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Geometry Level 3

sin ⁡ ( π 5 ) sin ⁡ ( 2 π 5 ) sin ⁡ ( 3 π 5 ) sin ⁡ ( 4 π 5 ) \sin \left( \frac{\pi}{5} \right) \sin \left( \frac{2\pi}{5} \right) \sin \left( \frac{3\pi}{5} \right) \sin \left( \frac{4\pi}{5} \right)

If the value of the expression is equals to A B \dfrac AB for coprime positive integers A A and B B , submit your answer as B − A B - A .


The answer is 11.

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Krishna Shankar by Krishna Shankar , 23, India

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5 solutions

Raj Rajput
Oct 8, 2015

9 Helpful 0 Interesting 0 Brilliant 0 Confused

sin ⁡ π 5 sin ⁡ 2 π 5 sin ⁡ 3 π 5 sin ⁡ 4 π 5 = sin ⁡ π 5 sin ⁡ 2 π 5 sin ⁡ 2 π 5 sin ⁡ π 5 [ sin ⁡ k π 5 = sin ⁡ 5 − k π 5 ] = sin ⁡ 2 π 5 sin ⁡ 2 2 π 5 [ cos ⁡ 2 k π 5 = 1 − sin ⁡ 2 k π 5 ] = 1 4 ( 1 − cos ⁡ 2 π 5 ) ( 1 − cos ⁡ 4 π 5 ) [ cos ⁡ 2 π 5 + cos ⁡ 4 π 5 = − 1 2 ] = 1 4 ( 1 − cos ⁡ 2 π 5 ) ( 3 2 + cos ⁡ 2 π 5 ) = 1 4 ( 3 2 − 1 2 cos ⁡ 2 π 5 − cos ⁡ 2 2 π 5 ) = 1 4 ( 3 2 − 1 2 cos ⁡ 2 π 5 − 1 2 − 1 2 cos ⁡ 4 π 5 ) = 1 4 ( 1 − 1 2 [ cos ⁡ 2 π 5 + cos ⁡ 4 π 5 ] ) = 1 4 ( 1 − 1 2 [ − 1 2 ] ) = 5 16 ⇒ B − A = 11 \begin{array} {rll} \sin \frac{\pi}{5} \sin \frac{2\pi}{5} \sin \color{#3D99F6} {\frac{3\pi}{5}} \sin \color{#D61F06}{\frac{4\pi}{5}} & = \sin \frac{\pi}{5} \sin \frac{2\pi}{5} \sin \color{#3D99F6} {\frac{2\pi}{5}} \sin \color{#D61F06}{\frac{\pi}{5}} & \quad \quad \small \color{#3D99F6} {\left[ \sin \frac{k\pi}{5} = \sin \frac{5-k\pi}{5} \right]} \\ & = \sin^2 \frac{\pi}{5} \sin^2 \frac{2\pi}{5} & \quad \quad \small \color{#3D99F6} {\left[ \cos \frac{2k\pi}{5} = 1 - \sin^2 \frac{k\pi}{5} \right]} \\ & = \frac{1}{4} \left(1-\cos \frac{2\pi}{5} \right) \left(1-\cos \frac{4\pi}{5} \right) & \quad \quad \small \color{#3D99F6} {\left[ \cos \frac{2\pi}{5} + \cos \frac{4\pi}{5} = - \frac{1}{2} \right]} \\ & = \frac{1}{4} \left(1-\cos \frac{2\pi}{5} \right) \left(\frac{3}{2} + \cos \frac{2\pi}{5} \right) \\ & = \frac{1}{4} \left(\frac{3}{2} - \frac{1}{2} \cos \frac{2\pi}{5} - \color{#3D99F6} {\cos^2 \frac{2\pi}{5}} \right) \\ & = \frac{1}{4} \left(\frac{3}{2} - \frac{1}{2} \cos \frac{2\pi}{5} - \color{#3D99F6} {\frac{1}{2} - \frac{1}{2} \cos \frac{4\pi}{5}} \right) \\ & = \frac{1}{4} \left(1 - \frac{1}{2} \left[\color{#3D99F6}{\cos \frac{2\pi}{5} + \cos \frac{4\pi}{5}} \right] \right) \\ & = \frac{1}{4} \left(1 - \frac{1}{2} \left[\color{#3D99F6}{-\frac{1}{2}} \right] \right) = \frac{5}{16} \\ & \\ \Rightarrow B - A & = \boxed{11} \end{array}

