sin(-11π/6)+cos12π/5·tan3π+cos(-π/4)

问题描述:

sin(-11π/6)+cos12π/5·tan3π+cos(-π/4)

解sin(-11π/6)=sin(2π-11π/6)=sinπ/6=1/2cos12π/5·tan3π=cos12π/5×tanπ=cos12π/5×0=0cos(-π/4)=cos(π/4)=√2/2故sin(-11π/6)+cos12π/5·tan3π+cos(-π/4)=1/2+0+√2/2=(√2+1)/2