高一数学必修四146页复习题a组答案极其过程(特别是6.7.8急需过程)
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6,1)∵π<θ<3π/2,cosθ=-3/5,∴sinθ=-4/5,
原式=(sinθ/2)^2-2sinθ/2*cosθ/2+(cosθ/2)^2=1-sinθ=1-(-4/5)=9/5,
2)∵sina/2-cosa/2=1/5,
(sina/2)^2-2sina/2cosa/2+(cosa/2)^2=1/25,
即1-sina=1/25,
sina=24/25,
3),配方,(sin^θ+cos^θ)^2-2sin^θcos^θ=5/9,
1-1/2*(4sin^θcos^θ)=5/9,
1-1/2*(sin2θ)^2=5/9,
sin2θ=8/9,
4)cos2θ=(cosθ)^2-(sinθ)^2=3/5,
(cosθ)^4+2(cosθ)^2(sinθ)^2+(sinθ)^4=[(cosθ)^2+(sinθ)^2]^2=1,
(cosθ)^4-2(cosθ)^2(sinθ)^2+(sinθ)^4=9/25,
4(cosθ)^2(sinθ)^2+(sinθ)^4=16/25,
2(cosθ)^2(sinθ)^2+(sinθ)^4=8/25,,
(cosθ)^4+(sinθ)^4=17/25,
原式=(sinθ/2)^2-2sinθ/2*cosθ/2+(cosθ/2)^2=1-sinθ=1-(-4/5)=9/5,
2)∵sina/2-cosa/2=1/5,
(sina/2)^2-2sina/2cosa/2+(cosa/2)^2=1/25,
即1-sina=1/25,
sina=24/25,
3),配方,(sin^θ+cos^θ)^2-2sin^θcos^θ=5/9,
1-1/2*(4sin^θcos^θ)=5/9,
1-1/2*(sin2θ)^2=5/9,
sin2θ=8/9,
4)cos2θ=(cosθ)^2-(sinθ)^2=3/5,
(cosθ)^4+2(cosθ)^2(sinθ)^2+(sinθ)^4=[(cosθ)^2+(sinθ)^2]^2=1,
(cosθ)^4-2(cosθ)^2(sinθ)^2+(sinθ)^4=9/25,
4(cosθ)^2(sinθ)^2+(sinθ)^4=16/25,
2(cosθ)^2(sinθ)^2+(sinθ)^4=8/25,,
(cosθ)^4+(sinθ)^4=17/25,
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