A complete ripoff

Geometry Level 5

sin ⁡ 6 0 ∘ + sin ⁡ 6 1 0 ∘ + sin ⁡ 6 2 0 ∘ + sin ⁡ 6 3 0 ∘ + … + sin ⁡ 6 18 0 ∘ cos ⁡ 6 0 ∘ + cos ⁡ 6 1 0 ∘ + cos ⁡ 6 2 0 ∘ + cos ⁡ 6 3 0 ∘ + … + cos ⁡ 6 18 0 ∘ \dfrac{\sin^6{0^{\circ}}+\sin^6{10^{\circ}}+\sin^6{20^{\circ}}+\sin^6{30^{\circ}}+\ldots+\sin^6{180^{\circ}}}{\cos^6{0^{\circ}}+\cos^6{10^{\circ}}+\cos^6{20^{\circ}}+\cos^6{30^{\circ}}+\ldots+\cos^6{180^{\circ}}} \\ \\

If the value of the above expression is equals to a b \frac{a}{b} where a a and b b are coprime positive integers, find the value of b − a b-a .

Bonus : For the general expression below, can you find a general formula for the answer, b n − a n b_{n} - a_{n} for all positive integers n n ?

sin ⁡ 2 n 0 ∘ + sin ⁡ 2 n 1 0 ∘ + sin ⁡ 2 n 2 0 ∘ + sin ⁡ 2 n 3 0 ∘ + … + sin ⁡ 2 n 18 0 ∘ cos ⁡ 2 n 0 ∘ + cos ⁡ 2 n 1 0 ∘ + cos ⁡ 2 n 2 0 ∘ + cos ⁡ 2 n 3 0 ∘ + … + cos ⁡ 2 n 18 0 ∘ \dfrac{\sin^{2n}{0^{\circ}}+\sin^{2n}{10^{\circ}}+\sin^{2n}{20^{\circ}}+\sin^{2n}{30^{\circ}}+\ldots+\sin^{2n}{180^{\circ}}}{\cos^{2n}{0^{\circ}}+\cos^{2n}{10^{\circ}}+\cos^{2n}{20^{\circ}}+\cos^{2n}{30^{\circ}}+\ldots+\cos^{2n}{180^{\circ}}}

Inspiration .


The answer is 8.

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Pi Han Goh by Pi Han Goh , 30, Malaysia

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2 solutions

I've found a general formula for S ( n ) = ∑ k = 0 18 sin ⁡ 2 n ( 1 0 ∘ k ) ∑ k = 0 18 cos ⁡ 2 n ( 1 0 ∘ k ) S(n)=\frac{\displaystyle \sum_{k=0}^{18} \sin^{2n}(10^\circ k)}{\displaystyle \sum_{k=0}^{18} \cos^{2n}(10^\circ k)} , and this space is big enough to contain it:

S ( n ) = 19 ( 2 n n ) − 2 ( 2 n − 1 n − 1 ) 2 2 n + 18 ( 2 n n ) \boxed{S(n)=\dfrac{19\displaystyle \binom{2n}{n}-2\displaystyle \binom{2n-1}{n-1}}{2^{2n}+18\displaystyle \binom{2n}{n}}}

Proof (quite long):

Let t = cis ( 1 0 ∘ ) t=\text{cis}(10^\circ) . Now, by De Moivre's theorem:

cos ⁡ ( 1 0 ∘ k ) = 1 2 ( t k + 1 t k ) sin ⁡ ( 1 0 ∘ k ) = 1 2 i ( t k − 1 t k ) \cos(10^\circ k)=\dfrac{1}{2}\left(t^k+\dfrac{1}{t^k}\right) \\ \sin(10^\circ k)=\dfrac{1}{2i}\left(t^k-\dfrac{1}{t^k}\right)

