已知4cosA·cosB=根号6.,4sinA·sinB=根号2,求(1-cos4A)(1-cos4B)的值

4cosAcosB=√6,4sinAsinB=√2
--->cosAcosB=√6/4,sinAsinB=√2/4.
(1-cos4A)(1-cos4B)
={1-[1-2(sin2A)^2]}*{1-[1-2(sin2B)^2]}
=4(sin2A)^2*(sin2B)^2
=4(2sinAcosA)^2*(2sinBcosB)^2
=4*4(sinA)^2*(sinB)^2*4(cosA)^2*(cosB)^2
=16*(√6/4)^2*4(√2/4)^2
=16*6/16*4*2/16
=3

(1-cos4A)(1-cos4B)=2sin²2A*2sin²2B=4(sin2A*sin2B)²=4(2sinAcosA*2sinBcosB)²
=64（sinAcosA sinBcosB）²
4cosA·cosB=√6.,4sinA·sinB=√2
∴ sinAcosA sinBcosB=√6* √2/16=√12/16
∴(1-cos4A)(1-cos4B)=3

由2倍角公式得cos4A=(cos2A)^2-(sin2A)^2
=((cosA)^2-(sinA)^2)^2-4(sinAcosA)^2
=((cosA)^2+(sinA)^2)^2-8(sinAcosA)^2
...