Question

The exact value of $\cos \dfrac{2\pi }{28}\csc \dfrac{3\pi }{28}+\cos \dfrac{6\pi }{28}\csc \dfrac{9\pi }{28}+\cos \dfrac{18\pi }{28}\csc \dfrac{27\pi }{28}$ is?

Answer 1

Hint:For solving this question we will assume $\dfrac{\pi }{28}=\theta $ and simply the given term with the help of trigonometric formulas like $2\cos A\cos B=\cos \left( A+B \right)+\cos \left( A-B \right)$ , $\sin \left( \dfrac{\pi }{2}-\theta \right)=\cos \theta $ , $2\sin A\cos B=\sin \left( A+B \right)+\sin \left( A-B \right)$ , $\sin \left( \pi -\theta \right)=\sin \theta $ . After that, we will try to analyse the result and try to get some numerical value of the given term with proper use of a suitable formula.

Complete step-by-step answer:
Given:
We have to find the exact value of $\cos \dfrac{2\pi }{28}\csc \dfrac{3\pi }{28}+\cos \dfrac{6\pi }{28}\csc \dfrac{9\pi }{28}+\cos \dfrac{18\pi }{28}\csc \dfrac{27\pi }{28}$ .
Now, before we proceed we should know the following formulas:
$\begin{align}
  & \sin \left( \dfrac{\pi }{2}-\theta \right)=\cos \theta .................\left( 1 \right) \\
 & \cos \left( \dfrac{\pi }{2}+\theta \right)=-\sin \theta ..............\left( 2 \right) \\
 & \sin \left( \pi -\theta \right)=\sin \theta ...................\left( 3 \right) \\
 & 2\cos A\cos B=\cos \left( A+B \right)+\cos \left( A-B \right).....................\left( 4 \right) \\
 & 2\sin A\sin B=\cos \left( A-B \right)-\cos \left( A+B \right).......................\left( 5 \right) \\
 & 2\sin A\cos B=\sin \left( A+B \right)+\sin \left( A-B \right)........................\left( 6 \right) \\
\end{align}$
Now, let $\dfrac{\pi }{28}=\theta $ . Then,
$\begin{align}
  & \cos \dfrac{2\pi }{28}\csc \dfrac{3\pi }{28}+\cos \dfrac{6\pi }{28}\csc \dfrac{9\pi }{28}+\cos \dfrac{18\pi }{28}\csc \dfrac{27\pi }{28} \\
 & \Rightarrow \cos 2\theta \csc 3\theta +\cos 6\theta \csc 9\theta +\cos 18\theta \csc 27\theta \\
\end{align}$
Now, we will simplify $\cos 2\theta \csc 3\theta +\cos 6\theta \csc 9\theta +\cos 18\theta \csc 27\theta $ and as we know that $\csc \theta =\dfrac{1}{\sin \theta }$ . Then,
$\begin{align}
  & \cos 2\theta \csc 3\theta +\cos 6\theta \csc 9\theta +\cos 18\theta \csc 27\theta \\
 & \Rightarrow \dfrac{\cos 2\theta }{\sin 3\theta }+\dfrac{\cos 6\theta }{\sin 9\theta }+\dfrac{\cos 18\theta }{\sin 27\theta } \\
 & \Rightarrow \dfrac{\cos 2\theta }{\sin 3\theta }+\dfrac{\cos 6\theta }{\sin \left( 14\theta -5\theta \right)}+\dfrac{\cos \left( 14\theta +4\theta \right)}{\sin \left( 28\theta -\theta \right)} \\
\end{align}$
Now, as per our assumption $\dfrac{\pi }{28}=\theta $ we can write $14\theta=\dfrac{\pi}{2}$ . Then,
$\begin{align}
  & \dfrac{\cos 2\theta }{\sin 3\theta }+\dfrac{\cos 6\theta }{\sin \left( 14\theta -5\theta \right)}+\dfrac{\cos \left( 14\theta +4\theta \right)}{\sin \left( 28\theta -\theta \right)} \\
