If u,v and w is a linearly independent system of vectors, examine the system p,q and r, where `p=(cosa)u+(cosb)v+(cosc)w` `q=(sina)u+(sinb)v+(sinc)w` `r=sin(x+a)u+sin(x+b)v+sin(x+c)w` for linearly dependent.

Explanation

Let l, m and n be scalars such that
`lp+mq+nr=0`
`implies{l(cosa)u+(cosb)v+(cosc)w}+m{(sina)u+(sinb)v+(sinc)w}`
`+n{sin(x+a)u+sin(x+b)v+sin(x+c)w}=0`
`implies {lcosa+msina+nsin(x+a)}u+{lsinb+msin(x+b)}v+{lcosc+msinc+nsin(x+v)w=0`
`implies lcosa+msina+nsin(x+a)=0` . .. (i)
`lcosb+msinb+nsin(x+b)=0` . .. (ii)
`lcosc+msinc+nsin(x+c)=0` . . . (iii)
this is a homogeneous system of linear equations in l, m and n.
the determinant of the coefficient matrix is
`Delta=|(cosa,sin,sin(x+a)),(cosb,sinb,sin(x+b)),(cosc,sinc,sin(x+c))|=|(cosa,sina,0),(cosb,sinb,0),(cosc,sinc,0)|=0`
(using `C_(3)toC_(3)-sinxC_(1)-cosxC_(2)`)
`implies`So, the above system of equastions have non-trivial solutions also. this means that l,m and n may attain non-zero values also.
Hence, the given system of vectors is a linearly dependennt system of vectors.