Raise both sides to 2 n 2n , using the binomial theorem:

cos ⁡ 2 n ( 1 0 ∘ k ) = 1 2 2 n ( t k + 1 t k ) 2 n sin ⁡ 2 n ( 1 0 ∘ k ) = ( − 1 ) n 2 2 n ( t k − 1 t k ) 2 n \cos^{2n}(10^\circ k)=\dfrac{1}{2^{2n}}\left(t^k+\dfrac{1}{t^k}\right)^{2n} \\ \sin^{2n}(10^\circ k)=\dfrac{(-1)^n}{2^{2n}}\left(t^k-\dfrac{1}{t^k}\right)^{2n}

cos ⁡ 2 n ( 1 0 ∘ k ) = 1 2 2 n ( ∑ m = 0 n − 1 ( 2 n m ) ( ( t 2 k ) n − m + 1 ( t 2 k ) n − m ) + ( 2 n n ) ) sin ⁡ 2 n ( 1 0 ∘ k ) = ( − 1 ) n 2 2 n ( ∑ m = 0 n − 1 ( − 1 ) m ( 2 n m ) ( ( t 2 k ) n − m + 1 ( t 2 k ) n − m ) + ( − 1 ) n ( 2 n n ) ) \cos^{2n}(10^\circ k)=\dfrac{1}{2^{2n}}\left(\displaystyle \sum_{m=0}^{n-1}\displaystyle \binom{2n}{m} \left((t^{2k})^{n-m}+\dfrac{1}{(t^{2k})^{n-m}}\right)+\displaystyle \binom{2n}{n}\right) \\ \sin^{2n}(10^\circ k)=\dfrac{(-1)^n}{2^{2n}}\left(\displaystyle \sum_{m=0}^{n-1}\left(-1\right)^m \displaystyle \binom{2n}{m} \left((t^{2k})^{n-m}+\dfrac{1}{(t^{2k})^{n-m}}\right)+\left(-1\right)^n \displaystyle \binom{2n}{n}\right)

Now, to find the summation of sin ⁡ 2 n \sin^{2n} and cos ⁡ 2 n \cos^{2n} , find the sums inside, using the G.P. sum formula and the fact that t 36 = 1 t^{36}=1 :

∑ k = 0 18 ( t 2 k ) n − m = ∑ k = 0 18 ( t 2 ( n − m ) ) k = ( t 2 ( n − m ) ) 19 − 1 t 2 ( n − m ) − 1 = t 38 ( n − m ) − 1 t 2 ( n − m ) − 1 = t 2 ( n − m ) − 1 t 2 ( n − m ) − 1 = 1 \displaystyle \sum_{k=0}^{18} \left(t^{2k}\right)^{n-m}=\displaystyle \sum_{k=0}^{18} \left(t^{2(n-m)}\right)^k=\dfrac{(t^{2(n-m)})^{19}-1}{t^{2(n-m)}-1}=\dfrac{t^{38(n-m)}-1}{t^{2(n-m)}-1}=\dfrac{t^{2(n-m)}-1}{t^{2(n-m)}-1}=1 \\

∑ k = 0 18 ( 1 t 2 k ) n − m = ∑ k = 0 18 ( 1 t 2 ( n − m ) ) k = ( 1 t 2 ( n − m ) ) 19 − 1 1 t 2 ( n − m ) − 1 = 1 t 38 ( n − m ) − 1 1 t 2 ( n − m ) − 1 = 1 t 2 ( n − m ) − 1 1 t 2 ( n − m ) − 1 = 1 \displaystyle \sum_{k=0}^{18} \left(\dfrac{1}{t^{2k}}\right)^{n-m}=\displaystyle \sum_{k=0}^{18} \left(\dfrac{1}{t^{2(n-m)}}\right)^k=\dfrac{\left(\dfrac{1}{t^{2(n-m)}}\right)^{19}-1}{\dfrac{1}{t^{2(n-m)}}-1}=\dfrac{\dfrac{1}{t^{38(n-m)}}-1}{\dfrac{1}{t^{2(n-m)}}-1}=\dfrac{\dfrac{1}{t^{2(n-m)}}-1}{\dfrac{1}{t^{2(n-m)}}-1}=1 \\

∑ k = 0 18 ( 2 n n ) = 19 ( 2 n n ) \displaystyle \sum_{k=0}^{18}\displaystyle \binom{2n}{n}=19\displaystyle \binom{2n}{n}

Hence:

∑ k = 0 18 cos ⁡ 2 n ( 1 0 ∘ k ) = 1 2 2 n ( ∑ m = 0 n − 1 ( 2 n m ) ( 1 + 1 ) + 19 ( 2 n n ) ) \displaystyle \sum_{k=0}^{18} \cos^{2n}(10^\circ k)=\dfrac{1}{2^{2n}} \left(\displaystyle \sum_{m=0}^{n-1} \displaystyle \binom{2n}{m}\left(1+1\right)+19\displaystyle \binom{2n}{n}\right)

