Question

若函数f(x)=e^xsinx，则此函数图像在点（4，f(4))处的切线的倾斜角为 A.π/2 B.0 C.钝角 D。锐角

f'x=e^xsinx+e^xcosx

在（4，f(4))处的切线的斜率是e^4(sin4+cos4)

设倾斜角为a，则有tana=e^4(sin4+cos4)

a=arctan[e^4(sin4+cos4)]
后面怎解？

Answer 1

tana=e^4(sin4+cos4)=tana=
=(根号2)e^4(sin[4+(pi)/4]
=(根号2)e^4(sin[pi-4-(pi)/4]
-(pi)/2<[pi-4-(pi)/4]<0
故tana<0
选 C.钝角

Answer 2

tana=e^4(sin4+cos4)=e^4sin(4+π/4)√2
4+π/4≈3π/2
sin(4+π/4)≈-1
e^4√2>1
∴tana<-1
所以是钝角.