Question

一道数学小题.

已知sinα=3/5,α属于(π/2,π),则cos(π/4+α)sin(π/4+α)的值为

Answer 1

cos(π/4+α)sin(π/4+α)=（1／2）[2cos(π/4+α)sin(π/4+α)]
=（1／2）sin(π／2+2α）
=（1／2）cos(2α)
∵α属于(π/2,π),∴2α属于(π,2π)
又∵sinα=3/5 ∴cos2α=1-2（sinα）²=7／25
∴原式等于：=（1／2）×（7／25）=7／50

Answer 2

sinα=3/5,α属于(π/2,π)则cosα=-4/5
cos(π/4+α)sin(π/4+α)=cos(π/4+α)+sin(π/4+α)
= √2 * [(√2 /2)√cos(π/4+α) + (√2/2)sin(π/4+α)]
= √2 * [cos(π/4)cos(π/4+α) + sin(π/4)sin(π/4+α)]
= √2 * cos[(π/4+α) - (π/4)]
= √2 cosα =-4√2/5

Answer 3

7/50

Answer 4

sinα=3/5
cos(π/4+α)sin(π/4+α)
=1/2*sin(π/2+2α)
=1/2*cos(2α)
=1/2*(1-2*sinα*sinα)
=1/2*(1-2*3/5*3/5)
=7/50