Question

已知tana=2,求sin^2+sin2a/cos^2a+cos2a和2sin^2a+sinacosa

Answer 1

解：∵tana=sina/cosa=2
∴cosa≠0

sin²a+sin2a/cos²a+cos2a
=(sin²a+2sinacosa)/(cos²a+cos²a-sin²a)
=(sin²a+2sinacosa)/(2cos²a-sin²a) 分子、分母同时乘1/cos²a
=(sin²a/cos²a+2sina/cosa)/(2-sin²a/cos²a)
=(tan²a+2tana)/(2-tan²a)
=(2²+2×2)/(2-2²)
=(4+4)/(2-4)
=8/(-2)
=-4

2sin²a+sinacosa
=(2sin²a+sinacosa)/(sin²a+cos²a) 分子、分母同时乘1/cos²a
=(2sin²a/cos²a+sina/cosa)/(sin²a/cos²a+1)
=(2tan²a+tana)/(tan²a+1)
=(2×2²+2)/(2²+1)
=(8+2)/(4+1)
=2