Ptlg !!!!

N

nguyenbahiep1

[TEX](sin^2x)^3 + (cos^2x)^3 = 1 - sin^22x + \frac{1}{16} \\ (cos^2x+sin^2x)(cos^4x+sin^4x - sin^2x.cos^2x) = \frac{17}{16} - sin^22x \\ (cos^2x+sin^2x)^2 - 3sin^2x.cos^2x = \frac{17}{16} - sin^22x \\ 1 - \frac{3}{4}.sin^22x = \frac{17}{16} - sin^22x \\ sin^22x = \frac{1}{4}\\ 1 - cos4x = \frac{1}{2} \\ cos4x =\frac{1}{2} [/TEX]
 
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C

cudat196hyno1

^^

cách này là dùng phương pháp hạ bậc kép e mới đk thầy giáo chỉ cho ^^
 
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