Question

How do you prove \[\cos \left( {\dfrac{\pi }{7}} \right)\cos \left( {\dfrac{{2\pi }}{7}} \right)\cos \left( {\dfrac{{3\pi }}{7}} \right) = \dfrac{1}{8}\] ?

Answer 1

Hint: The mathematical expressions which include the ratios of the sides of a right-angled triangle can be called trigonometric expressions. An algebraic equation, which involves trigonometric ratios and expressions is known as a trigonometric equation. To prove the given trigonometric functions, let us consider an isosceles triangle, with respect to the given terms of LHS and hence further simplifying the terms we get LHS = RHS.

Complete step-by-step answer:
To prove
 \[\cos \left( {\dfrac{\pi }{7}} \right)\cos \left( {\dfrac{{2\pi }}{7}} \right)\cos \left( {\dfrac{{3\pi }}{7}} \right) = \dfrac{1}{8}\]
Let ABC be an isosceles triangle, \[AC = BC\] , with \[\angle C = \dfrac{\pi }{7}\] . Mark point D on AC such that AB=AD.
We have defined, for convenience, \[AB = 1\] , \[BC = x\] , and \[BD = y\] and also \[AD = 1\] and \[CD = y\] .

In \[\Delta BCD,\] we have:
 \[\cos C = \cos \dfrac{\pi }{7}\]
From, the diagram for the point C i.e., \[\dfrac{\pi }{7}\] we get:
 \[\cos C = \dfrac{{\dfrac{x}{2}}}{y}\]
Hence, we get:
 \[ \Rightarrow \cos C = \dfrac{x}{{2y}}\] …………………. 1
In \[\Delta ABD,\] we have:
 \[\cos D = \cos \dfrac{{2\pi }}{7}\]
From, the diagram for the point D i.e., \[\dfrac{{2\pi }}{7}\] we get:
 \[\cos D = \dfrac{{\dfrac{y}{2}}}{1}\]
Hence, we get:
 \[ \Rightarrow \cos D = \dfrac{y}{2}\] …………………….. 2
In \[\Delta ABC,\] we have:
 \[\cos A = \cos \dfrac{{3\pi }}{7}\]
From, the diagram for the point A i.e., \[\dfrac{{3\pi }}{7}\] we get:
 \[\cos A = \dfrac{{\dfrac{1}{2}}}{x}\]
Hence, we get:
 \[ \Rightarrow \cos A = \dfrac{1}{{2x}}\] ……………….. 3
Thus, given trigonometric function we have: \[\cos \dfrac{\pi }{7} \times \cos \dfrac{{2\pi }}{7} \times \cos \dfrac{{3\pi }}{7}\]
Hence, substitute the value of each terms obtained, from equation 1, 2 and 3 as:
 \[\cos \dfrac{\pi }{7} \times \cos \dfrac{{2\pi }}{7} \times \cos \dfrac{{3\pi }}{7} = \dfrac{x}{{2y}} \times \dfrac{y}{2} \times \dfrac{1}{{2x}}\]
Multiplying the terms, we get:
 \[ \Rightarrow \cos \dfrac{\pi }{7} \times \cos \dfrac{{2\pi }}{7} \times \cos \dfrac{{3\pi }}{7} = \dfrac{{xy}}{{8xy}}\]
Now, combining the common terms, we get:
 \[ \Rightarrow \cos \dfrac{\pi }{7} \times \cos \dfrac{{2\pi }}{7} \times \cos \dfrac{{3\pi }}{7} = \dfrac{1}{8}\]
Hence, proved \[\cos \left( {\dfrac{\pi }{7}} \right)\cos \left( {\dfrac{{2\pi }}{7}} \right)\cos \left( {\dfrac{{3\pi }}{7}} \right) = \dfrac{1}{8}\] .

Note: The three basic functions in trigonometry are sine, cosine and tangent. Based on these three functions the other three functions that are cotangent, secant and cosecant are derived. All the trigonometrical concepts are based on these functions. Hence, to understand trigonometry further we need to learn these functions and their respective formulas at first. The key point to prove trigonometric functions is that we must know all the trigonometric identity functions, to solve the terms in the given expression such that to prove LHS = RHS we must consider any of the terms of side and must know all the basic identities and relation between the trigonometric functions.