Question

Prove that $\sin \left( -{{420}^{\circ }} \right)\cos \left( {{390}^{\circ }} \right)+\cos \left( -{{660}^{\circ }} \right)\sin \left( {{330}^{\circ }} \right)=-1$ .

Answer 1

Hint:For solving this question, we will use the formulas like $\sin \left( -\theta \right)=-\sin \theta $ to write $\sin \left( -{{420}^{\circ }} \right)=-\sin \left( {{420}^{\circ }} \right)$ and $\cos \left( -\theta \right)=\cos \theta $ to write $\cos \left( -{{660}^{\circ }} \right)=\cos \left( {{660}^{\circ }} \right)$ . After that, we will write the given trigonometric ratios into the form of $\sin {{420}^{\circ }}=\sin \left( 2\pi +{{60}^{\circ }} \right)$ , $\cos {{390}^{\circ }}=\cos \left( 2\pi +{{30}^{\circ }} \right)$ , $\cos {{660}^{\circ }}=\cos \left( 4\pi -{{60}^{\circ }} \right)$ and $\sin {{330}^{\circ }}=\sin \left( 2\pi -{{30}^{\circ }} \right)$ . Then, we will use the formulas like $\sin \left( 2\pi +\theta \right)=\sin \theta $ , $\cos \left( 2\pi +\theta \right)=\cos \theta $ , $\cos \left( 4\pi -\theta \right)=\cos \theta $ and $\sin \left( 2\pi -\theta \right)=-\sin \theta $ to simplify further. And finally, we will use the result of trigonometric ratios like $\sin {{60}^{\circ }}=\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}$ and $\cos {{60}^{\circ }}=\sin {{30}^{\circ }}=\dfrac{1}{2}$ to solve further and prove the desired result easily.

