[LTDH] phương trình lượng giác

K

kakashi_hatake

$ (sin^3. \dfrac{x}{2} + \dfrac{1}{sin^3. \dfrac{x}{2}})^2 + (cos^3. \dfrac{x}{2} + \dfrac{1}{cos^3. \dfrac{x}{2}})^2 = sin^6x+cos^6x+ \dfrac{1}{sin^6x} + \dfrac{1}{cos^6x} + 4 = (sin^6x+cos^6x)(1+\dfrac{1}{sin^6x.cos^6x})+4 = (1-3sin^2x.cos^2x)(1+\dfrac{1}{sin^6x.cos^6x})+4$
Mà $ sin^2x.cos^2x \le ( \dfrac{sin^2x+cos^2x}{2})^2= \dfrac{1}{4}$
$ sin^6 x.cos^6 x \le \dfrac{1}{4^3}$
Suy ra $ (1-3sin^2 x.cos^2 x)(1+ \dfrac{1}{sin^6 x. cos^6 x})+4 \ge (1-3. \dfrac{1}{4})(1+4^3)+4= \dfrac{81}{4}$
Dấu đẳng thức xảy ra khi $ sin^2x=cos^2x$
 
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