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Let f\left ( \theta \right ) = \begin{vmatrix} cos^{2}\theta & cos\theta sin\theta & -sin\theta\\cos\theta sin\theta &sin^{2}\theta &cos\theta \\ sin\theta & -cos\theta &0\end{vmatrix} =0

Then f\left ( \frac{\pi}{6} \right )=

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best_answer

As learned in concept

Value of determinants of order 3 -

-

 

 f(\frac{\pi}{6})= \begin{vmatrix} \frac{3}{4} &\frac{\sqrt3}{4} &\frac{-1}{2} \\ \frac{\sqrt3}{4}& \frac{1}{4} & \frac{\sqrt3}{2}\\ \frac{1}{2}&\frac{-\sqrt3}{2} &0 \end{vmatrix}

=\frac{3}{4}(\frac{3}{4})- \frac{\sqrt3}{4}(\frac{-\sqrt3}{4})-\frac{1}{2}(\frac{-3}{8}- \frac{1}{8})

\frac{9}{16}+\frac{3}{16}+\frac{3}{16}+\frac{1}{16}=1

 

 


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prateek

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