Which Formula Should I Apply? (Part 2)

Geometry Level 3

Evaluate

241 2 − cot ⁡ 2 ( 9 ∘ ) + 241 2 − cot ⁡ 2 ( 2 7 ∘ ) + 241 2 − cot ⁡ 2 ( 4 5 ∘ ) + 241 2 − cot ⁡ 2 ( 6 3 ∘ ) + 241 2 − cot ⁡ 2 ( 8 1 ∘ ) . \frac{241}{2-\cot^{2} (9^{\circ})}+\frac{241}{2-\cot^{2} (27^{\circ})}+\frac{241}{2-\cot^{2} (45^{\circ})}\\ +\frac{241}{2-\cot^{2} (63^{\circ})}+\frac{241}{2-\cot^{2} (81^{\circ})} .


The answer is 365.

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Pratik Shastri by Pratik Shastri , 35, India

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2 solutions

Chew-Seong Cheong
Dec 14, 2014

241 2 − cot ⁡ 2 9 ∘ + 241 2 − cot ⁡ 2 2 7 ∘ + 241 2 − cot ⁡ 2 4 5 ∘ + 241 2 − cot ⁡ 2 6 3 ∘ + 241 2 − cot ⁡ 2 8 1 ∘ = 241 2 − cot ⁡ 2 9 ∘ + 241 2 − cot ⁡ 2 8 1 ∘ + 241 2 − cot ⁡ 2 2 7 ∘ + 241 2 − cot ⁡ 2 6 3 ∘ + 241 = 241 ( sin ⁡ 2 9 ∘ 2 sin ⁡ 2 9 ∘ − cos ⁡ 2 9 ∘ + sin ⁡ 2 8 1 ∘ 2 sin ⁡ 2 8 1 ∘ − cos ⁡ 2 8 1 ∘ + sin ⁡ 2 2 7 ∘ 2 sin ⁡ 2 2 7 ∘ − cos ⁡ 2 2 7 ∘ + sin ⁡ 2 6 3 ∘ 2 sin ⁡ 2 6 3 ∘ − cos ⁡ 2 6 3 ∘ + 1 ) = 241 ( sin ⁡ 2 9 ∘ 3 sin ⁡ 2 9 ∘ − 1 + sin ⁡ 2 8 1 ∘ 3 sin ⁡ 2 8 1 ∘ − 1 + sin ⁡ 2 2 7 ∘ 3 sin ⁡ 2 2 7 ∘ − 1 + sin ⁡ 2 6 3 ∘ 3 sin ⁡ 2 6 3 ∘ − 1 + 1 ) = 241 ( sin ⁡ 2 9 ∘ 3 sin ⁡ 2 9 ∘ − 1 + cos ⁡ 2 9 ∘ 3 cos ⁡ 2 9 ∘ − 1 + sin ⁡ 2 2 7 ∘ 3 sin ⁡ 2 2 7 ∘ − 1 + cos ⁡ 2 2 7 ∘ 3 cos ⁡ 2 2 7 ∘ − 1 + 1 ) = 241 ( 3 sin ⁡ 2 9 ∘ cos ⁡ 2 9 ∘ − sin ⁡ 2 9 ∘ + 3 sin ⁡ 2 9 ∘ cos ⁡ 2 9 ∘ − cos ⁡ 2 9 ∘ 9 sin ⁡ 2 9 ∘ cos ⁡ 2 9 ∘ − 3 sin ⁡ 2 9 ∘ − 3 cos ⁡ 2 9 ∘ + 1 + . . . + 1 ) = 241 ( 6 sin ⁡ 2 9 ∘ cos ⁡ 2 9 ∘ − 1 9 sin ⁡ 2 9 ∘ cos ⁡ 2 9 ∘ − 2 + 6 sin ⁡ 2 2 7 ∘ cos ⁡ 2 2 7 ∘ − 1 9 sin ⁡ 2 2 7 ∘ cos ⁡ 2 2 7 ∘ − 2 + 1 ) = 241 ( 6 sin ⁡ 2 1 8 ∘ − 4 9 sin ⁡ 2 1 8 ∘ − 8 + 6 sin ⁡ 2 5 4 ∘ − 4 9 sin ⁡ 2 5 4 ∘ − 8 + 1 ) = 241 ( 6 ( 1 − cos ⁡ 3 6 ∘ 2 ) − 4 9 ( 1 − cos ⁡ 3 6 ∘ 2 ) − 8 + 6 ( 1 − cos ⁡ 10 8 ∘ 2 ) − 4 9 ( 1 − cos ⁡ 10 8 ∘ 2 ) − 8 + 1 ) = 241 ( 6 cos ⁡ 3 6 ∘ + 2 9 cos ⁡ 3 6 ∘ + 7 + 6 cos ⁡ 10 8 ∘ + 2 9 cos ⁡ 10 8 ∘ + 7 + 1 ) = 241 ( 108 cos ⁡ 3 6 ∘ cos ⁡ 10 8 ∘ + 60 ( cos ⁡ 3 6 ∘ + cos ⁡ 10 8 ∘ ) + 28 81 cos ⁡ 3 6 ∘ cos ⁡ 10 8 ∘ + 63 ( cos ⁡ 3 6 ∘ + cos ⁡ 10 8 ∘ ) + 49 + 1 ) See Note = 241 ( 108 ( − 1 4 ) + 60 ( 1 2 ) + 28 81 ( − 1 4 ) + 63 ( 1 2 ) + 49 + 1 ) = 241 ( 31 60.25 + 1 ) = 365 \color{#3D99F6}{\frac{241}{2-\cot^2{9^\circ}}} + \color{#D61F06}{\frac{241}{2-\cot^2{27^\circ}}} + \frac{241}{2-\cot^2{45^\circ}} + \color{#3D99F6}{\frac{241}{2-\cot^2 {63^ \circ }}} + \color{#D61F06}{\frac{241}{2-\cot^2{81^\circ}}} \\ = \color{#3D99F6}{\frac{241}{2-\cot^2{9^\circ}} + \frac{241}{2-\cot^2{81^\circ}}} + \color{#D61F06}{\frac{241}{2-\cot^2{27^\circ}} + \frac{241}{2-\cot^2 {63^ \circ }}} + 241 \\ = 241 \left(\color{#3D99F6}{\frac{\sin^2{9^\circ}}{2\sin^2{9^\circ}-\cos^2{9^\circ}} + \frac{\sin^2{81^\circ}}{2\sin^2{81^\circ}-\cos^2{81^\circ}}} + \color{#D61F06}{\frac{\sin^2{27^\circ}}{2\sin^2{27^\circ}-\cos^2{27^\circ}} + \frac{\sin^2{63^\circ}}{2\sin^2{63^\circ}-\cos^2{63^\circ}}} + 1 \right) \\ = 241 \left(\frac{\sin^2{9^\circ}}{3\sin^2{9^\circ}-1} + \color{#3D99F6}{\frac{\sin^2{81 ^\circ}}{3\sin^2{81^\circ}-1}} +\frac{\sin^2{27^\circ}}{3\sin^2{27^\circ}-1} + \color{#D61F06}{\frac{\sin^2{63^\circ}}{3\sin^2{63^\circ}-1}} + 1 \right) \\ = 241 \left(\frac{\sin^2{9^\circ}}{3\sin^2{9^\circ}-1} + \color{#3D99F6}{\frac{ \cos^2 {9^\circ}}{3\cos^2{9^\circ}-1}} + \frac{\sin^2{27^\circ}}{3\sin^2{27^\circ}-1} + \color{#D61F06}{\frac{\cos^2{27^\circ}}{3\cos^2{27^\circ}-1}} + 1 \right) \\ = 241 \left( \frac{3\sin^2{9^\circ}\cos^2{9^\circ} - \sin^2{9^\circ} + 3\sin^2 {9^\circ}\cos^2{9^\circ} - \cos^2{9^\circ}} {9\sin^2{9^\circ} \cos^2{9^\circ} - 3\sin^2{9^\circ} - 3\cos^2{9^\circ} + 1} +...+ 1 \right) \\ = 241 \left( \frac{6 \sin^2{9^\circ}\cos^2{9^\circ} - 1} {9\sin^2{9^\circ} \cos^2{9^\circ} - 2} +\frac{6 \sin^2{27^\circ}\cos^2{27^\circ} - 1} {9\sin^2{27^\circ} \cos^2{27^\circ} - 2} + 1 \right) \\ = 241 \left( \frac{6 \sin^2{18^\circ} - 4} {9\sin^2{18^\circ} - 8} +\frac{6 \sin^2{54^\circ} - 4} {9\sin^2{54^\circ} - 8} + 1 \right) \\ = 241 \left( \frac{6 \left(\frac{1-\cos{36^\circ}}{2} \right) - 4} {9\left(\frac{1-\cos{36^\circ}}{2} \right) - 8} +\frac{6 \left(\frac{1-\cos{108^\circ}}{2} \right) - 4} {9\left(\frac{1-\cos{108^\circ}}{2} \right) - 8} + 1 \right) \\ = 241 \left( \frac{6 \cos{36^\circ}+2} {9\cos{36^\circ} + 7} +\frac{6 \cos{108^\circ}+2} {9\cos{108^\circ} + 7} + 1 \right) \\ = 241 \left( \frac{108 \color{#3D99F6} {\cos{36^\circ}\cos{108^\circ}} + 60 \color{#3D99F6}{(\cos{36^\circ} + \cos{108^\circ} )} +28} {81 \color{#3D99F6}{ \cos{36^ \circ} \cos {108 ^\circ}} + 63 \color{#3D99F6} {(\cos{36^\circ} + \cos{108^\circ})}+49 } + 1 \right) \quad \quad \color{#3D99F6}{\text {See Note}} \\ = 241 \left( \frac{108 \color{#3D99F6} {(-\frac{1}{4})} + 60 \color{#3D99F6}{(\frac{1}{2} )} +28} {81 \color{#3D99F6} {(-\frac{1}{4})} + 63 \color{#3D99F6} {(\frac{1}{2})}+49 } + 1 \right) = 241 \left( \frac{31} {60.25 } + 1 \right) = \boxed{365}

