已知cos2ａcos2β是方程2X^2-3X+1=0的两根求sinａ^2×sinβ^2
热心网友
cos2ａcos2β是方程2X^2-3X+1=0的两根cos2ａ + cos2β = 3/2, cos2ａ * cos2β = 1/2cos2ａ = 1-2*(sinａ)^2, cos2β = 1 - 2*(sinβ)^2== [1-2*(sinａ)^2] + [1 - 2*(sinβ)^2] = 3/2 == 2*[(sinａ)^2 + (sinβ)^2] = 1/2[1-2*(sinａ)^2]*[1 - 2*(sinβ)^2] = 1/2== 1 - 2*[(sinａ)^2 + (sinβ)^2] + 4*(sinａ * sinβ)^2 = 1/2== sinａ^2×sinβ^2 = (sinａ * sinβ)^2 = 0
热心网友
2x^2-3x+1=0,x1=1/2;x2=1---cos2a=1/2;cos2b=1.---1-2(sina)^2=1/2;1-2(sinb)^2=1---(sina)^2=1/2;(sinb)^2=0---(sina)^2*(sinb)^2=0
热心网友
答案为1/4.你只需用倍角公式就能做出来了.