-π/2>f(x)+g(x)f(x)-g(x)sin{g(x)}
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-π/2>f(x)+g(x)f(x)-g(x)sin{g(x)}-π/2>f(x)+g(x)f(x)-g(x)sin{g(x)}-π/2>f(x)+g(x)f(x)-g(x)sin{g(x)}∵cos
-π/2>f(x)+g(x)f(x)-g(x)sin{g(x)}
-π/2>f(x)+g(x)<π/2 -π/2>f(x)-g(x)<π/2证cos{f(x)}>sin{g(x)}
-π/2>f(x)+g(x)f(x)-g(x)sin{g(x)}
∵cos{f(x)}-sin{g(x)}=sin{π/2-f(x)}-sin{g(x)}
=2cos{π/2-f(x)+g(x)}/2*sin{π/2-f(x)-g(x)}/2(和差化积)
又-π/2
'(x)又向右平移了π/2个单位。。。综述f(x)的图像先对y轴作对称变换,然后向右平移π/2个单位,就得到g(x)的图像。。
-π/2>f(x)+g(x)f(x)-g(x)sin{g(x)}
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