f(x)=αsinx+btanx+1
f(5)=asin5+btan5+1=7
asin5+btan5=6
你的题目有问题：是不是求已知f(x)=asinx+btanx+1满足f(兀/5)=7.求f(99兀/5)
f(99兀/5)=f(20兀-兀/5)=asin(20兀-兀/5)+btan(20兀-兀/5)+1
=-6+1=-5