Question

已知向量a=（sinx，-1），向量b=（cosx，3/2）

（1）当a∥b是求cos²x-3sin2x的值
（2）求f（x）=（a+b）*b的单调递增区间

a、b均为向量

Answer 1

(1)
a//b
=> a=kb
(sinx,-1)=k(cosx, 3/2)
=> sinx = kcosx and -1=3k/2
=> sinx = -(2/3)cosx
tanx = -2/3
cos²x-3sin2x
= 9/13 + 6(2/√13)(3/√13)
=9/13+36/13
=45/13
(2)
f(x)
=(a+b).b
=(sinx+cosx, 1/2).(cosx,3/2)
=sinxcosx +(cosx)^2 + 3/4
=(1/2)sin2x + (cos2x+1)/2 +3/4
= sin(2x+45°) + 5/4
单调递增
n360°-90°<=2x+45°<=n360°+90°
n180°-67.5°<=x<= n180°+22.5°