Complete step-by-step answer:
Given:
We have to prove the following equation:
$\sin \left( -{{420}^{\circ }} \right)\cos \left( {{390}^{\circ }} \right)+\cos \left( -{{660}^{\circ }} \right)\sin \left( {{330}^{\circ }} \right)=-1$
Now, we will simplify the term on the left-hand side and prove that it is equal to the term on the right-hand side.
Now, before we proceed we should know the following formulas:
$\begin{align}
  & \sin \left( -\theta \right)=-\sin \theta ...............\left( 1 \right) \\
 & \cos \left( -\theta \right)=\cos \theta .................\left( 2 \right) \\
 & \sin \left( 2n\pi +\theta \right)=\sin \theta ...........\left( 3 \right) \\
 & \cos \left( 2n\pi +\theta \right)=\cos \theta ..........\left( 4 \right) \\
 & \cos \left( 2n\pi -\theta \right)=\cos \theta ...........\left( 5 \right) \\
 & \sin \left( 2n\pi -\theta \right)=-\sin \theta ..........\left( 6 \right) \\
\end{align}$
Now, we will use the above six formulas to simplify the term on the left-hand side.
On the left-hand side, we have $\sin \left( -{{420}^{\circ }} \right)\cos \left( {{390}^{\circ }} \right)+\cos \left( -{{660}^{\circ }} \right)\sin \left( {{330}^{\circ }} \right)$ .
Now, we will use the formula from the equation (1) to write $\sin \left( -{{420}^{\circ }} \right)=-\sin \left( {{420}^{\circ }} \right)$ and formula from the equation (2) to write $\cos \left( -{{660}^{\circ }} \right)=\cos \left( {{660}^{\circ }} \right)$ in the term on the left-hand side. Then,
$\begin{align}
  & \sin \left( -{{420}^{\circ }} \right)\cos \left( {{390}^{\circ }} \right)+\cos \left( -{{660}^{\circ }} \right)\sin \left( {{330}^{\circ }} \right) \\
 & \Rightarrow -\sin \left( {{420}^{\circ }} \right)\cos \left( {{390}^{\circ }} \right)+\cos \left( {{660}^{\circ }} \right)\sin \left( {{330}^{\circ }} \right) \\
\end{align}$
Now, we will write $\sin \left( {{420}^{\circ }} \right)=\sin \left( {{360}^{\circ }}+{{60}^{\circ }} \right)$ , $\cos \left( {{390}^{\circ }} \right)=\cos \left( {{360}^{\circ }}+{{30}^{\circ }} \right)$ , $\cos \left( {{660}^{\circ }} \right)=\cos \left( {{720}^{\circ }}-{{60}^{\circ }} \right)$ and $\sin \left( {{330}^{\circ }} \right)=\sin \left( {{360}^{\circ }}-{{30}^{\circ }} \right)$ in the above term. Then,
$\begin{align}
  & -\sin \left( {{420}^{\circ }} \right)\cos \left( {{390}^{\circ }} \right)+\cos \left( {{660}^{\circ }} \right)\sin \left( {{330}^{\circ }} \right) \\
 & \Rightarrow -\sin \left( {{360}^{\circ }}+{{60}^{\circ }} \right)\cos \left( {{360}^{\circ }}+{{30}^{\circ }} \right)+\cos \left( {{720}^{\circ }}-{{60}^{\circ }} \right)\sin \left( {{360}^{\circ }}-{{30}^{\circ }} \right) \\
\end{align}$
Now, as we know that ${{\pi }^{c}}={{180}^{\circ }}$ so, we can write ${{360}^{\circ }}=2{{\pi }^{c}}$ and ${{720}^{\circ }}=4{{\pi }^{c}}$ in the above term. Then,
$\begin{align}
  & -\sin \left( {{360}^{\circ }}+{{60}^{\circ }} \right)\cos \left( {{360}^{\circ }}+{{30}^{\circ }} \right)+\cos \left( {{720}^{\circ }}-{{60}^{\circ }} \right)\sin \left( {{360}^{\circ }}-{{30}^{\circ }} \right) \\
 & \Rightarrow -\sin \left( 2\pi +{{60}^{\circ }} \right)\cos \left( 2\pi +{{30}^{\circ }} \right)+\cos \left( 4\pi -{{60}^{\circ }} \right)\sin \left( 2\pi -{{30}^{\circ }} \right) \\
\end{align}$
Now, when we put $n=1$ in equation (3), then $\sin \left( 2\pi +\theta \right)=\sin \theta $ so, we can write $\sin \left( 2\pi +{{60}^{\circ }} \right)=\sin {{60}^{\circ }}$ in the above equation. Then,
$\begin{align}
  & -\sin \left( 2\pi +{{60}^{\circ }} \right)\cos \left( 2\pi +{{30}^{\circ }} \right)+\cos \left( 4\pi -{{60}^{\circ }} \right)\sin \left( 2\pi -{{30}^{\circ }} \right) \\