Note: \color{#3D99F6}{\text{Note:}}

We note that z 5 = e 2 k π 5 = e 36 k ∘ = 1 z^5 = e^{\frac{2k\pi}{5}} = e^{36k^\circ} = 1 are the 5 t h 5^{th} roots of unity. From Argand's diagram, we have:

{ cos ⁡ 3 6 ∘ + cos ⁡ 10 8 ∘ = 1 2 . . . ( 1 ) cos ⁡ 7 2 ∘ + cos ⁡ 14 4 ∘ = − 1 2 . . . ( 2 ) \begin{cases} \cos{36^\circ} + \cos{108^\circ} = \frac{1}{2} & ...(1) \\ \cos{72^\circ} + \cos{144^\circ} = -\frac{1}{2} & ...(2) \end{cases}

( 2 ) : cos ⁡ 7 2 ∘ + 2 cos ⁡ 2 7 2 ∘ − 1 = − 1 2 4 cos ⁡ 2 7 2 ∘ + 2 cos ⁡ 7 2 ∘ − 1 = 0 \begin{aligned} (2): \quad \cos{72^\circ} + 2\cos^2{72^\circ} - 1 & = -\frac{1}{2} \\ 4\cos^2{72^\circ} + 2\cos{72^\circ} - 1 & = 0 \end{aligned}

⇒ cos ⁡ 7 2 ∘ = 5 − 1 4 ⇒ 2 cos ⁡ 2 3 6 ∘ − 1 = 5 − 1 4 cos ⁡ 2 3 6 ∘ = 5 + 3 8 = 2 5 + 6 16 = ( 5 + 1 ) 2 16 ⇒ cos ⁡ 3 6 ∘ = 5 + 1 4 \begin{aligned} \Rightarrow \cos{72^\circ} & = \frac{\sqrt{5}-1}{4} \\ \Rightarrow 2\cos^2{36^\circ} - 1 & = \frac{\sqrt{5}-1}{4} \\ \cos^2{36^\circ} & = \frac{\sqrt{5}+3}{8} = \frac{2 \sqrt{5} +6}{16} = \frac{(\sqrt{5}+1)^2}{16} \\ \Rightarrow \cos{36^\circ} & = \frac{\sqrt{5}+1}{4} \end{aligned}

( 1 ) : cos ⁡ 10 8 ∘ = 1 2 − cos ⁡ 3 6 ∘ = 1 2 − 5 + 1 4 = 1 − 5 4 \begin{aligned} (1): \quad \cos{108^\circ} & = \frac{1}{2} - \cos{36^\circ} = \frac{1}{2} - \frac{\sqrt{5}+1}{4} = \frac{1 - \sqrt{5}}{4} \end{aligned}

⇒ cos ⁡ 3 6 ∘ cos ⁡ 10 8 ∘ = 1 − 5 4 × 5 + 1 4 = − 1 4 \begin{aligned} \Rightarrow \cos{36^\circ} \cos{108^\circ} & = \frac{1 - \sqrt{5}}{4} \times \frac{\sqrt{5}+1}{4} = - \frac{1}{4} \end{aligned}

20 Helpful 1 Interesting 1 Brilliant 1 Confused

@Chew-Seong Cheong , we really liked your comment, and have converted it into a solution. If you subscribe to this solution, you will receive notifications about future comments.

Calvin Lin Staff - 5 years, 10 months ago

can you explain how you went from the 7th line to the 8th line. i thought you used the double angle formula on 6(sin9cos9)^2 - 1 in the 7th line but if that's the case, i don't see how the constant term can change to -4 in the following line. did you apply some other trick? thanks!!!

Willia Chang - 4 years, 12 months ago

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There is no trick. Write the calculation out.

Hint: sin ⁡ θ cos ⁡ θ = 1 2 sin ⁡ 2 θ \sin \theta \cos \theta = \frac{1}{2} \sin 2 \theta .

Calvin Lin Staff - 4 years, 12 months ago

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OH...i get it now, sorry i simplified the numerator and found that it didn't match the next line when i should have simplified numerator and denominator and change the fractional coefficients into integers. that way, i'll get the same form as in line 8. thanks for the help!!!!!!!

Willia Chang - 4 years, 12 months ago

can be solved in a much easier way if you use eulers identity, this corresponds to cos(10x)=0 which gives the roots as (2 n-1) pi/20 where n varies from 1 to 10

subharthi chowdhuri - 6 years, 6 months ago
汶良 林
Mar 18, 2016

8 Helpful 2 Interesting 3 Brilliant 2 Confused

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