已知sin^2a+sin^2b+sin^2c=1(a、b、c为锐角),则cosa*cosb*cosc的最大值高二的不等式题
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已知sin^2a+sin^2b+sin^2c=1(a、b、c为锐角),则cosa*cosb*cosc的最大值高二的不等式题已知sin^2a+sin^2b+sin^2c=1(a、b、c为锐角),则cos
已知sin^2a+sin^2b+sin^2c=1(a、b、c为锐角),则cosa*cosb*cosc的最大值高二的不等式题
已知sin^2a+sin^2b+sin^2c=1(a、b、c为锐角),则cosa*cosb*cosc的最大值
高二的不等式题
已知sin^2a+sin^2b+sin^2c=1(a、b、c为锐角),则cosa*cosb*cosc的最大值高二的不等式题
cos^2a*cos^2b*cos^2c
≤((cos^2a+cos^2b+cos^2c)/3)^3
=((1-sin^2a+1-sin^2b+1-sin^2c)/3)^3
=8/27
故cosa*cosb*cosc≤根号下(8/27)
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