Question

三角函数已知向量a为【sinx，3／2】b为【cosx。负1】【1】当两者共线时，求2cos²x减sin2x 。

[2]求[a向量加b向量】乘以b向量在【负二分之π。0】上的值域。。。谢
一共两问啊

Answer 1

sinx*(-1)=3/2*cosx
sinx/cosx=-3/2
tanx=-3/2

2(cosx)^2-sin2x
=cos2x+1-sin2x
=[1-(tanx)^2]/[1+(tanx)^2]-2(tanx)/[1+(tanx)^2]+1
=[1-(tanx)^2-2(tanx)]/[1+(tanx)^2]+1
=[1-9/4+3]/[1+9/4]+1
=(3/4)/(13/4)+1
=3/13+1
=16/13

(a+b)*b
=[sinx+cosx,3/2+(-1)]*b
=[sinx+cosx,1/2]*(cosx,-1）
=(sinx+cosx)*cosx+(-1)*1/2
=(sinx+cosx)*cosx-1/2
=sinxcosx+(cosx)^2-1/2
=1/2*sin2x+(1+cos2x)/2-1/2
=1/2*sin2x+1/2*cos2x
=1/2*(sin2x+cos2x)
=√2/2*(√2/2*sin2x+√2/2*cos2x)
=√2/2*sin(2x+π/4)
x∈[-π/2,0]
2x∈[-π,0]
2x+π/4∈[-3π/4,π/4]
-1<=sin(2x+π/4)<=√2/2
-√2/2<=√2/2sin(2x+π/4)<=1/2
[a向量加b向量】乘以b向量在【负二分之π。0】上的值域:[-√2/2,1/2]

Answer 2

(1)
当两者共线时，sinx*（-1）=3/2*cosx
sinx=-3/2cosx
(sinx)^2=(9/4cosx)^2
(9/4cosx)^2+(cosx)^2=1
(cosx)^2=4/13
2cos²x-sin2x =2cos²x-2cosx*(-3/2cosx)=5cos²x=20/13
(2)
(sinx+cosx,1/2)(cosx,-1)=cos²x+cosx*(-3/2cosx)-1/2=-1/2(cos²x+1)
令t=cosx(0<=t<=1)
y=-1/2(t^2+1)
-1<=y<=-1/2

Answer 3

向量a，b共线
sinx.cosx-3/2=0
所以sin2x=3/4
2cos²x-sin2x=1+cos2x+sin2x=（7加或减根号7）/4