Question

Solve the equation ${{z}^{7}}+1=0$ then
(a) $\cos \dfrac{\pi }{7}\cos \dfrac{3\pi }{7}\cos \dfrac{5\pi }{7}=-\dfrac{1}{8}$
(b) $\cos \dfrac{\pi }{14}\cos \dfrac{3\pi }{14}\cos \dfrac{5\pi }{14}=\dfrac{\sqrt{\pi }}{8}$
(c) $\sin \dfrac{\pi }{14}\sin \dfrac{3\pi }{14}\sin \dfrac{5\pi }{14}=\dfrac{1}{8}$
(d) ${{\tan }^{2}}\dfrac{\pi }{14}+{{\tan }^{2}}\dfrac{3\pi }{14}+{{\tan }^{2}}\dfrac{5\pi }{14}=5$

Answer 1

For solving this question you should know about the solving equations and then use these values for making new Functions. In this question first we will find the exponential values and then by using the values of the difference of z will calculate the trigonometric functions.

Complete step-by-step solution:
According to the question we have to solve the equation ${{z}^{7}}+1=0$ and then we have to deduce trigonometric equations which are given.
So, if we take our question then the equation is ${{z}^{7}}+1=0$.
So, we can write it as ${{z}^{7}}=-1\Rightarrow z={{\left( -1 \right)}^{{1}/{7}\;}}$

