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Geometry Level 3

Find the value of : 16 s i n 144 ° s i n 108 ° s i n 72 ° s i n 36 ° 16sin144°sin108°sin72°sin36°


The answer is 5.

Mehul Chaturvedi by Mehul Chaturvedi , 22, India

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1 solution

Chew-Seong Cheong
Nov 17, 2014

Since sin ⁡ 3 6 ∘ = sin ⁡ 14 4 ∘ \sin {36^\circ} = \sin {144^\circ} , sin ⁡ 7 2 ∘ = sin ⁡ 10 8 ∘ \sin {72^\circ} = \sin {108^\circ} and sin ⁡ 7 2 ∘ = cos ⁡ 1 8 ∘ \sin {72^\circ} = \cos {18^\circ} then,

16 sin ⁡ 14 4 ∘ sin ⁡ 10 8 ∘ sin ⁡ 7 2 ∘ sin ⁡ 3 6 ∘ = 16 sin ⁡ 2 3 6 ∘ sin ⁡ 2 7 2 ∘ = 16 sin ⁡ 2 3 6 ∘ cos ⁡ 2 1 8 ∘ 16 \sin {144^\circ} \sin {108^\circ} \sin {72^\circ} \sin {36^\circ} = 16 \sin^2 {36^\circ} \sin^2 {72^\circ} = 16 \sin^2 {36^\circ} \cos^2 {18^\circ}

Since sin ⁡ 3 6 ∘ = cos ⁡ 5 4 ∘ \sin {36^\circ} = \cos {54^\circ} ,

⇒ sin ⁡ 1 8 ∘ cos ⁡ 1 8 ∘ = cos ⁡ 3 6 ∘ cos ⁡ 1 8 ∘ − sin ⁡ 3 6 ∘ sin ⁡ 1 8 ∘ \Rightarrow \sin {18^\circ} \cos {18^\circ} = \cos {36^\circ} \cos {18^\circ} - \sin {36^\circ} \sin {18^\circ}

= ( 1 − 2 sin ⁡ 2 1 8 ∘ cos ⁡ 1 8 ∘ − ( sin ⁡ 1 8 ∘ cos ⁡ 1 8 ∘ ) sin ⁡ 1 8 ∘ \quad = (1 - 2 \sin^2 {18^\circ} \cos {18^\circ} - (\sin {18^\circ} \cos {18^\circ}) \sin {18^\circ}

⇒ sin ⁡ 1 8 ∘ = 1 − 2 sin ⁡ 2 1 8 ∘ − sin ⁡ 2 1 8 ∘ \Rightarrow \sin {18^\circ} = 1 - 2 \sin^2 {18^\circ} - \sin^2 {18^\circ}

⇒ 4 sin ⁡ 2 1 8 ∘ + 2 sin ⁡ 1 8 ∘ − 1 = 0 \Rightarrow 4 \sin^2 {18^\circ} + 2 \sin {18^\circ} - 1 = 0

⇒ sin ⁡ 1 8 ∘ = − 2 + 4 + 16 8 = 5 − 1 4 \Rightarrow \sin {18^\circ} = \dfrac {-2+\sqrt{4+16}} {8} = \dfrac {\sqrt{5} - 1} {4}

⇒ cos ⁡ 3 6 ∘ = 1 − 2 sin ⁡ 2 1 8 ∘ = 1 − 2 ( 5 − 1 4 ) 2 = 1 − 2 ( 6 − 2 5 16 ) = 5 + 1 4 \Rightarrow \cos {36^\circ} = 1 - 2 \sin^2 {18^\circ} = 1 - 2 \left( \dfrac {\sqrt{5} - 1} {4} \right) ^2 = 1 - 2 \left( \dfrac {6 - 2\sqrt{5}} {16} \right) = \dfrac {\sqrt{5} + 1} {4}

We note that:

16 sin ⁡ 14 4 ∘ sin ⁡ 10 8 ∘ sin ⁡ 7 2 ∘ sin ⁡ 3 6 ∘ = 16 sin ⁡ 2 3 6 ∘ cos ⁡ 2 1 8 ∘ 16 \sin {144^\circ} \sin {108^\circ} \sin {72^\circ} \sin {36^\circ} = 16 \sin^2 {36^\circ} \cos^2 {18^\circ}

= 16 ( 1 − cos ⁡ 2 3 6 ∘ ) ( 1 − sin ⁡ 2 1 8 ∘ ) = 16 ( 1 - \cos^2 {36^\circ} ) (1 - \sin^2 {18^\circ} )

= 16 ( 1 + cos ⁡ 3 6 ∘ ) ( 1 − cos ⁡ 3 6 ∘ ) ( 1 + sin ⁡ 1 8 ∘ ) ( 1 − sin ⁡ 1 8 ∘ ) = 16 ( 1 + \cos {36^\circ} ) ( 1 - \cos {36^\circ} ) (1 + \sin {18^\circ} ) (1 - \sin {18^\circ} )

= 16 ( 1 + 5 + 1 4 ) ( 1 − 5 + 1 4 ) ( 1 + 5 − 1 4 ) ( 1 − 5 − 1 4 ) = 16 \left( 1 + \dfrac {\sqrt{5} + 1} {4} \right) \left(1 - \dfrac {\sqrt{5} + 1} {4} \right) \left( 1 + \dfrac {\sqrt{5} - 1} {4} \right) \left( 1 - \dfrac {\sqrt{5} - 1} {4} \right)

= 16 ( 5 + 5 4 ) ( 5 − 5 4 ) ( 3 + 5 4 ) ( 3 − 5 4 ) = 16 \left( \dfrac {5+\sqrt{5}} {4} \right) \left(\dfrac {5-\sqrt{5}} {4} \right) \left(\dfrac {3+\sqrt{5}} {4}\right) \left(\dfrac {3-\sqrt{5}} {4}\right)

= 16 ( 25 − 5 16 ) ( 9 − 5 16 ) = 16 ( 20 16 ) ( 4 16 ) = 5 = 16 \left( \dfrac {25-5} {16} \right) \left(\dfrac {9-5} {16} \right) = 16 \left( \dfrac {20} {16} \right) \left(\dfrac {4} {16} \right) = \boxed {5}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

There are some typos in the solution, like in the 4 th 4^{\textrm{th}} line, it should be sin ⁡ 3 6 ∘ = 2 sin ⁡ 1 8 ∘ cos ⁡ 1 8 ∘ \sin36^{\circ} = 2\sin18^{\circ}\cos18^{\circ} . Following that, despite some errors, you calculated the value correctly because the final quadratic equation obtained was correct. Also, the transition from line sin ⁡ 1 8 ∘ = 1 − 2 sin ⁡ 2 1 8 ∘ − sin ⁡ 2 1 8 ∘ \sin {18^\circ} = 1 - 2 \sin^2 {18^\circ} - \sin^2 {18^\circ} to the line 4 sin ⁡ 2 1 8 ∘ + 2 sin ⁡ 1 8 ∘ − 1 = 0 4 \sin^2 {18^\circ} + 2 \sin {18^\circ} - 1 = 0 is incorrect. You should correct the minor mistakes ASAP!

Prasun Biswas - 6 years, 5 months ago

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