1+sin2x+cos2x=1+sin(2x+π/4),为什么?
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/24 05:40:42
![1+sin2x+cos2x=1+sin(2x+π/4),为什么?](/uploads/image/z/4327406-62-6.jpg?t=1%2Bsin2x%2Bcos2x%3D1%2Bsin%282x%2B%CF%80%2F4%29%2C%E4%B8%BA%E4%BB%80%E4%B9%88%3F)
1+sin2x+cos2x=1+sin(2x+π/4),为什么?
1+sin2x+cos2x=1+sin(2x+π/4),
为什么?
1+sin2x+cos2x=1+sin(2x+π/4),为什么?
1+sin2x+cos2x
=1+√2(sin2x*√2/2+cos2x*√2/2)
=1+√2(sin2xcosπ/4+cos2xsinπ/4)
=1+√2sin(2x+π/4)
1+sin2x+cos2x=1+sin(2x+π/4),——错误的!!
1+sin2x+cos2x=1+√2*sin(2x+π/4),
怎么会?应该不等于吧。。。
1-cos2x+sin2x=
怎样算Sin2X-2Sin^2X =Sin2X+COS2X-1
求证1+sin2x-cos2x/1+sin2x+cos2x=tanx
化简 (1+sin2x-cos2x)/(1+sin2x+cos2x)
tanx=4,则1+cos2x+8sin^2x/sin2x?tanx=4,则(1+cos2x+8sin^2 x)/sin2x=?
为什么f(x)=sin2x-2sin²x=sin2x+(1-2sin²x)-1=sin2x+cos2x-1?
若sin^x-sinxcosx-6cos^x=0,则[(cos2x-sin2x)/(1-cos2x)(1-tan2x)]=?
解方程:(sin2x+cos2x)/(1-sin^2x-2cos2x)=2X属于(π/2,π)
f(x)=sin2x-2sin²x 是怎么到下面这一步的?=sin2x-(1-cos2x)
tanx=-1/2,sin2x+cos2x/4cos^2x-3sin^2x+1=
1+sin2x+cos2x=1+sin(2x+π/4),为什么?
f(x)=1/2(cos2x-sin2x)+sin^2x 化简出来…
已知sin^2x+sin2x*sinx+cos2x=1,x属于(0,派/2),求tan2x
已知:sinx-2cosx=0,求sin2x-cos2x/1+sin平方x
函数f(x)=(1+cos2x+8sin^2x)/sin2x的值域为
sinx-2cosx=0.求sin2x-cos2x/1+sin^2x
2sinx^2-sinxcosx-3cosx^2=0 求sin(x+45)/(sin2x+cos2x+1)
cos2x/sin(x+45°)=1/2 那么sin2x等于多少