Do we need to construct decagons?

Geometry Level 3

2 tan ⁡ ( π 10 ) + 3 sec ⁡ ( π 10 ) − 4 cos ⁡ ( π 10 ) = ? 2\tan \left( \frac{\pi}{10} \right) + 3 \sec \left( \frac{\pi}{10} \right) - 4 \cos \left ( \frac{\pi}{10}\right) = \ ?


The answer is 0.

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Utkarsh Yadav by Utkarsh Yadav , 22, India

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3 solutions

Chew-Seong Cheong
Jun 12, 2015

2 tan ⁡ π 10 + 3 sec ⁡ π 10 − 4 cos ⁡ π 10 = 2 sin ⁡ π 10 cos ⁡ π 10 + 3 cos ⁡ π 10 − 4 cos ⁡ π 10 = 2 sin ⁡ π 10 + 3 − 4 cos ⁡ 2 π 10 cos ⁡ π 10 = 2 cos ⁡ 2 π 5 + 3 − 2 ( 2 cos ⁡ 2 π 10 − 1 + 1 ) cos ⁡ π 10 = 2 cos ⁡ 2 π 5 + 3 − 2 cos ⁡ π 5 − 2 cos ⁡ π 10 = 2 cos ⁡ 2 π 5 − 2 cos ⁡ π 5 + 1 cos ⁡ π 10 = 2 cos ⁡ 2 π 5 + 2 cos ⁡ 4 π 5 + 1 cos ⁡ π 10 = 2 ( cos ⁡ 2 π 5 + cos ⁡ 4 π 5 ) + 1 cos ⁡ π 10 = 2 ( − 1 2 ) + 1 cos ⁡ π 10 = 0 \begin{aligned} 2\tan{\frac{\pi}{10}} + 3\sec {\frac{\pi}{10}} - 4 \cos {\frac{\pi}{10}} & = \frac{2\sin{\frac{\pi}{10}}}{\cos{\frac{\pi}{10}}} + \frac{3}{\cos{\frac{\pi}{10}}} - 4\cos{\frac{\pi}{10}} \\ & = \frac {2\sin{\frac{\pi}{10}}+3 - 4\cos^2{\frac{\pi}{10}}} {\cos{\frac{\pi}{10}}} \\ & = \frac {2\cos{\frac{2\pi}{5}}+3 - 2\left( 2\cos^2{\frac{\pi}{10}}-1+1 \right)} {\cos{\frac{\pi}{10}}} \\ & = \frac {2\cos{\frac{2\pi}{5}}+3 - 2\cos{\frac{\pi}{5}} -2 } {\cos{\frac{\pi}{10}}} \\ & = \frac {2\cos{\frac{2\pi}{5}} - 2\cos{\frac{\pi}{5}} + 1 } {\cos{\frac{\pi}{10}}} \\ & = \frac {2\cos{\frac{2\pi}{5}} + 2\cos{\frac{4\pi}{5}} + 1 } {\cos{\frac{\pi}{10}}} \\ & = \frac {2 \color{#3D99F6}{\left( \cos{\frac{2\pi}{5}} + \cos{\frac{4\pi}{5}} \right)} + 1 } {\cos{\frac{\pi}{10}}} \\ & = \frac {2 \color{#3D99F6}{\left( -\frac{1}{2} \right)} + 1 } {\cos{\frac{\pi}{10}}} = \boxed{0} \end{aligned}

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Moderator note:

How can one tell that 4 sin ⁡ 2 π 10 + 2 sin ⁡ π 10 − 1 = 0 4 \sin ^2 \frac{\pi}{10} + 2 \sin \frac{ \pi}{10} - 1 = 0 ?

