[4cos2(-15pi/4)]/[tan(-11pi/3)+(根号2)*sin(25pi/4)]的值为?

[4cos²(-15π/4)]/[tan(-11π/3)+√2sin(25π/4)]
= [4cos²(-15π/4+4π)]/[tan(-11π/3+4π)+√2sin(25π/4-6π)]
= [4cos²(π/4)]/[tan(π/3)+√2sin(π/4)]
= [4(√2/2)²]/[√3+√2(√2/2)]
= 2/[√3+1]
= √3-1

∵1＞π/4，cos(π/4)=sin(π/4)=√2/2，tan(π/4)=1
∴cos1＜√2/2＜sin1＜1＜tan1

[4cos2(-15pi/4)]/[tan(-11pi/3)+(根号2)*sin(25pi/4)]的值为?
[4cos2(-15pi/4)]/[tan(-11pi/3)+(根号2)*sin(25pi/4)]的值为?

[4cos²(-15π/4)]/[tan(-11π/3)+√2sin(25π/4)]
= [4cos²(-15π/4+4π)]/[tan(-11π/3+4π)+√2sin(25π/4-6π)]
= [4cos²(π/4)]/[tan(π/3)+√2sin(π/4)]
= [4(√2/2)²]/[√3+√2(√2/2)]
= 2/[√3+1]
= √3-1

because 1＞π/4，cos(π/4)=sin(π/4)=√2/2，tan(π/4)=1
so that cos1＜√2/2＜sin1＜1＜tan1