函数=asin2X +btanX+1,f(2)=5,问f (派-2)+f(派)等于多少

问题描述:

函数=asin2X +btanX+1,f(2)=5,问f (派-2)+f(派)等于多少

f(x)=asin2x+btanx +1f(2)=asin4+btan2+1=5,所以 asin4+btan2=4f(π-2)=asin2(π-2)+btan(π-2)+1=asin(2π-4)-btan2+1=-(asin4+btan2)+1=-4+1=-3f(π)=asin2π+btanπ+1=1从而 f(π-2)+f(π)=-3+1=-2