sin(3π-α)=√2sin(2π+β),√3cos(-α)=-√2cos(π+β) ,且0
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sin(3π-α)=√2sin(2π+β),√3cos(-α)=-√2cos(π+β) ,且0
sin(3π-α)=√2sin(2π+β),
sinα=√2sinβ
√3cos(-α)=-√2cos(π+β)
√3cosα=√2cosβ
cosα=√2cosβ/(√3)
(sinα)^2+(cosα)^2=(√2sinβ)^2+[√2cosβ/(√3)]^2
2(sinβ)^2+2(cosβ)^2/3=1
6(sinβ)^2+2(cosβ)^2=3
4(sinβ)^2+2=3
(sinβ)^2=1/4
0
sinα=√2sinβ
√3cos(-α)=-√2cos(π+β)
√3cosα=√2cosβ
cosα=√2cosβ/(√3)
(sinα)^2+(cosα)^2=(√2sinβ)^2+[√2cosβ/(√3)]^2
2(sinβ)^2+2(cosβ)^2/3=1
6(sinβ)^2+2(cosβ)^2=3
4(sinβ)^2+2=3
(sinβ)^2=1/4
0
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