 & \Rightarrow \dfrac{\cos 2\theta }{\sin 3\theta }+\dfrac{\cos 6\theta }{\sin \left( \dfrac{\pi }{2}-5\theta \right)}+\dfrac{\cos \left( \dfrac{\pi }{2}+4\theta \right)}{\sin \left( \pi -\theta \right)} \\
\end{align}$
Now, using the formula from the equation (1) to write $\sin \left( \dfrac{\pi }{2}-5\theta \right)=\cos 5\theta $ , formula from the equation (2) to write $\cos \left( \dfrac{\pi }{2}+4\theta \right)=-\sin 4\theta $ and formula from the equation (3) to write $\sin \left( \pi -\theta \right)=\sin \theta $ in the above. Then,
$\begin{align}
  & \dfrac{\cos 2\theta }{\sin 3\theta }+\dfrac{\cos 6\theta }{\sin \left( \dfrac{\pi }{2}-5\theta \right)}+\dfrac{\cos \left( \dfrac{\pi }{2}+4\theta \right)}{\sin \left( \pi -\theta \right)} \\
 & \Rightarrow \dfrac{\cos 2\theta }{\sin 3\theta }+\dfrac{\cos 6\theta }{\cos 5\theta }-\dfrac{\sin 4\theta }{\sin \theta } \\
 & \Rightarrow \dfrac{\sin \theta \cos 2\theta \cos 5\theta +\cos 6\theta \sin 3\theta \sin \theta -\cos 5\theta \sin 3\theta \sin 4\theta }{\sin 3\theta \cos 5\theta \sin \theta }=\dfrac{N}{D}......................\left( 7 \right) \\
\end{align}$
Now, we will simplify the numerator $N$ . Then,
$\begin{align}
  & N=\sin \theta \cos 2\theta \cos 5\theta +\cos 6\theta \sin 3\theta \sin \theta -\cos 5\theta \sin 3\theta \sin 4\theta \\
 & \Rightarrow 2N=2\sin \theta \cos 2\theta \cos 5\theta +2\cos 6\theta \sin 3\theta \sin \theta -2\cos 5\theta \sin 4\theta \sin 3\theta \\
 & \Rightarrow 2N=\sin \theta \left( 2\cos 2\theta \cos 5\theta \right)+\cos 6\theta \left( 2\sin 3\theta \sin \theta \right)-\cos 5\theta \left( 2\sin 4\theta \sin 3\theta \right) \\
\end{align}$
Now, use the formula form the equation (4) to write $2\cos 2\theta \cos 5\theta =\cos 7\theta +\cos 3\theta $ and formula from the equation (5) to write $2\sin 3\theta \sin \theta =\cos 2\theta -\cos 4\theta $ and $2\sin 4\theta \sin 3\theta =\cos \theta -\cos 7\theta $ in the above equation. Then,
$\begin{align}
  & 2N=\sin \theta \left( 2\cos 2\theta \cos 5\theta \right)+\cos 6\theta \left( 2\sin 3\theta \sin \theta \right)-\cos 5\theta \left( 2\sin 4\theta \sin 3\theta \right) \\
 & \Rightarrow 2N=\sin \theta \left( \cos \left( 5\theta +2\theta \right)+\cos \left( 5\theta -2\theta \right) \right)+\cos 6\theta \left( \cos \left( 3\theta -\theta \right)-\cos \left( 3\theta +\theta \right) \right) \\
 & -\cos 5\theta \left( \cos \left( 4\theta -3\theta \right)-\cos \left( 4\theta +3\theta \right) \right) \\
 & \Rightarrow 2N=\sin \theta \left( \cos 7\theta +\cos 3\theta \right)+\cos 6\theta \left( \cos 2\theta -\cos 4\theta \right)-\cos 5\theta \left( \cos \theta -\cos 7\theta \right) \\
 & \Rightarrow 2N=\sin \theta \cos 7\theta +\sin \theta \cos 3\theta +\cos 6\theta \cos 2\theta -\cos 6\theta \cos 4\theta -\cos 5\theta \cos \theta +\cos 5\theta \cos 7\theta \\
