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
3个回答
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
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