That would have been the numerator if we replaced cos ⁡ 2 π 10 \cos^2 \frac{ \pi }{10} with 1 − sin ⁡ 2 π 10 1 - \sin ^2 \frac{ \pi}{10} in the 2nd line.

how you know that the blue expression is equal to -1/2

Aldo Mauricio - 5 years, 11 months ago

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z 10 = e k π 5 i = 1 z^{10} = e^{\frac{k\pi}{5}i} = 1 , the 1 0 t h 10^{th} roots of unity. Using Argand's diagram we can see that cos ⁡ π 5 + cos ⁡ 3 π 5 = 1 2 \cos{\frac{\pi}{5}} + \cos{\frac{3\pi}{5}} = \frac{1}{2} and cos ⁡ 2 π 5 + cos ⁡ 4 π 5 = − 1 2 \cos{\frac{2\pi}{5}} + \cos{\frac{4\pi}{5}} = -\frac{1}{2} . There are similar identities such as cos ⁡ π 3 = 1 2 \cos{\frac{\pi}{3}} = \frac{1}{2} and cos ⁡ 2 π 3 = − 1 2 \cos{\frac{2\pi}{3}} = -\frac{1}{2} ; cos ⁡ π 7 + cos ⁡ 3 π 7 + cos ⁡ 5 π 7 = 1 2 \cos{\frac{\pi}{7}} + \cos{\frac{3\pi}{7}} + \cos{\frac{5\pi}{7}} = \frac{1}{2} and cos ⁡ 2 π 7 + cos ⁡ 4 π 7 + cos ⁡ 6 π 7 = − 1 2 \cos{\frac{2\pi}{7}} + \cos{\frac{4\pi}{7}} + \cos{\frac{6\pi}{7}} = -\frac{1}{2} ; and for 9 , 11 , 13 , . . . 9,11,13,... .

Chew-Seong Cheong - 5 years, 11 months ago

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if e to pi by 5 is equal to one does it mean it doesn't have an imaginary part ?

Aldo Mauricio - 5 years, 11 months ago

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@Aldo Mauricio – Aldo, sorry, I missed out the k k earlier. When z 10 = 1 z^{10} = 1 means that there are altogether 10 10 complex roots e 2 k π i 10 = e k π i 5 = cos ⁡ k π 5 + i sin ⁡ k π 5 e^{\frac{2k\pi i}{10}} = e^{\frac{k \pi i}{5}} = \cos{\frac{k \pi}{5}} + i \sin{\frac{k \pi}{5}} , where k = 0 , 1 , 2 , . . . 9 k = 0,1,2,...9 . When k = 0 k=0 , then z = e 0 = 1 z=e^0=1 , the other 9 9 roots have imaginary part.

Chew-Seong Cheong - 5 years, 11 months ago

4 sin ⁡ 2 π 10 + 2 sin ⁡ π 10 − 1 = − 2 ( 1 − 2 sin ⁡ 2 π 10 ) + 2 + 2 cos ⁡ ( π 2 − π 10 ) − 1 = − 2 cos ⁡ π 5 + 2 cos ⁡ 2 π 5 + 1 = 2 cos ⁡ 4 π 5 + 2 cos ⁡ 2 π 5 + 1 = 2 ( − 1 2 ) + 1 = 0 \begin{aligned} 4\sin^2{\frac{\pi}{10}} + 2\sin{\frac{\pi}{10}} - 1 & = -2(1-2\sin^2{\frac{\pi}{10}}) +2 + 2\cos{\left(\frac{\pi}{2} - \frac{\pi}{10}\right)} - 1 \\ & = -2\cos{\frac{\pi}{5}} + 2\cos{\frac{2\pi}{5}} + 1 \\ & = 2\cos{\frac{4\pi}{5}} + 2\cos{\frac{2\pi}{5}} + 1 \\ & = 2\left(-\frac{1}{2}\right) + 1 \\ & = \boxed{0} \end{aligned}

Chew-Seong Cheong - 5 years, 11 months ago
Pi Han Goh
Jun 12, 2015

Let x = π 10 x = \frac{\pi}{10} , then cos ⁡ ( 2 x ) = sin ⁡ ( 3 x ) \cos(2x) = \sin \left( 3x \right) . By applying the double angle formula and triple angle formula, and letting y = sin ⁡ ( x ) y = \sin(x) , we have 1 − 2 y 2 = − 4 y 3 + 3 y ⇒ ( y − 1 ) ( 4 y 2 + 2 y − 1 ) = 0 1 - 2y^2 = -4y^3 + 3y \Rightarrow (y-1)(4y^2 + 2y-1) = 0 . It's easy to easy that y ≠ 1 y\ne 1 so we can cancel out the factor ( y − 1 ) (y-1) . Reverse the process:

