已知f(x)=asinx+btanx+1满足f(兀/5)=7.求f(99兀/5)

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已知f(x)=asinx+btanx+1满足f(兀/5)=7.求f(99兀/5)

已知f(x)=asinx+btanx+1满足f(兀/5)=7.求f(99兀/5)
已知f(x)=asinx+btanx+1满足f(兀/5)=7.求f(99兀/5)

已知f(x)=asinx+btanx+1满足f(兀/5)=7.求f(99兀/5)
asin(兀/5)+btan(兀/5)=7-1=6
f(99兀/5)=f(20兀-兀/5)=asin(20兀-兀/5)+btan(20兀-兀/5)+1
=-6+1=-5

f(99兀/5)=f(20兀-兀/5)
=asin(20兀-兀/5)+btan(20兀-兀/5)+1
=asin(-兀/5)+btan(-兀/5)+1
=-[asin(-兀/5)+btan(-兀/5)+1]+2
=-f(兀/5)+2
=-7+2=-5

sinx,tanx都是周期函数,则
f(x+2兀)=f(x)
99兀/5=20兀-兀/5
f(-兀/5+20兀)=f(-兀/5)
f(-兀/5)=-asin(兀/5)-btan(兀/5)+1
=-f(兀/5)+2=-5
所以f(99兀/5)=f(-兀/5)=-5