向量a=(-1,cosx),b=(3/2,sinx)
若ab平行
則tanx=-3/2
2cos²x-sin2x=(2cos²x-sin2x)/(sin²x+cos²x)=(2cos²x-cos²xtanx)/(sin²x+cos²x)同除cos²x得
=(2-tanx)/(tan²x+1)=1
求f（x）=（a+b）·b=3/4+cosx*sinx+sin²x=5/4+√2/2sin(2x+π/4)
x∈[-π/2,0],2x+π/4∈【-3π/4,π/4】
則最大值為7/4第二題中間(2cos²x-cos²xtanx)怎么得來(lái)的f（x）=（a+b）·b=3/4+cosx*sinx+sin²x=3/4+1/2sin2x+(1-cos2x)/2=5/4+1/2(sin2x+cos2x)=5/4+√2/2(sin2xcosπ/4+cos2xsinπ/4）=5/4+√2/2sin(2x+π/4)x∈[-π/2,0]，2x+π/4∈【-3π/4,π/4】此步畫(huà)圖后知sin函數(shù)在此區(qū)間單調(diào)遞增知π/2x+π/4=π/4時(shí)取得最大值sin2x=2sinxcosx=2cosxtanxcosx=cos²xtanx2cosxtanxcosx???????=2cos²xtanx?????????????????????????????