已知A=(2cos^2x-1)/[(tan(派/4)-x)(sin^2(派/4)+x)]
来源:学生作业帮 编辑:拍题作业网作业帮 分类:数学作业 时间:2024/04/28 20:45:10
已知A=(2cos^2x-1)/[(tan(派/4)-x)(sin^2(派/4)+x)]
(1)求A
(2)当A=tanB,求4B
A=(2cos^2x-1)/[(tan(派/4)-x)(sin^2[(派/4)+x)] ]
(1)求A
(2)当A=tanB,求4B
A=(2cos^2x-1)/[(tan(派/4)-x)(sin^2[(派/4)+x)] ]
(1)求A
[(tan(派/4-x)(sin^2[(派/4+x)]
=cot[派/4+x](sin^2[(派/4+x)]
=2sin(π/4+x)cos(π/4+x)
=sin2(π/4+x)
=sin(π/2+2x)
=cos2x
A=(2cos^2x-1)/[(tan(派/4)-x)(sin^2[(派/4)+x)] ]
=cos2x/cos2x
=1
(2)tanB=1,所以B=kπ+π/4
4B=4kπ+π,k整数.
再问: 谢谢啊~~采纳了!
[(tan(派/4-x)(sin^2[(派/4+x)]
=cot[派/4+x](sin^2[(派/4+x)]
=2sin(π/4+x)cos(π/4+x)
=sin2(π/4+x)
=sin(π/2+2x)
=cos2x
A=(2cos^2x-1)/[(tan(派/4)-x)(sin^2[(派/4)+x)] ]
=cos2x/cos2x
=1
(2)tanB=1,所以B=kπ+π/4
4B=4kπ+π,k整数.
再问: 谢谢啊~~采纳了!
已知A=(2cos^2x-1)/[(tan(派/4)-x)(sin^2(派/4)+x)]
化简sin(派-x)+sin(派+x)-cos(-x)+cos(2派-x)-tan(派+x)cot(派-x)
.若f(x)={sin(n派-x)cos(n派+x)/cos[(n+1)派-x]}*tan(x-n派)cot[(n派/2
f(x)=sin(派-x)cos(2派-x)/sin(派/2+x)tan(派+a) 求f(31派/3)
已知sin x=cos 2a,a∈(1/2派,派),则tan a=
化简:sin(2派-x)tan(派+x)cot(-x-派)/tan(3派-x)cos(派-x)
已知sinx=3/5,求sin(x-3/2派cos(派-x)tan(派+x)
已知函数f(x)=2根号3sin(x/2+派/4)cos(x/2+派/4)-sin(x+派).求f(x)的最小正周期
已知COS(X+派/4)=根号2/10,x属于(派/2,3派/4) 求sinx 求sin(2x+派\4)
已知函数f(x)=2根号3sin(x/2+派/4)cos(x/2+派/4)-sin(x+派).
已知函数f(x)=2√3sin(x/2+派/4)cos(x/2+派/4)-sin(x+派)
已知函数f(x)=2根3sin(x-派/4)cos(x-派/4)-sin(2x-派)求单调递增区间