1) cho sinx+cosx=1/2.tính a) sinx.cosx b) sin^3x+cos^3x c)sin^4x+cos^4x 30/09/2021 Bởi Jade 1) cho sinx+cosx=1/2.tính a) sinx.cosx b) sin^3x+cos^3x c)sin^4x+cos^4x
`\qquad sinx +cos x=1/ 2` `a)` Ta có: `\qquad (sinx+cosx)^2=(1/ 2)^2=1/ 4` `<=>sin^2 x +cos^2x +2sinxcosx=1/ 4` `<=>1+2sinxcosx=1/ 4` `<=>2sinxcosx =1/ 4 -1=-3/ 4` `<=>sinxcosx=-3/ 8` $\\$ `b)` `sin^3x+cos^3x` `=(sinx+cosx)(sin^2x+cos^2x-sinxcosx)` `=1/ 2 .[1-(-3/ 8)]` `=1/ 2 . {11}/8` `={11}/{16}` $\\$ `c)` `sin^4x+cos^4x` `=(sin^2x+cos^2x)^2-2sin^2xcos^2x` `=1^2-2.(sinxcosx)^2` `=1-2. (-3/ 8)^2` `=1-9/{32}={23}/{32}` Bình luận
`\qquad sinx +cos x=1/ 2`
`a)` Ta có:
`\qquad (sinx+cosx)^2=(1/ 2)^2=1/ 4`
`<=>sin^2 x +cos^2x +2sinxcosx=1/ 4`
`<=>1+2sinxcosx=1/ 4`
`<=>2sinxcosx =1/ 4 -1=-3/ 4`
`<=>sinxcosx=-3/ 8`
$\\$
`b)` `sin^3x+cos^3x`
`=(sinx+cosx)(sin^2x+cos^2x-sinxcosx)`
`=1/ 2 .[1-(-3/ 8)]`
`=1/ 2 . {11}/8`
`={11}/{16}`
$\\$
`c)` `sin^4x+cos^4x`
`=(sin^2x+cos^2x)^2-2sin^2xcos^2x`
`=1^2-2.(sinxcosx)^2`
`=1-2. (-3/ 8)^2`
`=1-9/{32}={23}/{32}`