已知fx=asinx+btanx+1满足f(pia/5)=7
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已知fx=asinx+btanx+1满足f(pia/5)=7已知fx=asinx+btanx+1满足f(pia/5)=7已知fx=asinx+btanx+1满足f(pia/5)=7最值相减=2A所以A
已知fx=asinx+btanx+1满足f(pia/5)=7
已知fx=asinx+btanx+1满足f(pia/5)=7
已知fx=asinx+btanx+1满足f(pia/5)=7
最值相减=2A
所以A=3/2
则最大=3/2+k=7/3
k=5/6
同一周期,最大最小相差半个周期
所以T=4π=2π/ω
ω=1/2
y=3/2*sin(x/2+φ)+5/6
x=5π/3
y=3/2*sin(5π/6+φ)+5/6=7/3
sin(5π/6+φ)=1
5π/6+φ=π/2
φ=-π/3
y=3/2*sin(x/2-π/3)+5/6
f(π/5)=asinπ/5+btanπ/5+1=7
asinπ/5+btanπ/5=6
所以f(-π/5)=asin(-π/5)+btan(-π/5)+1
=-(asinπ/5+btanπ/5)+1
=-6+1
=-5
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