由倍角公式:
tana
=(2tana/2)/[1+(tana/2)^2]
=(2*2)/(1+2^2)
=4/5
再由tan的和差化积公式
tan(a+π/4)
=(tana+tanπ/4)/(1-tanatanπ/4)
=(4/5+1)/(1-4/5)
=9
即tan(a+π/4)=9.
由倍角公式:
tana
=(2tana/2)/[1+(tana/2)^2]
=(2*2)/(1+2^2)
=4/5
再由tan的和差化积公式
tan(a+π/4)
=(tana+tanπ/4)/(1-tanatanπ/4)
=(4/5+1)/(1-4/5)
=9
即tan(a+π/4)=9.