Toán Lớp 11: Rút gọn vế trái bằng vế phải sin⁸+ cos⁸x= 35/64+7/16 cos4x+1/64cos8x

Question

Toán Lớp 11:  Rút gọn vế trái bằng vế phải sin⁸+ cos⁸x= 35/64+7/16 cos4x+1/64cos8x

phuongthuydung · Answer

Lời giải và giải thích chi tiết:   $ sin^8x+ cos^8x$   $=( sin^4x)^2+2 sin^4x cos^4x+( cos^4x)^2-2 sin^4x cos^4x$   $=( sin^4x+ cos^4x)^2- dfrac{ sin^42x}{8}$   $=( sin^4x+2 sin^2x cos^2x+ cos^4x-2 sin^2x cos^2x)^2- dfrac{ sin^42x}{8}$   $= bigg{[}( sin^2x+ cos^2x)^2- dfrac{ sin^22x}{2} bigg{]}^2- dfrac{ left( dfrac{1- cos4x}{2} right)^2}{8}$   $= left(1- dfrac{1- cos4x}{4} right)^2- dfrac{(1- cos4x)^2}{32}$   $= left( dfrac{3+ cos4x}{4} right)^2- dfrac{(1- cos4x)^2}{32}$   $= dfrac{(3+ cos4x)^2}{16}- dfrac{(1- cos4x)^2}{32}$   $= dfrac{9+6 cos4x+ cos^24x}{16}- dfrac{1-2 cos4x+ cos^24x}{32}$   $= dfrac{18+12 cos4x+2 cos^24x-1+2 cos4x- cos^24x}{32}$   $= dfrac{17+14 cos4x+ cos^24x}{32}$   $= dfrac{17+14 cos4x+ dfrac{1+ cos8x}{2}}{32}$   $= dfrac{ dfrac{35}{2}+14 cos4x+ dfrac{ cos8x}{2}}{32}$   $= dfrac{35}{64}+ dfrac{7}{16} cos4x+ dfrac{1}{64} cos8x$   V&acirc;̣y $ sin^8x+ cos^8x= dfrac{35}{64}+ dfrac{7}{16} cos4x+ dfrac{1}{64} cos8x$ (Đpcm).

haianhtu97 · Answer

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