Question

已知tanα=2,0<α<π/2,求下列各式的值 （1)(sinα+2cosα)/(4cosα-sinα)

(2)√2sin（2α+π/4)+1

Answer 1

(sinα+2cosα)/(4cosα-sinα)=(2+tana)/(4-tana)=2
记p=圆周率/4 t=根号2
tana=2 sina=2cosa 2sinacosa=4cosacosa sin2a=4cosacosa
sinasina=4cosacosa=sin2a cosacosa=0.2
tsin(2a+p)+1=t(sin2acosp+cos2asinp)+1=0.5(sin2a+cos2a)+1
=0.5(sinasina+cosacosa-sinasina)+1=0.5cosacosa+1=0.5*0.2+1=1.1