Which Formula Should I Apply?

Geometry Level 2

Find the value of E = tan ⁡ 2 π 40 + tan ⁡ 2 3 π 40 + tan ⁡ 2 5 π 40 + ⋯ + tan ⁡ 2 19 π 40 . E=\tan^2{\dfrac{\pi}{40}}+\tan^2{\dfrac{3\pi}{40}}+\tan^2{\dfrac{5\pi}{40}}+\cdots+\tan^2{\dfrac{19\pi}{40}}.


The answer is 190.

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Pratik Shastri by Pratik Shastri , 35, India

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1 solution

Dhruva Patil
Dec 12, 2014

E = tan ⁡ 2 π 40 + tan ⁡ 2 3 π 40 + tan ⁡ 2 5 π 40 + tan ⁡ 2 7 π 40 + . . . tan ⁡ 2 19 π 40 { 10 t e r m s } E = ( sec ⁡ 2 π 40 − 1 ) + ( sec ⁡ 2 3 π 40 − 1 ) + ( sec ⁡ 2 5 π 40 − 1 ) + ( sec ⁡ 2 7 π 40 − 1 ) + . . . E = 1 cos ⁡ 2 π 40 + 1 cos ⁡ 2 3 π 40 + 1 cos ⁡ 2 5 π 40 + 1 cos ⁡ 2 7 π 40 + . . . − 10 E = ( 1 cos ⁡ 2 π 40 + 1 sin ⁡ 2 ( π 2 − 19 π 40 ) ) + ( 1 cos ⁡ 2 3 π 40 + 1 sin ⁡ 2 ( π 2 − 17 π 40 ) ) + . . . − 10 E = 4 × ( sin ⁡ 2 π 40 + cos ⁡ 2 π 40 ( 2 sin ⁡ π 40 cos ⁡ π 40 ) 2 ) + 4 × ( sin ⁡ 2 3 π 40 + cos ⁡ 2 3 π 40 ( 2 sin ⁡ 3 π 40 cos ⁡ 3 π 40 ) 2 ) + . . . − 10 E = 4 × ( 1 sin ⁡ 2 π 20 + 1 sin ⁡ 2 3 π 20 + 1 sin ⁡ 2 5 π 20 + 1 sin ⁡ 2 7 π 20 + 1 sin ⁡ 2 9 π 20 ) − 10 E = 4 × ( ( 1 sin ⁡ 2 π 20 + 1 cos ⁡ 2 ( π 2 − 9 π 20 ) ) + ( 1 sin ⁡ 2 3 π 20 + 1 cos ⁡ 2 ( π 2 − 7 π 20 ) ) + 1 sin ⁡ 2 5 π 20 ) − 10 E = 16 ( 1 sin ⁡ 2 π 10 + 1 sin ⁡ 2 3 π 10 ) + 8 − 10 { sin ⁡ π 10 = cos ⁡ 2 π 5 = 5 − 1 4 , sin ⁡ 3 π 10 = cos ⁡ π 5 = 5 + 1 4 } E = 16 × 12 − 2 E = 190 E=\displaystyle\tan ^{ 2 }{ \frac { \pi }{ 40 } } +\tan ^{ 2 }{ \frac { 3\pi }{ 40 } } +\tan ^{ 2 }{ \frac { 5\pi }{ 40 } } +\tan ^{ 2 }{ \frac { 7\pi }{ 40 } } +...\tan ^{ 2 }{ \frac { 19\pi }{ 40 } } \quad \{ 10\quad terms\} \\ E=\displaystyle(\sec ^{ 2 }{ \frac { \pi }{ 40 } } -1)+(\sec ^{ 2 }{ \frac { 3\pi }{ 40 } } -1)+(\sec ^{ 2 }{ \frac { 5\pi }{ 40 } } -1)+(\sec ^{ 2 }{ \frac { 7\pi }{ 40 } } -1)+...\\ E=\displaystyle \frac { 1 }{ \cos ^{ 2 }{ \frac { \pi }{ 40 } } } +\frac { 1 }{ \cos ^{ 2 }{ \frac { 3\pi }{ 40 } } } +\frac { 1 }{ \cos ^{ 2 }{ \frac { 5\pi }{ 40 } } } +\frac { 1 }{ \cos ^{ 2 }{ \frac { 7\pi }{ 40 } } } +...-10\\ E=\displaystyle(\frac { 1 }{ \cos ^{ 2 }{ \frac { \pi }{ 40 } } } +\frac { 1 }{ \sin ^{ 2 }{ (\frac { \pi }{ 2 } -\frac { 19\pi }{ 40 } ) } } )+(\frac { 1 }{ \cos ^{ 2 }{ \frac { 3\pi }{ 40 } } } +\frac { 1 }{ \sin ^{ 2 }{ (\frac { \pi }{ 2 } -\frac { 17\pi }{ 40 } ) } } )+...