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

& \text{(A) 1} \\\ & \text{(B) }\sqrt{2} \\\ & (\text{C) 2} \\\ & \text{(D) }\sqrt{3} \\\ \end{aligned}$$

Answer 1

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).

& {{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{aligned}$$ We know that $$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{aligned} & \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{aligned}$$ From the question, we were given that $$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{aligned} & {{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{aligned}$$ We know that $$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{aligned} & \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{aligned}$$ Now we will add equation (1) and equation (2). $$\begin{aligned} & \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{aligned}$$ We know that $${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$$. $$\begin{aligned} & \Rightarrow {{A}^{2}}+{{B}^{2}}=3 \\\ & \Rightarrow \sqrt{{{A}^{2}}+{{B}^{2}}}=\sqrt{3}....(3) \\\ \end{aligned}$$ So, the value of $$\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{aligned} & {{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{aligned}$$ $$\begin{aligned} & {{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{aligned}$$ 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.