# Using properties of determinants, prove that: $\begin{vmatrix} sin\alpha&cos\alpha&cos(\alpha+\delta)\\ sin\beta&cos\beta&cos(\beta+\delta)\\ sin\gamma&cos\gamma&cos(\gamma+\delta)) \end{vmatrix}= 0.$

Toolbox:
• If each element of a row(or a column) of a determinant is multiplied by a constant k,then its value gets multiplied by k.
• By this property we can take out any common factor from any one row or any column .
• Elementary transformations can be done by
• 1. Interchanging any two rows or columns.
• 2. Mutiplication of the elements of any row or column by a non-zero number
• 3. The addition of any row or column , the corresponding elemnets of any other row or column multiplied by any non zero number.
$\Delta=\begin{vmatrix} sin\alpha&cos\alpha&cos(\alpha+\delta)\\ sin\beta&cos\beta&cos(\beta+\delta)\\ sin\gamma&cos\gamma&cos(\gamma+\delta)) \end{vmatrix}$

Multiply $C_1$ by $\sin\delta$ and $C_2$ by $\cos\delta$

$\Delta=\frac{1}{sin\delta cos\delta}\begin{vmatrix} sin\alpha sin\delta&cos\alpha cos\delta&cos(\alpha+\delta)\\ sin\beta sin\delta&cos\beta cos\delta&cos(\beta+\delta)\\ sin\gamma sin\delta&cos\gamma cos\delta&cos(\gamma+\delta)) \end{vmatrix}$

We know cos(A+B)=cos Acos B-sin Asin B.

$\Delta=\frac{1}{sin\delta cos\delta}\begin{vmatrix} sin\alpha sin\delta&cos\alpha cos\delta&cos\alpha cos\delta-sin\alpha sin\delta\\ sin\beta sin\delta&cos\beta cos\delta&cos\beta cos\delta-sin\beta sin\delta\\ sin\gamma sin\delta&cos\gamma cos\delta&cos\gamma cos\delta-sin\beta sin\delta \end{vmatrix}$

Apply $C_1\rightarrow C_1+C_3$

$\Delta=\frac{1}{sin\delta cos\delta}\begin{vmatrix} cos\alpha cos\delta&cos\alpha cos\delta&cos\alpha cos\delta-sin\alpha sin\delta\\ cos\beta cos\delta&cos\beta cos\delta&cos\beta cos\delta-sin\beta sin\delta\\ cos\gamma cos\delta&cos\gamma cos\delta&cos\gamma cos\delta-sin\beta sin\delta \end{vmatrix}$

Since two columns $C_1$ and $C_2$ the value of the determinant is 0.

Therefore $\Delta=0.$

Hence proved.

edited Mar 5, 2013