化简(1+sinx)/cosx*sin2x/2cos(π/4-x/2)
来源:学生作业帮助网 编辑:作业帮 时间:2024/08/02 18:01:44
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化简(1+sinx)/cosx*sin2x/2cos(π/4-x/2)
化简(1+sinx)/cosx*sin2x/2cos(π/4-x/2)
化简(1+sinx)/cosx*sin2x/2cos(π/4-x/2)
2. 因为,2{cos[(π/4)-(x/2)]}^2 =cos2[(π/4)-(x/2)] 1 =sinx 1 xtan[(π/4)-(x/2)] =xsin[(π/4)-(x/2)]/cos[(π/4)-(x/2)] =2xsin[(π/4)-(x/2)]*cos[(π/4)-(x/2)]/2{cos[(π/4)-(x/2)]}^2 =xsin2[(π/4)-(x/2)]/(sinx 1) =xcosx/(sinx 1) 所以, 原式=(sinx 1){3cosx/2{cos[(π/4)-1)]}^2-2tan[(π/4)-1]} =(sinx 1)[3cosx/(sinx 1)-2cosx/(sinx 1)] =3cosx-2cosx =cosx
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