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sin-2-x-cos-2-x-




Question Number 190993 by mathlove last updated on 16/Apr/23
sin^2 x ∙cos^2 x=?
sin2xcos2x=?
Answered by cortano12 last updated on 16/Apr/23
 = (1/4) sin^2 2x
=14sin22x
Answered by mustafazaheen last updated on 16/Apr/23
=[(2/2)sin(x)∙cos(x)][(2/2)sin(x)∙cos(x)]  =((sin(2x))/2)×((sin(2x))/2)  =(1/4)sin^2 (2x)
=[22sin(x)cos(x)][22sin(x)cos(x)]=sin(2x)2×sin(2x)2=14sin2(2x)
Answered by manxsol last updated on 16/Apr/23
y=sin^2 x ∙cos^2 x  4y=4sin^2 x ∙cos^2 x  4y=(2sinxcosx)^2   4y=(sin2x)^2   y=(((sin2x)^2 )/4)
y=sin2xcos2x4y=4sin2xcos2x4y=(2sinxcosx)24y=(sin2x)2y=(sin2x)24

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