-10\\ E=\displaystyle4\times (\frac { \sin ^{ 2 }{ \frac { \pi }{ 40 } } +\cos ^{ 2 }{ \frac { \pi }{ 40 } } }{ { (2\sin { \frac { \pi }{ 40 } } \cos { \frac { \pi }{ 40 } } ) }^{ 2 } } )+4\times (\frac { \sin ^{ 2 }{ \frac { 3\pi }{ 40 } } +\cos ^{ 2 }{ \frac { 3\pi }{ 40 } } }{ { (2\sin { \frac { 3\pi }{ 40 } } \cos { \frac { 3\pi }{ 40 } } ) }^{ 2 } } )+...-10\\ E=\displaystyle4\times (\frac { 1 }{ \sin ^{ 2 }{ \frac { \pi }{ 20 } } } +\frac { 1 }{ \sin ^{ 2 }{ \frac { 3\pi }{ 20 } } } +\frac { 1 }{ \sin ^{ 2 }{ \frac { 5\pi }{ 20 } } } +\frac { 1 }{ \sin ^{ 2 }{ \frac { 7\pi }{ 20 } } } +\frac { 1 }{ \sin ^{ 2 }{ \frac { 9\pi }{ 20 } } } )-10\\ E=\displaystyle4\times ((\frac { 1 }{ \sin ^{ 2 }{ \frac { \pi }{ 20 } } } +\frac { 1 }{ \cos ^{ 2 }{ (\frac { \pi }{ 2 } -\frac { 9\pi }{ 20 } ) } } )+(\frac { 1 }{ \sin ^{ 2 }{ \frac { 3\pi }{ 20 } } } +\frac { 1 }{ \cos ^{ 2 }{ (\frac { \pi }{ 2 } -\frac { 7\pi }{ 20 } ) } } )+\frac { 1 }{ \sin ^{ 2 }{ \frac { 5\pi }{ 20 } } } )-10\\ E=\displaystyle16(\frac { 1 }{ \sin ^{ 2 }{ \frac { \pi }{ 10 } } } +\frac { 1 }{ \sin ^{ 2 }{ \frac { 3\pi }{ 10 } } } )+8-10\quad \\ \displaystyle\{ \sin { \frac { \pi }{ 10 } } =\cos { \frac { 2\pi }{ 5 } } =\frac { \sqrt { 5 } -1 }{ 4 } ,\sin { \frac { 3\pi }{ 10 } } =\cos { \frac { \pi }{ 5 } } =\frac { \sqrt { 5 } +1 }{ 4 } \} \\ E=16\times 12-2\\ E=\boxed { 190 } \\

18 Helpful 0 Interesting 2 Brilliant 0 Confused

80 is the product of a power of 2 and distinct Fermat primes, so trigonometric functions of 2 n π / 80 = n π / 40 2n\pi/80=n\pi/40 can be evaluated. I did this problem this way.

bobby jim - 6 years, 5 months ago

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Can you add more details about how you made use of this fact?

Calvin Lin Staff - 6 years, 4 months ago

Exactly did the same!!!

Rudresh Tomar - 6 years, 6 months ago

@Dhruva Patil Just a suggestion,if you change \frac to \dfrac,your solution would be a bit more clear.Here is an exmaple, 1 sin ⁡ 2 7 π 20 , \frac{1}{\sin^2{\frac{7\pi}{20}}}, 1 sin ⁡ 2 7 π 20 \dfrac{1}{\sin^2{\dfrac{7\pi}{20}}} .

Adarsh Kumar - 6 years, 6 months ago

exactly the same!

Asim Das - 6 years, 5 months ago

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