Question

Consider two functions given by \[A=\sin \dfrac{2\pi }{7}+\sin \dfrac{4\pi }{7}+\sin \dfrac{8\pi }{7}\] and \[B=\cos \dfrac{2\pi }{7}+\cos \dfrac{4\pi }{7}+\cos \dfrac{8\pi }{7}\] then \[\sqrt{{{A}^{2}}+{{B}^{2}}}\] is equal to
\[\begin{align}
  & \text{(A) 1} \\
 & \text{(B) }\sqrt{2} \\
 & (\text{C) 2} \\
 & \text{(D) }\sqrt{3} \\
\end{align}\]

Answer 1

Hint: Let us \[A=\sin \dfrac{2\pi }{7}+\sin \dfrac{4\pi }{7}+\sin \dfrac{8\pi }{7}\] as equation (1). Now let us square the equation (1) on both sides. Now by using the formula \[2\sin C\sin D=\cos \left( \dfrac{C-D}{2} \right)-\cos \left( \dfrac{C+D}{2} \right)\] we should solve the problem. While solving the problem, we should also write \[\cos (\pi +\theta )=-\cos \theta \] and \[\cos (\pi -\theta )=-\cos \theta \]. Let us \[B=\cos \dfrac{2\pi }{7}+\cos \dfrac{4\pi }{7}+\cos \dfrac{8\pi }{7}\] as equation (2). Now let us square the equation (2) on both sides. Now by using the formula \[2\sin C\sin D=\cos \left( \dfrac{C-D}{2} \right)-\cos \left( \dfrac{C+D}{2} \right)\] we should solve the problem. While solving the problem, we should also write \[\cos (\pi +\theta )=-\cos \theta \] and \[\cos (\pi -\theta )=-\cos \theta \]. Now we should add equation (1) and equation (2). Now by using the formula \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\], we can find the value of \[\sqrt{{{A}^{2}}+{{B}^{2}}}\].

Complete step-by-step answer:
From the question, we were given that \[A=\sin \dfrac{2\pi }{7}+\sin \dfrac{4\pi }{7}+\sin \dfrac{8\pi }{7}\].
Let us assume \[A=\sin \dfrac{2\pi }{7}+\sin \dfrac{4\pi }{7}+\sin \dfrac{8\pi }{7}.....(1)\].
We know that \[{{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ca\].
Now we squaring on both sides of equation (1).
\[\begin{align}
  & {{A}^{2}}={{\left( \sin \dfrac{2\pi }{7}+\sin \dfrac{4\pi }{7}+\sin \dfrac{8\pi }{7} \right)}^{2}} \\
 & \Rightarrow {{A}^{2}}={{\sin }^{2}}\dfrac{2\pi }{7}+{{\sin }^{2}}\dfrac{4\pi }{7}+{{\sin }^{2}}\dfrac{8\pi }{7}+2\sin \dfrac{4\pi }{7}\sin \dfrac{2\pi }{7}+2\sin \dfrac{8\pi }{7}\sin \dfrac{4\pi }{7}+2\sin \dfrac{8\pi }{7}\sin \dfrac{2\pi }{7} \\
\end{align}\]
We know that \[2\sin C\sin D=\cos \left( \dfrac{C-D}{2} \right)-\cos \left( \dfrac{C+D}{2} \right)\].
\[\Rightarrow {{A}^{2}}={{\sin }^{2}}\dfrac{2\pi }{7}+{{\sin }^{2}}\dfrac{4\pi }{7}+{{\sin }^{2}}\dfrac{8\pi }{7}+\left( \cos \dfrac{2\pi }{7}-\cos \dfrac{6\pi }{7} \right)+\left( \cos \dfrac{4\pi }{7}-\cos \dfrac{12\pi }{7} \right)+\left( \cos \dfrac{6\pi }{7}-\cos \dfrac{10\pi }{7} \right)\]
Now let us write \[\cos \dfrac{12\pi }{7}=\cos \left( \pi +\dfrac{5\pi }{7} \right)\] and \[\cos \dfrac{10\pi }{7}=\cos \left( \pi +\dfrac{3\pi }{7} \right)\].
