Question

tan(45°+α)=根号2,则sin2α=

Answer 1

tan(45°+α)=（tan45°+tanα）/（1-tan45°tanα）=(1+tanα)/(1-tanα)=√2

√2（1-tanα)=1+tanα

√2tanα+tanα=√2-1

tanα=（√2-1）/（√2+1）=3-2√2

sin2α=2tanα/(1+tan²α)=（6-4√2）/（1+17-12√2）=1/3[（6-4√2）/（6-4√2）]=1/3