由已知2sina=sinΘ+cosΘ (1)
(sinβ)^2=sinΘcosΘ (2)
(1)^2 4(sina)^2=1+2sinΘcosΘ=1+2(sinβ)^2
2cos2α=cos2β
平方得4(cos2α)^2=(cos2β)^2
2(cos4α+1)=(1/2)(cos4β+1)
故cos4β-4cos4a=3
由已知2sina=sinΘ+cosΘ (1)
(sinβ)^2=sinΘcosΘ (2)
(1)^2 4(sina)^2=1+2sinΘcosΘ=1+2(sinβ)^2
2cos2α=cos2β
平方得4(cos2α)^2=(cos2β)^2
2(cos4α+1)=(1/2)(cos4β+1)
故cos4β-4cos4a=3