\[\Rightarrow {{A}^{2}}={{\sin }^{2}}\dfrac{2\pi }{7}+{{\sin }^{2}}\dfrac{4\pi }{7}+{{\sin }^{2}}\dfrac{8\pi }{7}+\left( \cos \dfrac{2\pi }{7}-\cos \left( \dfrac{6\pi }{7} \right) \right)+\left( \cos \dfrac{4\pi }{7}-\cos \left( \pi +\dfrac{5\pi }{7} \right) \right)+\left( \cos \left( \dfrac{6\pi }{7} \right)-\cos \left( \pi +\dfrac{3\pi }{7} \right) \right)\]
We know that \[\cos (\pi +\theta )=-\cos \theta \].
\[\Rightarrow {{A}^{2}}={{\sin }^{2}}\dfrac{2\pi }{7}+{{\sin }^{2}}\dfrac{4\pi }{7}+{{\sin }^{2}}\dfrac{8\pi }{7}+\left( \cos \dfrac{2\pi }{7}-\cos \dfrac{6\pi }{7} \right)+\left( \cos \dfrac{4\pi }{7}+\cos \dfrac{5\pi }{7} \right)+\left( \cos \left( \dfrac{6\pi }{7} \right)+\cos \dfrac{3\pi }{7} \right)\]
We know that
\[\Rightarrow {{A}^{2}}={{\sin }^{2}}\dfrac{2\pi }{7}+{{\sin }^{2}}\dfrac{4\pi }{7}+{{\sin }^{2}}\dfrac{8\pi }{7}+\left( cos\dfrac{2\pi }{7}-\cos \dfrac{6\pi }{7} \right)+\left( \cos \left( \pi -\dfrac{3\pi }{7} \right)+\cos \left( \pi -\dfrac{2\pi }{7} \right) \right)+\left( \cos \dfrac{6\pi }{7}+\cos \left( \dfrac{3\pi }{7} \right) \right)\]
We know that \[\cos (\pi -\theta )=-\cos \theta \].
\[\begin{align}
  & \Rightarrow {{A}^{2}}={{\sin }^{2}}\dfrac{2\pi }{7}+{{\sin }^{2}}\dfrac{4\pi }{7}+{{\sin }^{2}}\dfrac{8\pi }{7}+\left( \cos \dfrac{2\pi }{7}-\cos \dfrac{6\pi }{7} \right)+\left( -\cos \dfrac{3\pi }{7}-\cos \dfrac{2\pi }{7} \right)+\left( \cos \dfrac{6\pi }{7}+\cos \dfrac{3\pi }{7} \right) \\
 & \Rightarrow {{A}^{2}}={{\sin }^{2}}\dfrac{2\pi }{7}+{{\sin }^{2}}\dfrac{4\pi }{7}+{{\sin }^{2}}\dfrac{8\pi }{7}.....(1) \\
\end{align}\]
From the question, we were given that \[B=\cos \dfrac{2\pi }{7}+\cos \dfrac{4\pi }{7}+\cos \dfrac{8\pi }{7}\].
Let us assume \[B=\cos \dfrac{2\pi }{7}+\cos \dfrac{4\pi }{7}+\cos \dfrac{8\pi }{7}.......(2)\]
Now squaring on both sides on equation (2).
We know that \[{{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ca\].
\[\begin{align}
  & {{B}^{2}}={{\left( \cos \dfrac{2\pi }{7}+\cos \dfrac{4\pi }{7}+\cos \dfrac{8\pi }{7} \right)}^{2}} \\
 & \Rightarrow {{B}^{2}}={{\cos }^{2}}\dfrac{2\pi }{7}+{{\cos }^{2}}\dfrac{4\pi }{7}+{{\cos }^{2}}\dfrac{8\pi }{7}+2\cos \dfrac{4\pi }{7}\cos \dfrac{2\pi }{7}+2\cos \dfrac{8\pi }{7}\cos \dfrac{4\pi }{7}+2\cos \dfrac{8\pi }{7}\cos \dfrac{2\pi }{7} \\
\end{align}\]
We know that \[2\cos C\cos D=\cos \left( \dfrac{C+D}{2} \right)+\cos \left( \dfrac{C-D}{2} \right)\].
