三角函数的题 Tg2α= - 4/3 ,2α∈(π/2.π) 求√2sin(α+π/4) / 1+sinα-2cos²π/2
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三角函数的题 Tg2α= - 4/3 ,2α∈(π/2.π) 求√2sin(α+π/4) / 1+sinα-2cos²π/2
三角函数的题
Tg2α= - 4/3 ,2α∈(π/2.π)
求
√2sin(α+π/4) / 1+sinα-2cos²π/2
三角函数的题 Tg2α= - 4/3 ,2α∈(π/2.π) 求√2sin(α+π/4) / 1+sinα-2cos²π/2
注释:这道题√2sin(α+π/4) / 1+sinα-2cos²π/2很可能写掉了括号.
所以原题有两种可能的写法:
(1)√2sin(α+π/4) / (1+sinα)-2cos²π/2;
(2)√2sin(α+π/4) / (1+sinα-2cos²π/2).
∵cos²π/2=0,
∴无论哪种写法结果都一样.于是,我只解(1).
∵Tg2α= - 4/3 , 2α∈(π/2.π),
∴sin2α/cos2α=tg2α=-4/3.
∴cos2α=-3/5
∴sinα=√((1-cos2α)/2)=2/√5,
cosα==√((1+cos2α)/2)=1/√5.
∴√2sin(α+π/4) / (1+sinα)-2cos²π/2
=√2sin(α+π/4) / (1+sinα)
=√2(sinα*cosπ/4+cosα*sinπ/4)/(1+sinα)
=(sinα+cosα)/(1+sinα)
=(2/√5+1/√5)/(1+2/√5)
=3(√5-2).