If `x+y+z=pi/2,` then prove that `|[sinx,siny,sinz],[cosx,cosy,cosz],[cos^3x,cos^3 y,cos^3z]|=0`

Explanation

`L.H.S. = |[sinx,siny,sinz],[cosx,cosy,cosz],[cos^3x,cos^3y,cos^3z]|`

By expending the determinant along `R_3`,

`=sum cos^3x[sinycosz - cosysinz]`

`=sum cos^3xsin(y-z)`

As, `x+y+z = pi/2`, so the determinant becomes,

`=sum cos^3(pi/2-(y+z))sin(y-z)`

`=sum sin^3(y+z)sin(y-z)`

`=sum sin^2(y+z) sin(y+z)sin(y-z)`

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