1-cos2x+sin2x= -2sin^x+2sinxcosx应该是等于 2sin^x+2sinxcosx吧?

怎样算Sin2X-2Sin^2X =Sin2X+COS2X-1 tanx=4,则1+cos2x+8sin^2x/sin2x?tanx=4,则(1+cos2x+8sin^2 x)/sin2x=? 1-cos2x+sin2x= 为什么sin2x+cos2x=2sin(2x+π/4) 为什么f(x)=sin2x-2sin²x=sin2x+（1-2sin²x）-1=sin2x+cos2x-1? 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= 2sinx^2-sinxcosx-3cosx^2=0 求sin(x+45)/(sin2x+cos2x+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 sinx-2cosx=0.求sin2x-cos2x/1+sin^2x 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 cos2x/sin(x+45°)=1/2 那么sin2x等于多少 (1-cos2x)+sin2x=根号2(sin2x cosπ/4-cos2x sinπ/4)+1 这个是怎么化简得的? (1-cos2x)+sin2x=根号2(sin2x cosπ/4-cos2x sinπ/4)+1这一步骤是怎样得出的呢 请说明