Question

已知tan=-4/3求2sin平方a+sinacosa减3cos平方a

Answer 1

∵tana=tan(π+a)=-4/3,

∴2sin²a+sinacosa-3cos²a=(2sin²a+sinacosa-3cos²a)/(sin²a+cos²a)

=(2tan²a+tana-3)/(tan²a+1)

=[2×(-4/3)²-4/3-3]/[(-4/3)²+1]

=-7/25.