Question

The value of $\sin {{100}^{\circ }}+\sin {{200}^{\circ }}+\sin {{290}^{\circ }}+\sin {{380}^{\circ }}$ is
(A) 0
(B) $-2\cos {{10}^{\circ }}$
(C) $-2\sin {{10}^{\circ }}$
(D) None

Answer 1

Hint: We start solving this question by taking the trigonometric identities such as $\sin \left( {{360}^{\circ }}+x \right)=\sin x$, $\sin \left( {{360}^{\circ }}-x \right)=-\sin x$ and $\sin \left( {{90}^{\circ }}-x \right)=\cos x$. Then we use them to find the values of $\sin {{100}^{\circ }}$, $\sin {{200}^{\circ }}$, $\sin {{290}^{\circ }}$ and $\sin {{380}^{\circ }}$ in terms of $\sin {{10}^{\circ }}$ and $\cos {{10}^{\circ }}$. Then which of the options has the answer that we got and mark it.

Complete step-by-step answer:
Before solving the question, we need to go through some trigonometric identities.
$\begin{align}
  & \sin \left( {{360}^{\circ }}+x \right)=\sin x \\
 & \sin \left( {{360}^{\circ }}-x \right)=-\sin x \\
 & \sin \left( {{180}^{\circ }}+x \right)=-\sin x \\
 & \sin \left( {{180}^{\circ }}-x \right)=\sin x \\
 & \sin \left( {{90}^{\circ }}+x \right)=-\cos x \\
 & \sin \left( {{90}^{\circ }}-x \right)=\cos x \\
\end{align}$
Now let us consider the given expression $\sin {{100}^{\circ }}+\sin {{200}^{\circ }}+\sin {{290}^{\circ }}+\sin {{380}^{\circ }}$.
Let us consider $\sin {{100}^{\circ }}$, we can apply the above discussed identity $\sin \left( {{90}^{\circ }}+x \right)=-\cos x$ to it.
$\sin \left( {{100}^{\circ }} \right)=\sin \left( {{90}^{\circ }}+{{10}^{\circ }} \right)=-\cos {{10}^{\circ }}$
Now let us consider $\sin {{200}^{\circ }}$. Let us apply the property $\sin \left( {{180}^{\circ }}+x \right)=-\sin x$ discussed above to it.
$\sin {{200}^{\circ }}=\sin \left( {{180}^{\circ }}+{{20}^{\circ }} \right)=-\sin {{20}^{\circ }}$
Now let us consider $\sin {{290}^{\circ }}$. Let us apply the property $\sin \left( {{270}^{\circ }}+x \right)=-\cos x$ discussed above to it.
$\sin {{290}^{\circ }}=\sin \left( {{270}^{\circ }}+{{20}^{\circ }} \right)=-\cos {{20}^{\circ }}$
Now let us consider $\sin {{380}^{\circ }}$. Let us apply the property $\sin \left( {{360}^{\circ }}+x \right)=\sin x$ discussed above to it.
$\sin {{380}^{\circ }}=\sin \left( {{360}^{\circ }}+{{20}^{\circ }} \right)=\sin {{20}^{\circ }}$
So, by adding them we can find our required value. So, we get
$\begin{align}
  & \Rightarrow \sin {{100}^{\circ }}+\sin {{200}^{\circ }}+\sin {{290}^{\circ }}+\sin {{380}^{\circ }}=-\cos {{10}^{\circ }}-\sin {{20}^{\circ }}-\cos {{20}^{\circ }}+\sin {{20}^{\circ }} \\
 & \Rightarrow \sin {{100}^{\circ }}+\sin {{200}^{\circ }}+\sin {{290}^{\circ }}+\sin {{380}^{\circ }}=-\cos {{10}^{\circ }}-\cos {{20}^{\circ }} \\
\end{align}$
Hence the value we get is $-\left( \cos {{10}^{\circ }}+\cos {{20}^{\circ }} \right)$.
Now let us consider the formula,
$\cos A+\cos B=2\cos \dfrac{A+B}{2}\cos \dfrac{A-B}{2}$
Using this formula, we can write $-\left( \cos {{10}^{\circ }}+\cos {{20}^{\circ }} \right)$ as
$\begin{align}
  & \Rightarrow -\left( \cos {{10}^{\circ }}+\cos {{20}^{\circ }} \right)=-2\cos \dfrac{{{10}^{\circ }}+{{20}^{\circ }}}{2}\cos \dfrac{{{10}^{\circ }}-{{20}^{\circ }}}{2} \\
 & \Rightarrow -\left( \cos {{10}^{\circ }}+\cos {{20}^{\circ }} \right)=-2\cos {{15}^{\circ }}\cos \left( -{{5}^{\circ }} \right) \\
\end{align}$
As $\cos \left( -\theta \right)=\cos \theta $
Hence the value we get is $-2\cos {{15}^{\circ }}\cos {{5}^{\circ }}$.

So, the correct answer is “Option D”.

Note: The common mistake that one does while solving this type of problem is one might take the trigonometric identities wrong by taking the wrong sign like taking $\sin \left( {{180}^{\circ }}-x \right)=-\sin x$ while the actual one is $\sin \left( {{180}^{\circ }}-x \right)=-\sin x$ or by taking sine instead of cosine like taking the identity as $\sin \left( {{90}^{\circ }}-x \right)=\sin x$ and $\sin \left( {{360}^{\circ }}-x \right)=\sin x$ which are also wrong. So, one needs to be careful while applying the trigonometric identities.