若tanα=√2,求值：(1-sin∧6x-cos∧6x)/(1-sin∧4x-cos∧4x)
问题描述：
若tanα=√2,求值：(1-sin∧6x-cos∧6x)/(1-sin∧4x-cos∧4x)
答
3C=2A+B
=2(a2+b2-c2)+(-4a2+2b2+3c2)
=2a2+2b2-2c2-4a2+2b2+3c2
=-2a2+4b2+c2
所以C=-2a2/3+4b2/3+c2/3
10351035
答
sin^6x+cos^6x=(sin^2x+cos^2x)(sin^4x-sin^2xcos^2x+cos^4x)
=sin^4x-sin^2xcos^2x+cos^4x
=(sin^2x+cos^2x)^2-3sin^2xcos^2x
=1-3sin^2xcos^2x
所以分子=3sin^2xcos^2x
sin^4x+cos^4x=(sin^2x+cos^2x)^2-2sin^2xcos^2x
=1-2sin^2xcos^2x
所以分母=2sin^2xcos^2x
所以原式=3/2