﻿ Find exact value of \sin\dfrac{\pi}{7} \sin\dfrac{2\pi}{7} \sin\dfrac{3\pi}{7} or sin(pi/7)sin(2pi/7)sin(3pi/7)

#### Question

Find exact value of \sin\dfrac{\pi}{7} \sin\dfrac{2\pi}{7} \sin\dfrac{3\pi}{7} or sin(pi/7)sin(2pi/7)sin(3pi/7)

Collected in the board: Trigonometry

Steven Zheng posted 3 weeks ago

Let

x= \sin\dfrac{\pi}{7} \sin\dfrac{2\pi}{7} \sin\dfrac{3\pi}{7}
(1)

and

y = \cos\dfrac{\pi}{7} \cos\dfrac{2\pi}{7} \cos\dfrac{3\pi}{7}

Multiply x by y and apply double angle identity for sine function

xy =\dfrac{1}{8} \Big(2 \sin\dfrac{\pi}{7}\cos\dfrac{\pi}{7}\Big) \Big(2\sin\dfrac{2\pi}{7}\cos\dfrac{2\pi}{7}\Big)\Big(2 \sin\dfrac{3\pi}{7}\cos\dfrac{3\pi}{7}\Big)

=\dfrac{1}{8}\sin\dfrac{2\pi}{7}\sin\dfrac{4\pi}{7} \sin\dfrac{6\pi}{7}

=\dfrac{1}{8}\sin\dfrac{2\pi}{7}\sin\Big(\pi-\dfrac{3\pi}{7}\Big) \sin\Big(\pi-\dfrac{\pi}{7}\Big) .

=\dfrac{1}{8} \sin\dfrac{\pi}{7} \sin\dfrac{2\pi}{7} \sin\dfrac{3\pi}{7}

=\dfrac{x}{8}

Therefore,

y = \cos\dfrac{\pi}{7} \cos\dfrac{2\pi}{7} \cos\dfrac{3\pi}{7} =\dfrac{1}{8}
(2)

To simplify the process, let's denote a=\cos\dfrac{\pi}{7}, b=\cos\dfrac{2\pi}{7}, and c=\cos\dfrac{3\pi}{7}. Then the equation (2) can be expressed as

y = abc = \dfrac{1}{8}
(3)

Now let's square the equation (1) and apply Power-Reducing Identity for sine function. And express the derived cosine functions in terms of their supplements for two of larger angles.

x^2= \sin^2\dfrac{\pi}{7} \sin^2\dfrac{2\pi}{7} \sin^2\dfrac{3\pi}{7}

=\dfrac{1}{8}\Big(1-\cos\dfrac{2\pi}{7} \Big)\Big(1-\cos\dfrac{4\pi}{7} \Big)\Big(1-\cos\dfrac{6\pi}{7} \Big)

=\dfrac{1}{8}\Big(1-\cos\dfrac{2\pi}{7} \Big)\Big(1-\cos\Big( \pi-\dfrac{3\pi}{7}\Big) \Big)\Big(1-\cos\Big(\pi-\dfrac{\pi}{7}\Big) \Big)

=\dfrac{1}{8}\Big(1-\cos\dfrac{2\pi}{7} \Big)\Big(1-\cos\dfrac{3\pi}{7} \Big)\Big(1-\cos\dfrac{\pi}{7} \Big)

Replacing the cosine functions with our denotions gives the following equation.

x^2 = \dfrac{1}{8}(1-a)(1-b)(1-c)

Expand brackets and simplify.

8x^2 = 1+a−b+c+ac−ab−bc−abc

Apply the conclusion (3), abc = \dfrac{1}{8} . The equation is further simplified to

8x^2 = a−b+c+ac−ab−bc+\dfrac{7}{8}
(4)

In this expression, if we could show that trigonometric terms (a-b-c+ac-ab-bc) is equal to zero, then we will reach the end of our evaluation. Let's start with product terms and apply the product identity for three pairs of the cosine functions.

ac-ab-bc

=\cos\dfrac{\pi}{7}\cos\dfrac{3\pi}{7}-\cos\dfrac{\pi}{7}\cos\dfrac{2\pi}{7}-\cos\dfrac{2\pi}{7}\cos\dfrac{3\pi}{7}

=\dfrac{1}{2}\Big(\cos\dfrac{4\pi}{7}+\cos\dfrac{2\pi}{7}-\cos\dfrac{3\pi}{7}-\cos\dfrac{\pi}{7}-\cos\dfrac{5\pi}{7}-\cos\dfrac{\pi}{7} \Big)

=\dfrac{1}{2}\Big(\cos\dfrac{4\pi}{7}+\cos\dfrac{2\pi}{7}-\cos\dfrac{3\pi}{7}-\cos\dfrac{5\pi}{7}-2\cos\dfrac{\pi}{7} \Big)

=\dfrac{1}{2}\Big(\cos\dfrac{4\pi}{7}-\cos\dfrac{5\pi}{7}+b-c-2a \Big)

=\dfrac{1}{2}\Big(-\cos\dfrac{3\pi}{7}+\cos\dfrac{2\pi}{7}+b-c-2a \Big)

=\dfrac{1}{2}\Big(-c+b+b-c-2a \Big)

=b-a-c

which shows a-b-c+ac-ab-bc is indeed equals to zero

Therefore, equaiton (4) is simplified to

8x^2 = \dfrac{7}{8}

Solving for x yields

x=\dfrac{\sqrt{7} }{8}

Now, the exact value of \sin\dfrac{\pi}{7} \sin\dfrac{2\pi}{7} \sin\dfrac{3\pi}{7} or sin(pi/7)sin(2pi/7)sin(3pi/7) is concluded to be equal to \dfrac{\sqrt{7} }{8}

Steven Zheng posted 3 weeks ago

Scroll to Top