Questions & Answers
Question
Answers

Find the value of \[\sin {{330}^{\circ }}\cos {{120}^{\circ }}+\cos {{210}^{\circ }}\sin {{300}^{\circ }}\].

Answer Verified Verified
Hint: Let us assume the value of \[\sin {{330}^{\circ }}\cos {{120}^{\circ }}+\cos {{210}^{\circ }}\sin {{300}^{\circ }}\] is equal to I. Now we should first express 330 in terms of \[2\pi -\theta \]. We know that \[\sin \left( n\pi -\theta \right)=-\sin \theta \] if n is even. Now by using this concept, we will find the value of \[\sin {{330}^{\circ }}\]. Now we should express 120 in terms of \[\pi -\theta \]. We know that \[\sin \left( n\pi -\theta \right)=-\sin \theta \] if n is even. Now by using this concept, we will find the value of \[\cos {{120}^{\circ }}\]. Now we should express 210 in terms of \[\pi +\theta \]. We know that \[\cos \left( n\pi +\theta \right)=-\cos \theta \] if n is odd. Now by using this concept, we will find the value of \[\cos {{210}^{\circ }}\]. Now we should first express 300 in terms of \[2\pi -\theta \]. We know that \[\sin \left( n\pi -\theta \right)=-\sin \theta \] if n is even. Now by using this concept, we will find the value of \[\sin {{300}^{\circ }}\]. In this way, we can find the value of \[\sin {{330}^{\circ }}\cos {{120}^{\circ }}+\cos {{210}^{\circ }}\sin {{300}^{\circ }}\].

Complete step-by-step answer:
From the question, it is clear that we should find the value of \[\sin {{330}^{\circ }}\cos {{120}^{\circ }}+\cos {{210}^{\circ }}\sin {{300}^{\circ }}\].
Let us assume the value of \[\sin {{330}^{\circ }}\cos {{120}^{\circ }}+\cos {{210}^{\circ }}\sin {{300}^{\circ }}\] is equal to I.
\[I=\sin {{330}^{\circ }}\cos {{120}^{\circ }}+\cos {{210}^{\circ }}\sin {{300}^{\circ }}.....(1)\]
Now we have to find the value of \[\sin {{330}^{\circ }}\].
We know that \[\sin \left( n\pi -\theta \right)=-\sin \theta \] if n is even.
\[\begin{align}
  & \Rightarrow sin{{330}^{\circ }}=\sin \left( 2\pi -{{30}^{\circ }} \right) \\
 & \Rightarrow \sin {{330}^{\circ }}=-\sin {{30}^{\circ }} \\
\end{align}\]
We know that \[\sin {{30}^{\circ }}=\dfrac{1}{2}\].
\[\Rightarrow \sin {{330}^{\circ }}=-\dfrac{1}{2}......(2)\]
Now we should find the value of \[\cos {{120}^{\circ }}\].
We know that \[\cos \left( n\pi -\theta \right)=-\cos \theta \] if n is odd.
\[\begin{align}
  & \Rightarrow \cos {{120}^{\circ }}=\cos \left( \pi -{{60}^{\circ }} \right) \\
 & \Rightarrow \cos {{120}^{\circ }}=-\cos \left( {{60}^{\circ }} \right) \\
\end{align}\]
We know that \[\cos {{60}^{\circ }}=\dfrac{1}{2}\].
\[\Rightarrow \cos {{120}^{\circ }}=-\dfrac{1}{2}.......(3)\]
Now we should find the value of \[\cos {{210}^{\circ }}\].
We know that \[\cos \left( n\pi +\theta \right)=-\cos \theta \] if n is odd.
\[\begin{align}
  & \Rightarrow \cos {{210}^{\circ }}=\cos \left( \pi +{{30}^{\circ }} \right) \\
 & \Rightarrow \cos {{210}^{\circ }}=-\cos \left( {{30}^{\circ }} \right) \\
\end{align}\]
We know that \[\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}\].
\[\Rightarrow \cos {{210}^{\circ }}=-\dfrac{\sqrt{3}}{2}.......(4)\]
Now we have to find the value of \[\sin {{300}^{\circ }}\].
We know that \[\sin \left( n\pi -\theta \right)=-\sin \theta \] if n is even.
\[\begin{align}
  & \Rightarrow sin{{300}^{\circ }}=\sin \left( 2\pi -{{60}^{\circ }} \right) \\
 & \Rightarrow \sin {{300}^{\circ }}=-\sin {{60}^{\circ }} \\
\end{align}\]
We know that \[\sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2}\].
\[\Rightarrow \sin {{300}^{\circ }}=-\dfrac{\sqrt{3}}{2}......(5)\]
Now let us substitute equation (2), equation (3), equation (4) and equation (5) in equation (1), then we get
\[\begin{align}
  & I=\sin {{330}^{\circ }}\cos {{120}^{\circ }}+\cos {{210}^{\circ }}\sin {{300}^{\circ }} \\
 & \Rightarrow I=\left( \dfrac{-1}{2} \right)\left( \dfrac{-1}{2} \right)+\left( \dfrac{-\sqrt{3}}{2} \right)\left( \dfrac{-\sqrt{3}}{2} \right) \\
 & \Rightarrow I=\dfrac{1}{4}+\dfrac{3}{4} \\
 & \Rightarrow I=1.....(6) \\
\end{align}\]
From equation (6), it is clear that the value of \[\sin {{330}^{\circ }}\cos {{120}^{\circ }}+\cos {{210}^{\circ }}\sin {{300}^{\circ }}\] is equal to 1.

Note: Students may have a misconception that \[\sin \left( n\pi -\theta \right)=\sin \theta \] if n is even. If this misconception is followed, then the final answer may get interrupted. In the same way, students may have a misconception that \[\cos \left( n\pi +\theta \right)=\cos \theta \] if n is odd. If even this misconception is followed, then also the final answer will get interrupted. So, these misconceptions should get avoided.