Question

Using properties of determinants prove the following questions.

$\left|\begin{array}{ccc}sin\alpha & cos\alpha & cos(\alpha +\delta )\\ sin\beta & cos\beta & cos(\beta +\delta )\\ sin\gamma & cos\gamma & cos(\gamma +\delta )\end{array}\right|=0$

Answer 1

Let $A=\left|\begin{array}{ccc}sin\alpha & cos\alpha & cos(\alpha +\delta )\\ sin\beta & cos\beta & cos(\beta +\delta )\\ sin\gamma & cos\gamma & cos(\gamma +\delta )\end{array}\right|$
$=\left|\begin{array}{ccc}sin\alpha & cos\alpha & cos\alpha cos\delta -sin\alpha sin\beta \\ sin\beta & cos\beta & cps\beta cos\delta -sin\beta sin\delta \\ sin\gamma & cos\gamma & cos\gamma cos\delta -sin\gamma sin\delta \end{array}\right|$
$[\because cos(A+B)=coaAcosB-sinAsinB]$
$\left|\begin{array}{ccc}sin\alpha & cos\alpha & cos\alpha cos\delta \\ sin\beta & cos\beta & cos\beta cos\delta \\ sin\gamma & cos\gamma & cos\gamma cos\delta \end{array}\right|+\left|\begin{array}{ccc}sin\alpha & cos\alpha & -sin\alpha sin\delta \\ sin\beta & cos\beta & -sin\beta sin\delta \\ sin\gamma & cos\gamma & sin\delta \end{array}\right|$
Taking common $cos\delta$ from $C_3$ in first determinant and taking common $-sin\delta$ form $C_3$ in second deteminant
$=cos\delta \left|\begin{array}{ccc}sin\alpha & cos\alpha & cos\alpha \\ sin\beta & cos\beta & cos\beta \\ sin\gamma & cos\gamma & cos\gamma \end{array}\right|-sin\delta \left|\begin{array}{ccc}sin\alpha & cos\alpha & sin\alpha \\ sin\beta & cos\beta & sin\beta \\ sin\gamma & cos\gamma & sin\gamma \end{array}\right|$
$=cos\delta .0-sin\delta .0=0$
[Since, $C_2$ and $C_3$ in first determinant $C_2$ and $C_3$ in second determinant are identical so values determinants are zero.]