-π/2>f(x)+g(x)f(x)-g(x)sin{g(x)}
问题描述:
-π/2>f(x)+g(x)f(x)-g(x)sin{g(x)}
答
∵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