-c) = -1/2
同样可以得到:
cos(c-a) = -1/2
cos(a-b) = -1/2
(1) - (2),得
cos2a = cos2b + cos2c + 2*cos(b+c)
= 2*cos(b+c)*cos(b-c) + 2*cos(b+c) .cos(b-c) = -1/2
= cos(b+c) .(3)
同样可以得到:
cos2b = cos(c+a) .(4)
cos2c = cos(a+b) .(5)
(cosa)^2 + (cosb)^2 + (cosc)^2
= (cos2a + cos2b + cos2c)/2 + 3/2
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A = cos2a + cos2b + cos2c
= [(cos2a + cos2b) + (cos2b + cos2c) + (cos2c +cos2a)]/2
= cos(a+b)*cos(a-b) + cos(b+c)*cos(b-c) + cos(c+a)*cos(c-a)
= -[cos(a+b) + cos(b+c) + cos(c+a)]/2
由(3)、(4)、(5)得到
A = cos(a+b) + cos(b+c) + cos(c+a)
所以,A = 0
(cosa)^2 + (cosb)^2 + (cosc)^2 = 3/2
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