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Evaluate the value of $\cos \left( {\dfrac{{3\pi }}{2} + \theta } \right)$.
A) $ - \cos \theta $
B) $\cos \theta $
C) \[\sin \theta \]
D) $ - \sin \theta $

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Last updated date: 29th Feb 2024
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Answer
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Hint: We know a formula $\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B$ . By applying this formula we have to expand the $\cos \left( {\dfrac{{3\pi }}{2} + \theta } \right)$ and by putting the known value of $\cos ine$ function we will get the required value.

Complete step-by-step solution:
Here, We have to evaluate the value of $\cos \left( {\dfrac{{3\pi }}{2} + \theta } \right)$.
By applying the above written, formula we can expand $\cos \left( {\dfrac{{3\pi }}{2} + \theta } \right)$ as
$ \Rightarrow \cos \left( {\dfrac{{3\pi }}{2} + \theta } \right) = \cos \dfrac{{3\pi }}{2}\cos \theta - \sin \dfrac{{3\pi }}{2}\sin \theta $
By studying trigonometric value table, we will get the value of $\cos \dfrac{{3\pi }}{2} = 0$ and $\sin \dfrac{{3\pi }}{2} = - 1$. And put these values in the above equation. This will give
$ \Rightarrow \cos \left( {\dfrac{{3\pi }}{2} + \theta } \right) = \left( 0 \right)\cos \theta - \left( { - 1} \right)\sin \theta $
Solving this we get,
$ \Rightarrow \cos \left( {\dfrac{{3\pi }}{2} + \theta } \right) = 0 + \sin \theta $
$\therefore \cos \left( {\dfrac{{3\pi }}{2} + \theta } \right) = \sin \theta $
Thus, the required value of $\cos \left( {\dfrac{{3\pi }}{2} + \theta } \right)$ is equal to $\sin \theta $.

Hence, Option (C) is correct for this question.

Note: Alternatively, this problem can be solved by writing the trigonometric expression $\cos \left( {\dfrac{{3\pi }}{2} + \theta } \right)$ as $\cos \left( {\pi + \left( {\dfrac{\pi }{2} + \theta } \right)} \right)$ because adding $\pi $and $\left( {\dfrac{\pi }{2} + \theta } \right)$ we get the same value as $\left( {\dfrac{{3\pi }}{2} + \theta } \right)$. it is clearly visible that this angle lies in the third quadrant and we know that the value of $\cos ine$ is negative in the third quadrant so, the value of$\cos \left( {\dfrac{{3\pi }}{2} + \theta } \right) = - \cos \left( {\dfrac{\pi }{2} + \theta } \right)$. And the angle $\left( {\dfrac{\pi }{2} + \theta } \right)$ is in the second quadrant of coordinate system and we know that the value of $\cos ine$ function is also negative in the second quadrant. So, the value of $\cos \left( {\dfrac{\pi }{2} + \theta } \right) = - sin\theta $. Thus, the required value of $\cos \left( {\dfrac{{3\pi }}{2} + \theta } \right)$ is equal to $\sin \theta $.