Question

How do you simplify $\cos \left( x+\dfrac{3\pi }{2} \right)$ ?

Answer 1

Hint: To simplify $\cos \left( x+\dfrac{3\pi }{2} \right)$, we are going to use the following trigonometric property: $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$. In this formula, we will substitute “A” as $''x''$ and “B” as $\dfrac{3\pi }{2}$. After substituting these values, we will encounter $\cos \dfrac{3\pi }{2}\And \sin \dfrac{3\pi }{2}$ of which we will use the following values: $\cos \dfrac{3\pi }{2}=0\And \sin \dfrac{3\pi }{2}=-1$.

Complete step by step answer:
The cosine expression given above is as follows:
$\cos \left( x+\dfrac{3\pi }{2} \right)$
The above cosine expression is in the form of $\cos \left( A+B \right)$ and we know that the expansion of $\cos \left( A+B \right)$ is equal to:
$\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$
Using the above identity to expand $\cos \left( x+\dfrac{3\pi }{2} \right)$ we get,
$\cos \left( x+\dfrac{3\pi }{2} \right)=\cos x\cos \left( \dfrac{3\pi }{2} \right)-\sin x\sin \left( \dfrac{3\pi }{2} \right)$
In the above equation, we are going to substitute the following values of $\cos \dfrac{3\pi }{2}\And \sin \dfrac{3\pi }{2}$ is equal to:
$\cos \dfrac{3\pi }{2}=0,\sin \dfrac{3\pi }{2}=-1$
$\begin{align}
  & \cos \left( x+\dfrac{3\pi }{2} \right)=\cos x\left( 0 \right)-\sin x\left( -1 \right) \\
 & \Rightarrow \cos \left( x+\dfrac{3\pi }{2} \right)=0+\sin x \\
 & \Rightarrow \cos \left( x+\dfrac{3\pi }{2} \right)=\sin x \\
\end{align}$
From the above solution, we have evaluated the expression $\cos \left( x+\dfrac{3\pi }{2} \right)$ as $\sin x$.

Hence, the simplification of $\cos \left( x+\dfrac{3\pi }{2} \right)$ is equal to $\sin x$.

Note: The another way of approaching the above problem is as follows:
The expression given above is as follows:
$\cos \left( x+\dfrac{3\pi }{2} \right)$
We know that, $\cos \left( \dfrac{3\pi }{2}+\theta \right)=\sin \theta $ from the trigonometric relations so using this relation in the above expression we get,
$\sin x$
If you remember these trigonometric relations with odd multiples of $\dfrac{\pi }{2}$ being added to the angle then your time will be saved in the examination.
In the below, we have shown the different odd multiples of $\dfrac{\pi }{2}$ in cosine:
$\begin{align}
  & \cos \left( \dfrac{\pi }{2}+x \right)=-\sin x \\
 & \cos \left( \dfrac{3\pi }{2}+x \right)=\sin x \\
\end{align}$
In the above, the first equation has negative sign because the angle of cosine lies in the second quadrant and we know that cosine is negative in the second quadrant. And in the second equation, there is no negative sign before $\sin x$ because the angle of cosine is in the fourth quadrant and cosine is positive in the fourth quadrant.
You may also be confused with the below trigonometric relations:
$\begin{align}
  & \cos \left( \pi +x \right)=-\cos x \\
 & \cos \left( 2\pi +x \right)=\cos x \\
\end{align}$
When we add multiples of $\pi $ in the angles of cosine then no change in trigonometric function has occurred which you can see from the above that cosine remains cosine.
While when we add odd multiples of $\dfrac{\pi }{2}$ in the angles of cosine then the trigonometric function has changed from cosine to sine.