& \Rightarrow {{z}^{7}}=\left( -1 \right){{e}^{\arg \left( -1 \right)i}} \\\ & \Rightarrow {{z}^{7}}=1\cdot {{e}^{\pi i}} \\\ & \Rightarrow {{z}^{7}}={{e}^{\pi i}} \\\ & \Rightarrow z={{\left( {{e}^{\left( \pi +2\pi k \right)i}} \right)}^{\dfrac{1}{7}}} \\\ & \Rightarrow z={{e}^{\dfrac{1}{7}\left( \pi +2\pi k \right)i}} \\\ \end{aligned}$$ So, we can write all the values of z till $${{7}^{th}}$$ power as $$\begin{aligned} & {{z}_{0}}={{e}^{\dfrac{1}{7}\left( \pi +2\pi \times 0 \right)i}}={{e}^{\dfrac{i\pi }{7}}} \\\ & {{z}_{1}}={{e}^{\dfrac{1}{7}\left( \pi +2\pi \times 1 \right)i}}={{e}^{\dfrac{3i\pi }{7}}} \\\ & {{z}_{2}}={{e}^{\dfrac{1}{7}\left( \pi +2\pi \times 2 \right)i}}={{e}^{\dfrac{5i\pi }{7}}} \\\ & {{z}_{3}}={{e}^{\dfrac{1}{7}\left( \pi +2\pi \times 3 \right)i}}={{e}^{\dfrac{7i\pi }{7}}} \\\ & {{z}_{4}}={{e}^{\dfrac{1}{7}\left( \pi +2\pi \times 4 \right)i}}={{e}^{\dfrac{9i\pi }{7}}} \\\ & {{z}_{5}}={{e}^{\dfrac{1}{7}\left( \pi +2\pi \times 5 \right)i}}={{e}^{\dfrac{11i\pi }{7}}} \\\ & {{z}_{6}}={{e}^{\dfrac{1}{7}\left( \pi +2\pi \times 6 \right)i}}={{e}^{\dfrac{13i\pi }{7}}} \\\ \end{aligned}$$ Therefore, we can further write a few terms as $${{z}_{0}}.{{z}_{1}}={{e}^{\dfrac{i\pi }{7}}}.{{e}^{\dfrac{i3\pi }{7}}}={{e}^{i\left( \dfrac{3\pi }{7}+\dfrac{\pi }{7} \right)}}={{e}^{\dfrac{i4\pi }{7}}}-\left( 1 \right)$$ $${{z}_{0}}-{{z}_{6}}={{e}^{\dfrac{i\pi }{7}}}-{{e}^{\dfrac{i3\pi }{7}}}=2\sin \dfrac{\pi }{7}-\left( 2 \right)$$ $${{z}_{1}}-{{z}_{5}}=2\sin \dfrac{3\pi }{7}-\left( 3 \right)$$ And $${{z}_{2}}-{{z}_{4}}=2\sin \dfrac{5\pi }{7}-\left( 4 \right)$$ Now, if we multiply to the equations (2), (3), and (4) then, we get the expressions as $$\begin{aligned} & \left( {{z}_{0}}-{{z}_{6}} \right)\left( {{z}_{1}}-{{z}_{5}} \right)\left( {{z}_{2}}-{{z}_{4}} \right)=8\sin \dfrac{\pi }{7}\sin \dfrac{3\pi }{7}\sin \dfrac{5\pi }{7} \\\ & \left( {{z}_{0}}{{z}_{1}}-{{z}_{0}}{{z}_{5}}-{{z}_{6}}{{z}_{1}}+{{z}_{6}}{{z}_{5}} \right)\left( {{z}_{2}}-{{z}_{4}} \right)=8\sin \dfrac{\pi }{7}\sin \dfrac{3\pi }{7}\sin \dfrac{5\pi }{7} \\\ \end{aligned}$$ Now, if we solve this then it will be equal to -1 $$\begin{aligned} & \Rightarrow -1=8\sin \dfrac{\pi }{7}\sin \dfrac{3\pi }{7}\sin \dfrac{5\pi }{7} \\\ & \Rightarrow \sin \dfrac{\pi }{7}\sin \dfrac{3\pi }{7}\sin \dfrac{5\pi }{7}=\dfrac{-1}{8} \\\ \end{aligned}$$ Now, we also know that $${{z}_{0}}+{{z}_{6}}={{e}^{i\dfrac{\pi }{7}}}+{{e}^{-i\dfrac{\pi }{7}}}=2\cos \dfrac{\pi }{7}$$ And similarly, we can say that $${{z}_{1}}+{{z}_{5}}=2\cos \dfrac{3\pi }{7}$$ $${{z}_{2}}+{{z}_{4}}=2\cos \dfrac{5\pi }{7}$$ According to the question, we know that $${{z}^{7}}=-1$$ $${{z}^{7}}=\cos \pi +i\sin \pi $$ And by the concept of complex numbers, we can write $$\begin{aligned} & \Rightarrow {{z}^{7}}=\cos \left( 4n+2 \right)\pi +i\sin \left( 4n+2 \right)\pi \\\ & \Rightarrow z={{\left[ \cos \left( 4n+2 \right)\pi +i\sin \left( 4n+2 \right)\pi \right]}^{\dfrac{1}{7}}} \\\ & \Rightarrow z=\left[ \cos \left( \dfrac{4n+2}{7} \right)\pi +i\sin \left( 4n+2 \right)\dfrac{\pi }{7} \right] \\\ \end{aligned}$$ Or we can solve it as: $$\begin{aligned} & \cos \dfrac{\pi }{7}\cos \dfrac{3\pi }{7}\cos \dfrac{5\pi }{7} \\\ & =\left( -\cos \dfrac{3\pi }{7} \right)\left( -\cos \dfrac{4\pi }{7} \right)\left( -\cos \dfrac{2\pi }{7} \right) \\\ & =-\cos \dfrac{2\pi }{7}.\cos \dfrac{4\pi }{7}.\cos \dfrac{8\pi }{7} \\\ \end{aligned}$$ $$={-2\sin \dfrac{2\pi }{7}\cos \dfrac{2\pi }{7}\cos \dfrac{4\pi }{7}}/{\left( 2\sin \left( 2\pi /7 \right) \right)}\;$$ $$\begin{aligned} & =\dfrac{-2\times \sin {4\pi }/{7}\;.\cos {4\pi }/{7}\;\cos {8\pi }/{7}\;}{2\times 2\sin {2\pi }/{7}\;} \\\ & =\dfrac{-2\times \sin {3\pi }/{7}\;\cos {8\pi }/{7}\;}{2\times 4\sin {2\pi }/{7}\;} \\\ & =\dfrac{-\sin {16\pi }/{7}\;}{8\sin {2\pi }/{7}\;}=-\dfrac{1}{8} \\\ \end{aligned}$$ **So, option (a) is the correct option.** **Note:** For solving this question you should be careful for finding the value of $${{z}_{n}}$$. And then all $${{z}_{n}}$$ values will add, subtract, multiply with each other and these will provide us the exponential values to trigonometric values. And then solve this very much carefully.