 & \Rightarrow 4N=2\sin \theta \cos 7\theta +2\sin \theta \cos 3\theta +2\cos 6\theta \cos 2\theta -2\cos 6\theta \cos 4\theta -2\cos 5\theta \cos \theta +2\cos 7\theta \cos 5\theta \\
\end{align}$
Now, use the formula from the equation (6) to write $2\sin \theta \cos 7\theta =\sin 8\theta -\sin 6\theta $ , $2\sin \theta \cos 3\theta =\sin 4\theta -\sin 2\theta $ and equation (4) to write $2\cos 6\theta \cos 2\theta =\cos 8\theta +\cos 4\theta $ , $2\cos 6\theta \cos 4\theta =\cos 10\theta +\cos 2\theta $ , $2\cos 5\theta \cos \theta =\cos 6\theta +\cos 4\theta $ and $2\cos 7\theta \cos 5\theta =\cos 12\theta +\cos 2\theta $ in the above equation. Then,
$\begin{align}
  & 4N=2\sin \theta \cos 7\theta +2\sin \theta \cos 3\theta +2\cos 6\theta \cos 2\theta -2\cos 6\theta \cos 4\theta -2\cos 5\theta \cos \theta +2\cos 7\theta \cos 5\theta \\
 & \Rightarrow 4N=\sin \left( \theta +7\theta \right)+\sin \left( \theta -7\theta \right)+\sin \left( \theta +3\theta \right)+\sin \left( \theta -3\theta \right)+\cos \left( 6\theta +2\theta \right)+\cos \left( 6\theta -2\theta \right) \\
 & -\cos \left( 6\theta +4\theta \right)-\cos \left( 6\theta -4\theta \right)-\cos \left( 5\theta +\theta \right)-\cos \left( 5\theta -\theta \right)+\cos \left( 7\theta +5\theta \right)+\cos \left( 7\theta -5\theta \right) \\
 & \Rightarrow 4N=\sin 8\theta +\sin \left( -6\theta \right)+\sin 4\theta +\sin \left( -2\theta \right)+\cos 8\theta +\cos 4\theta -\cos 10\theta -\cos 2\theta -\cos 6\theta -\cos 4\theta \\
 & +\cos 12\theta +\cos 2\theta \\
\end{align}$
Now, as we know that $\sin \left( -\theta \right)=-\sin \theta $ . Then,
$\begin{align}
  & 4N=\sin 8\theta +\sin \left( -6\theta \right)+\sin 4\theta +\sin \left( -2\theta \right)+\cos 8\theta +\cos 4\theta -\cos 10\theta -\cos 2\theta -\cos 6\theta -\cos 4\theta \\
 & +\cos 12\theta +\cos 2\theta \\
 & \Rightarrow 4N=\sin 8\theta -\sin 6\theta +\sin 4\theta -\sin 2\theta +\cos 8\theta +\cos 4\theta -\cos 10\theta -\cos 2\theta -\cos 6\theta -\cos 4\theta \\
 & +\cos 12\theta +\cos 2\theta \\
 & \Rightarrow 4N=\left( \sin 8\theta -\cos 6\theta \right)+\left( \cos 8\theta -\sin 6\theta \right)+\left( \sin 4\theta -\cos 10\theta \right)+\left( \cos 12\theta -\sin 2\theta \right)+\left( \cos 4\theta -\cos 4\theta \right) \\
 & +\left( \cos 2\theta -\cos 2\theta \right) \\
 & \Rightarrow 4N=\left( \sin \left( 14\theta -6\theta \right)-\cos 6\theta \right)+\left( \cos 8\theta -\sin \left( 14\theta -8\theta \right) \right)+\left( \sin \left( 14\theta -10\theta \right)-\cos 10\theta \right) \\
 & +\left( \cos 12\theta -\sin \left( 14\theta -12\theta \right) \right) \\
\end{align}$
Now, as per our assumption $\dfrac{\pi }{28}=\theta $ . Then,
$\begin{align}
  & 4N=\left( \sin \left( 14\theta -6\theta \right)-\cos 6\theta \right)+\left( \cos 8\theta -\sin \left( 14\theta -8\theta \right) \right)+\left( \sin \left( 14\theta -10\theta \right)-\cos 10\theta \right) \\
