Question

已知向量a=（sinx-cosx,2cosx),b=(sinx+cosx,sinx). （1）若a⊥b,求tan2x的值

Answer 1

a⊥b x1x2+y1y2=0

即 (sinx-cosx)(sinx+cosx) + 2cosxsinx = 0

sin^2 (x) - cos^2 (x) + 2cosxsinx = 0

-cos(2x) + sin(2x) = 0

cos(2x) = sin(2x)

tan(2x) = sin(2x)/cos(2x) = 1

a*b = x1x2 + y1y2 = 3/5

即 (sinx-cosx)(sinx+cosx) + 2cosxsinx = 3/5

sin(2x) - cos(2x) = 3/5

[sin(2x) - cos(2x) ]^2 = 9/25

sin^2(2x) + cos^2(2x) - 2sin(2x)cos(2x) = 9/25

1 - sin(4x) = 9/25

sin(4x) = 16/25