Question

The exact value of $$\cos {\displaystyle \frac{2\pi }{28}}\cos ec{\displaystyle \frac{3\pi }{28}}+\cos {\displaystyle \frac{6\pi }{28}}\cos ec{\displaystyle \frac{9\pi }{28}}+\cos {\displaystyle \frac{18\pi }{28}}\cos ec{\displaystyle \frac{27\pi }{28}}$$ is?
$$\begin{gathered} A.-{\displaystyle \frac{1}{2}} \\ B.{\displaystyle \frac{1}{2}} \\ C.1 \\ D.0 \end{gathered}$$

Answer 1

Hint: Suitable uses of trigonometry identities is must here. Also apply complementary formulas of sin and cos values. Inverse ratios are also applicable here. Many sequential steps will be needed to reach the required result. Formula like sin(A+B) , cos(A+B) etc will be needed.

Complete step-by-step solution:
GIven $$\cos {\displaystyle \frac{2\pi }{28}}\cos ec{\displaystyle \frac{3\pi }{28}}+\cos {\displaystyle \frac{6\pi }{28}}\cos ec{\displaystyle \frac{9\pi }{28}}+\cos {\displaystyle \frac{18\pi }{28}}\cos ec{\displaystyle \frac{27\pi }{28}}$$ ……….…(1)
Let us assume that ${\displaystyle \frac{\pi }{28}}=\theta$ ……….…(2)
 Then putting above value from equation (2), in equation (1), expression will be now,
$$\cos 2\theta \cos ec3\theta +\cos 6\theta \cos ec9\theta +\cos 18\theta \cos ec27\theta$$ ……...…(3)
Now, replace cosec terms by their reciprocal sine terms, as follows
$${\displaystyle \frac{\cos 2\theta }{\sin 3\theta }}+{\displaystyle \frac{\cos 6\theta }{\sin 9\theta }}+{\displaystyle \frac{\cos 18\theta }{\sin 27\theta }}$$
Further, we get with simplification,
$${\displaystyle \frac{\cos 2\theta }{\sin 3\theta }}+{\displaystyle \frac{\cos 6\theta }{\sin (14\theta -5\theta )}}+{\displaystyle \frac{\cos (14\theta +4\theta )}{\sin (28\theta -\theta )}}$$ ………....(4)
As ${\displaystyle \frac{\pi }{28}}=\theta$ , then we get
${\displaystyle \frac{\pi }{2}}=14\theta$ ……...….(5)
Using the value from equations (5) and (2) in equation (4), we get
$${\displaystyle \frac{\cos 2\theta }{\sin 3\theta }}+{\displaystyle \frac{\cos 6\theta }{\sin ({\displaystyle \frac{\pi }{2}}-5\theta )}}+{\displaystyle \frac{\cos ({\displaystyle \frac{\pi }{2}}+4\theta )}{\sin (\pi -\theta )}}$$
Since we know that, $\sin \left({\displaystyle \frac{\pi }{2}}-A\right)=\cos A$ and $\sin \left(\pi -A\right)=\cos A$ and $\cos \left({\displaystyle \frac{\pi }{2}}+A\right)=-\sin A$ .
So above expression becomes,
$\Rightarrow {\displaystyle \frac{\cos 2\theta }{\sin 3\theta }}+{\displaystyle \frac{\cos 6\theta }{\cos 5\theta }}+{\displaystyle \frac{-\sin 4\theta }{\sin \theta }}$
$\Rightarrow {\displaystyle \frac{cos2\theta cos5\theta sin\theta +cos6\theta sin3\theta sin\theta -sin4\theta cos5\theta sin3\theta }{\sin 5\theta \cos 5\theta \sin \theta }}$ …………..(6)
Now, we will simplify the numerator of above expression, as below,
 $cos2\theta cos5\theta sin\theta +cos6\theta sin3\theta sin\theta -sin4\theta cos5\theta sin3\theta$
$$\begin{gathered} \Rightarrow {\displaystyle \frac{1}{2}}[(2\cos 2\theta \cos 5\theta )\sin \theta +\cos 6\theta (2\sin 3\theta \sin \theta )] \\ \Rightarrow {\displaystyle \frac{1}{2}}[(\cos 7\theta +\cos 3\theta )\sin \theta +\cos 6\theta (-\cos 4\theta +\cos 2\theta )-\cos 5\theta (-\cos 2\theta +\cos \theta )] \\ \Rightarrow {\displaystyle \frac{1}{4}}[2sin\theta cos7\theta +2sin\theta cos3\theta -2cos6\theta cos4\theta +2cos6\theta cos2\theta +2cos5\theta cos7\theta -2cos5\theta cos\theta ] \end{gathered}$$
Now, we do further simplification, then as below,
$\Rightarrow {\displaystyle \frac{1}{4}}[sin(4\theta )-sin(2\theta )+sin8\theta -sin6\theta +cos8\theta +cos4\theta -cos10\theta -cos2\theta -cos6\theta -cos4\theta +cos12\theta +cos2\theta ]$
Then
$\Rightarrow {\displaystyle \frac{1}{4}}[sin(14\theta -10\theta )-sin(14\theta -12\theta )+sin(14\theta -6\theta )-sin(14\theta -8\theta )+cos8\theta +cos4\theta -cos10\theta -cos2\theta -cos6\theta -cos4\theta +cos12\theta +cos2\theta ]$since we have the value of $14\theta$ as ${\displaystyle \frac{\pi }{2}}$ from equation (5). So above expression will become,
$$\begin{gathered} \Rightarrow {\displaystyle \frac{1}{4}}[cos10\theta -cos12\theta +cos6\theta -cos8\theta +cos8\theta -cos10\theta -cos2\theta -cos6\theta +cos12\theta +cos2\theta ] \\ \Rightarrow {\displaystyle \frac{1}{4}}\times 0 \\ \Rightarrow 0 \end{gathered}$$
Thus with the help of above simplification we substitute this value in equation (6) , then we get
$$\begin{gathered} \Rightarrow {\displaystyle \frac{cos2\theta cos5\theta sin\theta +cos6\theta sin3\theta sin\theta -sin4\theta cos5\theta sin3\theta }{\sin 5\theta \cos 5\theta \sin \theta }}={\displaystyle \frac{0}{\sin 5\theta \cos 5\theta \sin \theta }} \\ \Rightarrow 0 \end{gathered}$$

$\therefore$ The correct option is D.

Note: Trigonometry is the study of relationships between angles, lengths, and heights of triangles. Also, it shows the relationship between different parts of circles and other geometrical figures. Trigonometric identities are useful and hence its learning is very much required for solving the problems in a better way. There are many fields from science also, where these identities of trigonometry and formula of trigonometry are used.
One must know the difference between Trigonometric identities and Trigonometric Ratios. Trigonometric Identities are the formulas involving the trigonometric functions. Whereas, trigonometric Ratio is known for the relationship between the angles and the length of the side of the right triangle.