 & +\left( \cos 12\theta -\sin \left( 14\theta -12\theta \right) \right) \\
 & \Rightarrow 4N=\left( \sin \left( \dfrac{\pi }{2}-6\theta \right)-\cos 6\theta \right)+\left( \cos 8\theta -\sin \left( \dfrac{\pi }{2}-8\theta \right) \right)+\left( \sin \left( \dfrac{\pi }{2}-10\theta \right)-\cos 10\theta \right) \\
 & +\left( \cos 12\theta -\sin \left( \dfrac{\pi }{2}-12\theta \right) \right) \\
\end{align}$
Now, using the formula from the equation (1) to write $\sin \left( \dfrac{\pi }{2}-6\theta \right)=\cos 6\theta $ , $\sin \left( \dfrac{\pi }{2}-8\theta \right)=\cos 8\theta $ , $\sin \left( \dfrac{\pi }{2}-10\theta \right)=\cos 10\theta $ and $\sin \left( \dfrac{\pi }{2}-12\theta \right)=\cos 12\theta $ in the above equation. Then,
$\begin{align}
  & 4N=\left( \sin \left( \dfrac{\pi }{2}-6\theta \right)-\cos 6\theta \right)+\left( \cos 8\theta -\sin \left( \dfrac{\pi }{2}-8\theta \right) \right)+\left( \sin \left( \dfrac{\pi }{2}-10\theta \right)-\cos 10\theta \right) \\
 & +\left( \cos 12\theta -\sin \left( \dfrac{\pi }{2}-12\theta \right) \right) \\
 & \Rightarrow 4N=\left( \cos 6\theta -\cos 6\theta \right)+\left( \cos 8\theta -\sin 8\theta \right)+\left( \cos 10\theta -\cos 10\theta \right)+\left( \cos 12\theta -\cos 12\theta \right) \\
 & \Rightarrow 4N=0 \\
 & \Rightarrow N=0 \\
\end{align}$
Now, put $N=0$ from the above equation in the equation (7). Then,
$\begin{align}
  & \dfrac{\sin \theta \cos 2\theta \cos 5\theta +\cos 6\theta \sin 3\theta \sin \theta -\cos 5\theta \sin 3\theta \sin 4\theta }{\sin 3\theta \cos 5\theta \sin \theta }=\dfrac{N}{D} \\
 & \Rightarrow \dfrac{\sin \theta \cos 2\theta \cos 5\theta +\cos 6\theta \sin 3\theta \sin \theta -\cos 5\theta \sin 3\theta \sin 4\theta }{\sin 3\theta \cos 5\theta \sin \theta }=0 \\
\end{align}$
Now, as we have simplified the given term into $\dfrac{\sin \theta \cos 2\theta \cos 5\theta +\cos 6\theta \sin 3\theta \sin \theta -\cos 5\theta \sin 3\theta \sin 4\theta }{\sin 3\theta \cos 5\theta \sin \theta }$ and its value is zero. Then,
$\cos \dfrac{2\pi }{28}\csc \dfrac{3\pi }{28}+\cos \dfrac{6\pi }{28}\csc \dfrac{9\pi }{28}+\cos \dfrac{18\pi }{28}\csc \dfrac{27\pi }{28}=0$
Thus, the exact value of the given term will be 0.

Note: Here, the student should first understand what is asked in the question and then proceed in the right direction to get the correct answer. And we should proceed in a stepwise manner and use each formula with proper values and make assumptions like $\theta =\dfrac{\pi }{28}$ for smooth calculation. Moreover, we should avoid calculation mistakes while solving to get the correct answer.Students should remember important trigonometric identities ,formulas and standard trigonometric angles for solving these types of questions.