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Question: The exact value of \[\cos \dfrac{{2\pi }}{{28}}\cos ec\dfrac{{3\pi }}{{28}} + \cos \dfrac{{6\pi }}{{...

The exact value of cos2π28cosec3π28+cos6π28cosec9π28+cos18π28cosec27π28\cos \dfrac{{2\pi }}{{28}}\cos ec\dfrac{{3\pi }}{{28}} + \cos \dfrac{{6\pi }}{{28}}\cos ec\dfrac{{9\pi }}{{28}} + \cos \dfrac{{18\pi }}{{28}}\cos ec\dfrac{{27\pi }}{{28}} is?
A.  12 B.  12 C.  1 D.  0 A.\; - \dfrac{1}{2} \\\ B.\;\dfrac{1}{2} \\\ C.\;1 \\\ D.\;0

Explanation

Solution

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 cos2π28cosec3π28+cos6π28cosec9π28+cos18π28cosec27π28\cos \dfrac{{2\pi }}{{28}}\cos ec\dfrac{{3\pi }}{{28}} + \cos \dfrac{{6\pi }}{{28}}\cos ec\dfrac{{9\pi }}{{28}} + \cos \dfrac{{18\pi }}{{28}}\cos ec\dfrac{{27\pi }}{{28}} ……….…(1)
Let us assume that π28=θ\dfrac{\pi }{{28}} = \theta ……….…(2)
Then putting above value from equation (2), in equation (1), expression will be now,
cos2θcosec3θ+cos6θcosec9θ+cos18θcosec27θ\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
cos2θsin3θ+cos6θsin9θ+cos18θsin27θ\dfrac{{\cos 2\theta }}{{\sin 3\theta }} + \dfrac{{\cos 6\theta }}{{\sin 9\theta }} + \dfrac{{\cos 18\theta }}{{\sin 27\theta }}
Further, we get with simplification,
cos2θsin3θ+cos6θsin(14θ5θ)+cos(14θ+4θ)sin(28θθ)\dfrac{{\cos 2\theta }}{{\sin 3\theta }} + \dfrac{{\cos 6\theta }}{{\sin (14\theta - 5\theta )}} + \dfrac{{\cos (14\theta + 4\theta )}}{{\sin (28\theta - \theta )}} ………....(4)
As π28=θ\dfrac{\pi }{{28}} = \theta , then we get
π2=14θ\dfrac{\pi }{2} = 14\theta ……...….(5)
Using the value from equations (5) and (2) in equation (4), we get
cos2θsin3θ+cos6θsin(π25θ)+cos(π2+4θ)sin(πθ)\dfrac{{\cos 2\theta }}{{\sin 3\theta }} + \dfrac{{\cos 6\theta }}{{\sin (\dfrac{\pi }{2} - 5\theta )}} + \dfrac{{\cos (\dfrac{\pi }{2} + 4\theta )}}{{\sin (\pi - \theta )}}
Since we know that, sin(π2A)=cosA\sin \left( {\dfrac{\pi }{2} - A} \right) = \cos A and sin(πA)=cosA\sin \left( {\pi - A} \right) = \cos A and cos(π2+A)=sinA\cos \left( {\dfrac{\pi }{2} + A} \right) = - \sin A .
So above expression becomes,
cos2θsin3θ+cos6θcos5θ+sin4θsinθ\Rightarrow \dfrac{{\cos 2\theta }}{{\sin 3\theta }} + \dfrac{{\cos 6\theta }}{{\cos 5\theta }} + \dfrac{{ - \sin 4\theta }}{{\sin \theta }}
cos2θcos5θsinθ+cos6θsin3θsinθsin4θcos5θsin3θsin5θcos5θsinθ\Rightarrow \dfrac{{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,
cos2θcos5θsinθ+cos6θsin3θsinθsin4θcos5θsin3θcos2\theta cos5\theta sin\theta + cos6\theta sin3\theta sin\theta - sin4\theta cos5\theta sin3\theta
12[(2cos2θcos5θ)sinθ+cos6θ(2sin3θsinθ)] 12[(cos7θ+cos3θ)sinθ+cos6θ(cos4θ+cos2θ)cos5θ(cos2θ+cosθ)] 14[2sinθcos7θ+2sinθcos3θ2cos6θcos4θ+2cos6θcos2θ+2cos5θcos7θ2cos5θcosθ] \Rightarrow \dfrac{1}{2}[(2\cos 2\theta \cos 5\theta )\sin \theta + \cos 6\theta (2\sin 3\theta \sin \theta )] \\\ \Rightarrow \dfrac{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 \dfrac{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 ]
Now, we do further simplification, then as below,
14[sin(4θ)sin(2θ)+sin8θsin6θ+cos8θ+cos4θcos10θcos2θcos6θcos4θ+cos12θ+cos2θ]\Rightarrow \dfrac{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
14[sin(14θ10θ)sin(14θ12θ)+sin(14θ6θ)sin(14θ8θ)+cos8θ+cos4θcos10θcos2θcos6θcos4θ+cos12θ+cos2θ]\Rightarrow \dfrac{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 14θ14\theta as π2\dfrac{\pi }{2} from equation (5). So above expression will become,
14[cos10θcos12θ+cos6θcos8θ+cos8θcos10θcos2θcos6θ+cos12θ+cos2θ] 14×0 0 \Rightarrow \dfrac{1}{4}[cos10\theta - cos12\theta + cos6\theta - cos8\theta + cos8\theta - cos10\theta - cos2\theta - cos6\theta + cos12\theta + cos2\theta ] \\\ \Rightarrow \dfrac{1}{4} \times 0 \\\ \Rightarrow 0
Thus with the help of above simplification we substitute this value in equation (6) , then we get
cos2θcos5θsinθ+cos6θsin3θsinθsin4θcos5θsin3θsin5θcos5θsinθ=0sin5θcos5θsinθ 0 \Rightarrow \dfrac{{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 }} = \dfrac{0}{{\sin 5\theta \cos 5\theta \sin \theta }} \\\ \Rightarrow 0

\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.