Question

已知a属于(0,兀),且sin(兀/4-a)=根号2/10,则tan2a=

Answer 1

解：∵属于(0,π)
∴sina>0
∵sin(π/4-a)=√2/10
==>(cosa-sina)/√2=√2/10 (应用正弦差角公式)
==>cosa-sina=1/5
==>cosa=sina+1/5
==>(sina+1/5)²+sin²a=1
==>25sin²a+5sina-12=0
==>(5sina-3)(5sina+4)=0
==>5sina-3=0 (∵sina>0)
∴sina=3/5，则cosa=4/5
==>tana=sina/cosa=3/4
故 tan(2a)=2tana/(1-tan²a) (应用正切倍角公式)
=2(3/4)/(1-(3/4)²)
=24/7。