4 y 2 + 2 y − 1 = 0 2 y + 3 − 4 ( 1 − y 2 ) = 0 2 sin ⁡ ( x ) + 3 − 4 cos ⁡ 2 ( x ) = 0 2 tan ⁡ ( x ) + 3 sec ⁡ ( x ) − 4 cos ⁡ ( x ) = 0 \begin{aligned} 4y^2 + 2y - 1 &=& 0 \\ 2y + 3 - 4(1-y^2) &=& 0 \\ 2\sin(x) + 3 - 4\cos^2(x) &=& 0 \\ 2\tan(x) + 3\sec(x) - 4\cos(x) &=& \boxed 0 \end{aligned}

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Moderator note:

That's one way to find the minimal polynomial of sin ⁡ π 10 \sin \frac{ \pi }{10} .

Alternatively, you can use sin ⁡ 5 θ \sin 5 \theta , and factorize it from there.

Here's a dumb alternative approach:

Claim : 2 tan ⁡ ( π 10 ) + 3 sec ⁡ ( π 10 ) − 4 cos ⁡ ( π 10 ) = 0 2\tan \left( \frac{\pi}{10} \right) + 3 \sec \left( \frac{\pi}{10} \right) - 4 \cos \left ( \frac{\pi}{10}\right) = 0 .

Proof : Suppose otherwise, that is the equation is false. We begin by letting x = π 10 x =\frac{\pi}{10} , then 2 sin ⁡ ( x ) + 3 − 4 cos ⁡ 2 ( x ) ≠ 0 2\sin(x) + 3 - 4\cos^2(x) \ne 0 , applying the Pytharoas identites and set sin ⁡ ( x ) \sin(x) as the subject and via quadratic formula, we get sin ⁡ ( x ) ≠ 5 ± 1 4 \sin(x) \ne \frac{\sqrt5 \pm 1}{4} . However one can easily rule out that sin ⁡ ( x ) ≠ 5 + 1 4 > 1 2 = sin ⁡ ( π 6 ) \sin(x) \ne \frac{\sqrt5 + 1}{4} > \frac12 = \sin\left(\frac\pi 6 \right) . So it's just left to show that the equation sin ⁡ ( x ) ≠ 5 − 1 4 \sin(x) \ne \frac{\sqrt5 - 1}{4} must be false. Consider sin ⁡ ( 4 x ) = cos ⁡ ( x ) \sin(4x) = \cos(x) , apply the double angle formula repeatedly, we get 4 sin ⁡ ( x ) cos ⁡ ( 2 x ) = 1 ⇒ 4 sin ⁡ ( x ) ( 1 − 2 sin ⁡ 2 ( x ) ) = 1 4\sin(x) \cos(2x) = 1 \Rightarrow 4\sin(x) (1-2\sin^2 (x)) = 1 , then ( 5 − 1 ) ( 1 − 2 ( 5 − 1 4 ) 2 ) ≠ 1 (\sqrt5 - 1) \left(1 - 2\left(\frac{\sqrt 5-1}4\right)^2 \right) \ne 1 which is absurd. Hence, our assumption must be wrong therefore the answer must be 0.

Pi Han Goh - 6 years ago

Challenge Master: I've considered solving for sin ⁡ ( 5 θ ) − 1 = 0 \sin(5\theta) - 1 = 0 by quintuple angle formula, but the further factorization of ( x − 1 ) ( 16 x 4 + 16 x 3 − 4 x 2 − 4 x + 1 ) (x-1)(16x^4 + 16x^3-4x^2 -4x + 1) to ( x − 1 ) ( 4 x 2 + 2 x − 1 ) 2 (x-1)(4x^2+2x-1)^2 is not something that is obvious. Is there a quick way to determine whether this quartic factor is simply a square of a quadratic factor? And I also noticed a pattern that by converting the expression sin ⁡ ( ( 2 n + 1 ) θ ) − 1 \sin((2n+1)\theta) - 1 via Tchebyshev polynomials, we are always left with a factor of ( x − 1 ) (x-1) and a polynomial of degree n n with multiplicity of 2. Is that a coincidence? For prime ( n n ).

Pi Han Goh - 6 years ago
Noel Lo
Sep 3, 2015

Interesting question!

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