 & \Rightarrow -\sin {{60}^{\circ }}\cos \left( 2\pi +{{30}^{\circ }} \right)+\cos \left( 4\pi -{{60}^{\circ }} \right)\sin \left( 2\pi -{{30}^{\circ }} \right) \\
\end{align}$
Now, when we put $n=1$ in equation (4), then $\cos \left( 2\pi +\theta \right)=\cos \theta $ so, we can write $\cos \left( 2\pi +{{30}^{\circ }} \right)=\cos {{30}^{\circ }}$ in the above term. Then,
$\begin{align}
  & -\sin {{60}^{\circ }}\cos \left( 2\pi +{{30}^{\circ }} \right)+\cos \left( 4\pi -{{60}^{\circ }} \right)\sin \left( 2\pi -{{30}^{\circ }} \right) \\
 & \Rightarrow -\sin {{60}^{\circ }}\cos {{30}^{\circ }}+\cos \left( 4\pi -{{60}^{\circ }} \right)\sin \left( 2\pi -{{30}^{\circ }} \right) \\
\end{align}$
Now, when we put $n=2$ in equation (5), then $\cos \left( 4\pi -\theta \right)=\cos \theta $ so, we can write $\cos \left( 4\pi -{{60}^{\circ }} \right)=\cos {{60}^{\circ }}$ in the above term. Then,
$\begin{align}
  & -\sin {{60}^{\circ }}\cos {{30}^{\circ }}+\cos \left( 4\pi -{{60}^{\circ }} \right)\sin \left( 2\pi -{{30}^{\circ }} \right) \\
 & \Rightarrow -\sin {{60}^{\circ }}\cos {{30}^{\circ }}+\cos {{60}^{\circ }}\sin \left( 2\pi -{{30}^{\circ }} \right) \\
\end{align}$
Now, when we put $n=1$ in equation (6), then $\sin \left( 2\pi -\theta \right)=-\sin \theta $ so, we can write $\sin \left( 2\pi -{{30}^{\circ }} \right)=-\sin {{30}^{\circ }}$ in the above term. Then,
$\begin{align}
  & -\sin {{60}^{\circ }}\cos {{30}^{\circ }}+\cos {{60}^{\circ }}\sin \left( 2\pi -{{30}^{\circ }} \right) \\
 & \Rightarrow -\sin {{60}^{\circ }}\cos {{30}^{\circ }}-\cos {{60}^{\circ }}\sin {{30}^{\circ }} \\
 & \Rightarrow -\left( \sin {{60}^{\circ }}\cos {{30}^{\circ }}+\cos {{60}^{\circ }}\sin {{30}^{\circ }} \right) \\
\end{align}$
Now, as we know that, $\sin {{60}^{\circ }}=\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}$ and $\cos {{60}^{\circ }}=\sin {{30}^{\circ }}=\dfrac{1}{2}$ . Then,
$\begin{align}
  & -\left( \sin {{60}^{\circ }}\cos {{30}^{\circ }}+\cos {{60}^{\circ }}\sin {{30}^{\circ }} \right) \\
 & \Rightarrow -\left( \dfrac{\sqrt{3}}{2}\times \dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\times \dfrac{1}{2} \right) \\
 & \Rightarrow -\left( \dfrac{3}{4}+\dfrac{1}{4} \right) \\
 & \Rightarrow -1 \\
\end{align}$
Now, from the above result, we conclude that the value of the term $\sin \left( -{{420}^{\circ }} \right)\cos \left( {{390}^{\circ }} \right)+\cos \left( -{{660}^{\circ }} \right)\sin \left( {{330}^{\circ }} \right)$ will be equal to $-1$ . Then,
$\sin \left( -{{420}^{\circ }} \right)\cos \left( {{390}^{\circ }} \right)+\cos \left( -{{660}^{\circ }} \right)\sin \left( {{330}^{\circ }} \right)=-1$
Now, from the above result, we conclude that the term on the left-hand side is equal to the term on the right-hand side.
Thus, $\sin \left( -{{420}^{\circ }} \right)\cos \left( {{390}^{\circ }} \right)+\cos \left( -{{660}^{\circ }} \right)\sin \left( {{330}^{\circ }} \right)=-1$ .
Hence, proved.

Note: Here, the student should first understand what we have to prove in the question. After that, we should proceed in a stepwise manner and apply trigonometric formulas like $\sin \left( 2\pi +\theta \right)=\sin \theta $ , $\cos \left( 2\pi +\theta \right)=\cos \theta $ , $\cos \left( 4\pi -\theta \right)=\cos \theta $ and $\sin \left( 2\pi -\theta \right)=-\sin \theta $ correctly. Moreover, we could have also used the formula $\sin \left( A+B \right)=\sin A\cos B+\sin B\cos A$ to write $-\left( \sin {{60}^{\circ }}\cos {{30}^{\circ }}+\cos {{60}^{\circ }}\sin {{30}^{\circ }} \right)=-\sin \left( {{60}^{\circ }}+{{30}^{\circ }} \right)=-\sin {{90}^{\circ }}$ directly and then, used the result $\sin {{90}^{\circ }}=1$ to prove the desired result.