tanx=2,sin²x+sinxcosx-2cos²=(sin²x+sinxcosx-2cos²)/(sin²x+cos²)
=(tan^2x+tanx-2)/(tan^2x+1)
=(4+2-2)/(4+1)=4/5
tanx=2,sin²x+sinxcosx-2cos²=?
4个回答
相关问题
-
tanx=2,2sin²x+sinxcosx+cos²x=?
-
1-2sinxcosx/cos2x-sin2x=1-tanx/1+tanx
-
tanx=2 求sin^2x+sinxcosx-3cos^2x
-
(2sin^2x-sinxcosx+cos^2x)/sin^2x+cos^2x=(2tan^x-tanx+1)/(tan
-
tanx=-1/2,计算1/sin^2x-sinxcosx-2cos^2x
-
已知tanx=2,求2sin2x-sinxcosx cos2x
-
已知tanX=2求2sin^2X+sinXcosX-3cos^2X
-
证明：sin^2x(1+cotx)+cos^2x(1+tanx)=1-sinxcosx
-
求证1-2sinxcosx/cos2次方x-sin2次方x=1-tanx/1+tanx
-
已知tanx=2,则2sin2x-sinxcosx+cos2x 的值为_____