tan(45°+α)=根号2,则sin2α=

题目

tan(45°+α)=根号2,则sin2α=

答案

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