\[\Rightarrow {{B}^{2}}={{\cos }^{2}}\dfrac{2\pi }{7}+{{\cos }^{2}}\dfrac{4\pi }{7}+{{\cos }^{2}}\dfrac{8\pi }{7}+\left( \cos \dfrac{6\pi }{7}-\cos \dfrac{2\pi }{7} \right)+\left( \cos \dfrac{12\pi }{7}-\cos \dfrac{4\pi }{7} \right)+\left( \cos \dfrac{10\pi }{7}-\cos \dfrac{6\pi }{7} \right)\]
Now let us write \[\cos \dfrac{12\pi }{7}=\cos \left( \pi +\dfrac{5\pi }{7} \right)\] and \[\cos \dfrac{10\pi }{7}=\cos \left( \pi +\dfrac{3\pi }{7} \right)\].
\[\Rightarrow {{B}^{2}}={{\cos }^{2}}\dfrac{2\pi }{7}+{{\cos }^{2}}\dfrac{4\pi }{7}+{{\cos }^{2}}\dfrac{8\pi }{7}+\left( \cos \dfrac{6\pi }{7}-\cos \left( \dfrac{2\pi }{7} \right) \right)+\left( \cos \left( \pi +\dfrac{5\pi }{7} \right)-\cos \dfrac{4\pi }{7} \right)+\left( \cos \left( \pi +\dfrac{3\pi }{7} \right)-\cos \left( \dfrac{6\pi }{7} \right) \right)\]
We know that \[\cos (\pi +\theta )=-\cos \theta \] and \[\cos (\pi -\theta )=-\cos \theta \].
\[\Rightarrow {{B}^{2}}={{\cos }^{2}}\dfrac{2\pi }{7}+{{\cos }^{2}}\dfrac{4\pi }{7}+{{\cos }^{2}}\dfrac{8\pi }{7}+\cos \dfrac{6\pi }{7}-\cos \dfrac{2\pi }{7}-\cos \dfrac{5\pi }{7}-\cos \dfrac{4\pi }{7}-\cos \dfrac{3\pi }{7}-\cos \dfrac{6\pi }{7}\]
Now let us write \[\cos \dfrac{5\pi }{7}=\cos \left( \pi -\dfrac{2\pi }{7} \right)\] and \[\cos \dfrac{4\pi }{7}=\cos \left( \pi -\dfrac{3\pi }{7} \right)\].
\[\Rightarrow {{B}^{2}}={{\cos }^{2}}\dfrac{2\pi }{7}+{{\cos }^{2}}\dfrac{4\pi }{7}+{{\cos }^{2}}\dfrac{8\pi }{7}+\cos \dfrac{6\pi }{7}-\cos \dfrac{2\pi }{7}-\cos \left( \pi -\dfrac{2\pi }{7} \right)-\cos \left( \pi -\dfrac{3\pi }{7} \right)-\cos \dfrac{3\pi }{7}-\cos \dfrac{6\pi }{7}\]
We know that \[\cos (\pi +\theta )=-\cos \theta \] and \[\cos (\pi -\theta )=-\cos \theta \].
\[\begin{align}
  & \Rightarrow {{B}^{2}}={{\cos }^{2}}\dfrac{2\pi }{7}+{{\cos }^{2}}\dfrac{4\pi }{7}+{{\cos }^{2}}\dfrac{8\pi }{7}+\cos \dfrac{6\pi }{7}-\cos \dfrac{2\pi }{7}+\cos \dfrac{2\pi }{7}+\cos \dfrac{3\pi }{7}-\cos \dfrac{3\pi }{7}-\cos \dfrac{6\pi }{7} \\
 & \Rightarrow {{B}^{2}}={{\cos }^{2}}\dfrac{2\pi }{7}+{{\cos }^{2}}\dfrac{4\pi }{7}+{{\cos }^{2}}\dfrac{8\pi }{7}.....(2) \\
\end{align}\]
Now we will add equation (1) and equation (2).
