已知向量A=(1,cosx/2)向量B=(根3sinx/2+cosx/2,y)共线且有函数y=f(x)若f(x)=1求cos(2π/3-2x)的值
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![已知向量A=(1,cosx/2)向量B=(根3sinx/2+cosx/2,y)共线且有函数y=f(x)若f(x)=1求cos(2π/3-2x)的值](/uploads/image/z/4335100-52-0.jpg?t=%E5%B7%B2%E7%9F%A5%E5%90%91%E9%87%8FA%3D%281%2Ccosx%2F2%29%E5%90%91%E9%87%8FB%3D%28%E6%A0%B93sinx%2F2%2Bcosx%2F2%2Cy%29%E5%85%B1%E7%BA%BF%E4%B8%94%E6%9C%89%E5%87%BD%E6%95%B0y%3Df%28x%29%E8%8B%A5f%28x%29%3D1%E6%B1%82cos%282%CF%80%2F3-2x%29%E7%9A%84%E5%80%BC)
已知向量A=(1,cosx/2)向量B=(根3sinx/2+cosx/2,y)共线且有函数y=f(x)若f(x)=1求cos(2π/3-2x)的值
已知向量A=(1,cosx/2)向量B=(根3sinx/2+cosx/2,y)共线且有函数y=f(x)若f(x)=1求cos(2π/3-2x)的值
已知向量A=(1,cosx/2)向量B=(根3sinx/2+cosx/2,y)共线且有函数y=f(x)若f(x)=1求cos(2π/3-2x)的值
向量A=(1,cosx/2)向量B=(根3sinx/2+cosx/2,y)共线,则得到两向量之间有如下关系:存在实数k,使得根3sinx/2+cosx/2=k*1;y=k*cos(x/2),则有y=[根3*sin(x/2)+cos(x/2)]*cos(x/2)=
sin(x+pai/6)+1/2,所以f(x)=sin(x+pai/6)+1/2,当f(x)=1时,x+pai/6=pai/6+2k*pai,所以x=2k*pai,此时cos(2π/3-2x)=-1/2
A=(1,cosx/2) B=(根3sinx/2+cosx/2,y)
实数k,使得根3sinx/2+cosx/2=k*1;y=k*cos(x/2),
y=[根3*sin(x/2)+cos(x/2)]*cos(x/2)=sin(x+pai/6)+1/2,
所以f(x)=sin(x+pai/6)+1/2,
当f(x)=1时,x+pai/6=pai/6+2k*pai,所以x=2k*pai,此时cos(2π/3-2x)=-1/2