已知tanα=-½,则sin²α-2sinαcosα+1

已知tanα=-½,则sin²α-2sinαcosα+1
已知tanα=-½,则sin²α-2sinαcosα+1

已知tanα=-½,则sin²α-2sinαcosα+1
sin²α-2sinαcosα+1
=(sin²α-2sinαcosα+sin²a+cos²a)/(sin²a+cos²a)
=(2tan²a-2tana+1)/(tan²a+1) 分子分母同时除以cos²a
=(1/2+1+1)/(1/4+1)
=2

解
tana=-1/2
∴sin²a-2sinacosa+1
=sin²a-2sinacosa+cos²a+sin²a
=(2sin²a-2sinacosa+cos²a)/(sin²a+cos²a)
=(2tan²a-2tana+1)/(tan²a+1)
=[2×(-1/2)²-2×(-1/2)+1]/(1/4+1)
=(1/2+1+1)/(5/4)
=5/2×4/5
=2