Toán Lớp 11: Rút gọn vế trái bằng vế phải cos⁶ x- sin⁶ x= 15/16cos2x+1/16cos6x

Question

Toán Lớp 11:  Rút gọn vế trái bằng vế phải  cos⁶ x- sin⁶ x= 15/16cos2x+1/16cos6x

giahan44 · Answer

Lời giải và giải thích chi tiết:   $ cos^6x- sin^6x$   $=( cos^2x)^3-( sin^2x)^3$   $=( cos^2x- sin^2x)( cos^4x+ cos^2x sin^2x+ sin^4x)$   $= cos2x.( cos^4x+2 cos^2x sin^2x+ sin^4x- cos^2x sin^2x)$   $= cos2x.[( cos^x+ sin^2x)^2- cos^2x sin^2x]$   $= cos2x.(1- cos^2x sin^2x)$   $= cos2x. left(1- dfrac{ sin^22x}{4} right))$   $= cos2x. left((1- dfrac{1- cos4x}{8} right)$   $= cos2x- dfrac{ cos2x(1- cos4x)}{8}$   $= cos2x- dfrac{ cos2x- cos2x. cos4x}{8}$   $= cos2x- dfrac{ cos2x- dfrac{1}{2}( cos6x+ cos2x)}{8}$   $= cos2x- dfrac{ dfrac{1}{2} cos2x- dfrac{1}{2} cos6x}{8}$   $= cos2x- dfrac{1}{16} cos2x+ dfrac{1}{16} cos6x$   $= dfrac{15}{16} cos2x+ dfrac{1}{16} cos6x$   V&acirc;̣y $ cos^6x- sin^6x= dfrac{15}{16} cos2x+ dfrac{1}{16} cos6x$  (Đpcm).

tonhu12 · Answer

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