\[\begin{align}
  & \Rightarrow {{A}^{2}}+{{B}^{2}}={{\sin }^{2}}\dfrac{2\pi }{7}+{{\sin }^{2}}\dfrac{4\pi }{7}+{{\sin }^{2}}\dfrac{8\pi }{7}+{{\cos }^{2}}\dfrac{2\pi }{7}+{{\cos }^{2}}\dfrac{4\pi }{7}+{{\cos }^{2}}\dfrac{8\pi }{7} \\
 & \Rightarrow {{A}^{2}}+{{B}^{2}}={{\sin }^{2}}\dfrac{2\pi }{7}+{{\cos }^{2}}\dfrac{2\pi }{7}+{{\sin }^{2}}\dfrac{4\pi }{7}+{{\sin }^{2}}\dfrac{4\pi }{7}+{{\sin }^{2}}\dfrac{8\pi }{7}+{{\cos }^{2}}\dfrac{8\pi }{7} \\
\end{align}\]
We know that \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\].
\[\begin{align}
  & \Rightarrow {{A}^{2}}+{{B}^{2}}=3 \\
 & \Rightarrow \sqrt{{{A}^{2}}+{{B}^{2}}}=\sqrt{3}....(3) \\
\end{align}\]
So, the value of \[\sqrt{{{A}^{2}}+{{B}^{2}}}=\sqrt{3}\].

So, the correct answer is “Option D”.

Note: Students must be careful while solving using the formula mentioned in the above solution. Students may have a misconception that \[\cos (\pi +\theta )=\cos \theta \] and \[\cos (\pi -\theta )=\cos \theta \]. If this misconception is followed, the whole solution will go wrong. So, students should have a correct concept while solving this problem. Students may also assume that \[2\sin C\sin D=\cos \left( \dfrac{C+D}{2} \right)+\cos \left( \dfrac{C-D}{2} \right)\] and \[2\cos C\cos D=\cos \left( \dfrac{C+D}{2} \right)-\cos \left( \dfrac{C-D}{2} \right)\]. This misconception should be avoided to solve this problem. Students can also try to directly add the results below,
\[\begin{align}
  & {{A}^{2}}={{\left( \sin \dfrac{2\pi }{7}+\sin \dfrac{4\pi }{7}+\sin \dfrac{8\pi }{7} \right)}^{2}} \\
 & \Rightarrow {{A}^{2}}={{\sin }^{2}}\dfrac{2\pi }{7}+{{\sin }^{2}}\dfrac{4\pi }{7}+{{\sin }^{2}}\dfrac{8\pi }{7}+2\sin \dfrac{4\pi }{7}\sin \dfrac{2\pi }{7}+2\sin \dfrac{8\pi }{7}\sin \dfrac{4\pi }{7}+2\sin \dfrac{8\pi }{7}\sin \dfrac{2\pi }{7} \\
\end{align}\]
\[\begin{align}
  & {{B}^{2}}={{\left( \cos \dfrac{2\pi }{7}+\cos \dfrac{4\pi }{7}+\cos \dfrac{8\pi }{7} \right)}^{2}} \\
 & \Rightarrow {{B}^{2}}={{\cos }^{2}}\dfrac{2\pi }{7}+{{\cos }^{2}}\dfrac{4\pi }{7}+{{\cos }^{2}}\dfrac{8\pi }{7}+2\cos \dfrac{4\pi }{7}\cos \dfrac{2\pi }{7}+2\cos \dfrac{8\pi }{7}\cos \dfrac{4\pi }{7}+2\cos \dfrac{8\pi }{7}\cos \dfrac{2\pi }{7} \\
\end{align}\]
This method might get difficult to solve as eliminating terms will be difficult and students will find it hard to reach a final answer, so it is advised to proceed with the method in solution.