4 Helpful 0 Interesting 0 Brilliant 0 Confused

sin ⁡ π 5 sin ⁡ 2 π 5 sin ⁡ 3 π 5 sin ⁡ 4 π 5 = sin ⁡ π 5 sin ⁡ 2 π 5 sin ⁡ 2 π 5 sin ⁡ π 5 = sin ⁡ 2 π 5 ∗ sin ⁡ 2 2 π 5 sin ⁡ π 5 = 5 − 5 8 sin ⁡ 2 π 5 = 5 + 5 8 ∴ sin ⁡ π 5 sin ⁡ 2 π 5 sin ⁡ 3 π 5 sin ⁡ 4 π 5 = 5 − 5 8 ∗ 5 − 5 8 20 8 2 = 5 16 = A B B − A = 11 \sin \frac{\pi}{5} \sin \frac{2\pi}{5} \sin \color{#3D99F6} {\frac{3\pi}{5}} \sin \color{#D61F06}{\frac{4\pi}{5}} \\ = \sin \frac{\pi}{5} \sin \frac{2\pi}{5} \sin \color{#3D99F6} {\frac{2\pi}{5}} \sin \color{#D61F06}{\frac{\pi}{5}}\\ =\sin^2 \frac{\pi}{5} *\sin^2 \frac{2\pi}{5}\\ \color{#3D99F6} {\sin \frac{\pi}{5} =\large \sqrt{\dfrac{5-\sqrt5 }8} }\\ \color{#3D99F6} { \sin \frac{2\pi}{5} = \large \sqrt{\dfrac{5+\sqrt5 }8}} \therefore\ \sin \frac{\pi}{5} \sin \frac{2\pi}{5} \sin \color{#3D99F6} {\frac{3\pi}{5}} \sin \color{#D61F06}{\frac{4\pi}{5}}\\ = \dfrac{5-\sqrt5 }8*\dfrac{5-\sqrt5 }8\\ \dfrac{20}{8^2}= \dfrac{5}{16} = \dfrac A B\\ B-A=11\\\ \ \\

It is basically same as other two solutions. My apology and thanks to Chew-Seong Cheong, for using part of his Latex codes.

3 Helpful 0 Interesting 0 Brilliant 0 Confused
Mas Mus
Oct 10, 2015

sin ⁡ ( π 5 ) sin ⁡ ( 2 π 5 ) sin ⁡ ( 3 π 5 ) sin ⁡ ( 4 π 5 ) = sin ⁡ ( π 5 ) sin ⁡ ( 2 π 5 ) sin ⁡ ( 2 π 5 ) sin ⁡ ( π 5 ) = [ sin ⁡ ( π 5 ) sin ⁡ ( 2 π 5 ) ] 2 = [ − 1 2 ( cos ⁡ 3 π 5 − cos ⁡ π 5 ) ] 2 = 1 4 [ − sin ⁡ 1 8 ∘ − sin ⁡ 5 4 ∘ ] 2 \sin \left( \frac{\pi}{5} \right) \sin \left( \frac{2\pi}{5} \right) \sin \left( \frac{3\pi}{5} \right) \sin \left( \frac{4\pi}{5} \right) =\sin \left( \frac{\pi}{5} \right) \sin \left( \frac{2\pi}{5} \right) \sin \left( \frac{2\pi}{5} \right) \sin \left( \frac{\pi}{5} \right)\\=\left[\sin \left( \frac{\pi}{5} \right) \sin \left( \frac{2\pi}{5} \right)\right]^{2}=\left[-\frac{1}{2}\left(\cos\frac{3\pi}{5}-\cos\frac{\pi}{5}\right)\right]^{2}=\frac{1}{4}\left[-\sin18^{\circ}-\sin54^{\circ}\right]^{2} .

From this we have sin ⁡ 1 8 ∘ = − 1 + 5 4 \sin18^{\circ}=\frac{-1+\sqrt{5}}{4} and sin ⁡ 5 4 ∘ = 1 + 5 4 \sin54^{\circ}=\frac{1+\sqrt{5}}{4}

Then, 1 4 [ − sin ⁡ 1 8 ∘ − sin ⁡ 5 4 ∘ ] 2 = 1 4 [ 5 2 ] 2 = 5 16 = A B \frac{1}{4}\left[-\sin18^{\circ}-\sin54^{\circ}\right]^{2}=\frac{1}{4}\left[\frac{\sqrt{5}}{2}\right]^{2}=\frac{5}{16}=\frac{A}{B}

Thus, the required answer is B − A = 11 B-A=\large{11}

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Ben Habeahan
Oct 9, 2015

Use formula ( with induction):

∏ k = 1 n − 1 sin ⁡ ( k π n ) = 2 ( 1 − n ) n \boxed{ \large\prod_{k=1}^{n-1} \sin(\frac{k\pi}{n})=2^{(1-n)}n }

∏ k = 1 4 sin ⁡ ( k π 5 ) = 2 ( 1 − 5 ) 5 = 5 16 \large\prod_{k=1}^{4} \sin(\frac{k\pi}{5})=2^{(1-5)}5=\frac{5}{16}

⟹ 16 − 5 = 11 \implies16-5=\boxed{11}

N O T E : \large{NOTE:}

∏ k = 1 n − 1 cos ⁡ ( k π n ) = 2 ( 1 − n ) sin ⁡ ( n π 2 ) \large\prod_{k=1}^{n-1} \cos(\frac{k\pi}{n})=2^{(1-n)}\sin(\frac{n\pi}{2})

∏ k = 1 n − 1 tan ⁡ ( k π n ) = n csc ⁡ ( n π 2 ) \large\prod_{k=1}^{n-1} \tan(\frac{k\pi}{n})=n \csc (\frac{n\pi}{2})

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