题目不完整 ,应该是f'(pi/4）的值吧
令f'(pi/4）=t
则f(x)=t*cosx+sinx
f'(x)=cosx-t*sinx
f'(pi/4）=cos(pi/4)-t*sin(pi/4)
即t=cos(pi/4)-t*sin(pi/4)
解得t=f'(pi/4）=根号2 - 1