∑ k = 0 18 sin ⁡ 2 n ( 1 0 ∘ k ) = ( − 1 ) n 2 2 n ( ∑ m = 0 n − 1 ( 2 n m ) ( 1 + 1 ) ( − 1 ) m + 19 ( − 1 ) n ( 2 n n ) ) \displaystyle \sum_{k=0}^{18} \sin^{2n}(10^\circ k)=\dfrac{(-1)^n}{2^{2n}} \left(\displaystyle \sum_{m=0}^{n-1} \displaystyle \binom{2n}{m}\left(1+1\right)\left(-1\right)^m+19\left(-1\right)^n\displaystyle \binom{2n}{n}\right)

First simplify the expression for the sine using the following identity: ∑ m = 0 n − 1 ( − 1 ) m ( 2 n m ) = ( − 1 ) n + 1 ( 2 n − 1 n − 1 ) \displaystyle \sum_{m=0}^{n-1}\left(-1\right)^m\displaystyle \binom{2n}{m}=\left(-1\right)^{n+1}\displaystyle \binom{2n-1}{n-1} :

∑ k = 0 18 sin ⁡ 2 n ( 1 0 ∘ k ) = ( − 1 ) n 2 2 n ( 2 ( − 1 ) n + 1 ( 2 n − 1 n − 1 ) + 19 ( − 1 ) n ( 2 n n ) ) ∑ k = 0 18 sin ⁡ 2 n ( 1 0 ∘ k ) = 1 2 2 n ( 19 ( 2 n n ) − 2 ( 2 n − 1 n − 1 ) ) \displaystyle \sum_{k=0}^{18} \sin^{2n}(10^\circ k)=\dfrac{(-1)^n}{2^{2n}} \left(\displaystyle 2(-1)^{n+1}\binom{2n-1}{n-1}+19\left(-1\right)^n\displaystyle \binom{2n}{n}\right) \\ \displaystyle \sum_{k=0}^{18} \sin^{2n}(10^\circ k)=\dfrac{1}{2^{2n}} \left(19\displaystyle \binom{2n}{n}-\displaystyle 2\binom{2n-1}{n-1}\right)

And simplify the expression for the cosine with this other useful identity: 2 ∑ m = 0 n − 1 ( 2 n m ) + ( 2 n n ) = 2 2 n 2\displaystyle \sum_{m=0}^{n-1} \displaystyle \binom{2n}{m}+\displaystyle \binom{2n}{n}=2^{2n} :

∑ k = 0 18 cos ⁡ 2 n ( 1 0 ∘ k ) = 1 2 2 n ( 2 2 n − ( 2 n n ) + 19 ( 2 n n ) ) ∑ k = 0 18 cos ⁡ 2 n ( 1 0 ∘ k ) = 1 2 2 n ( 2 2 n + 18 ( 2 n n ) ) \displaystyle \sum_{k=0}^{18} \cos^{2n}(10^\circ k)=\dfrac{1}{2^{2n}} \left(2^{2n}-\displaystyle \binom{2n}{n}+19\displaystyle \binom{2n}{n}\right) \\ \displaystyle \sum_{k=0}^{18} \cos^{2n}(10^\circ k)=\dfrac{1}{2^{2n}} \left(2^{2n}+18\displaystyle \binom{2n}{n}\right)

Finally, find S ( n ) S(n) :

S ( n ) = 1 2 2 n ( 19 ( 2 n n ) − 2 ( 2 n − 1 n − 1 ) ) 1 2 2 n ( 2 2 n + 18 ( 2 n n ) ) S ( n ) = 19 ( 2 n n ) − 2 ( 2 n − 1 n − 1 ) 2 2 n + 18 ( 2 n n ) S(n)=\dfrac{\dfrac{1}{2^{2n}} \left(19\displaystyle \binom{2n}{n}-\displaystyle 2\binom{2n-1}{n-1}\right)}{\dfrac{1}{2^{2n}} \left(2^{2n}+18\displaystyle \binom{2n}{n}\right)} \\ S(n)=\dfrac{19\displaystyle \binom{2n}{n}-2\displaystyle \binom{2n-1}{n-1}}{2^{2n}+18\displaystyle \binom{2n}{n}}

And we are done. Link to my proof by hand

And for this case, n = 3 n=3 :

S ( 3 ) = 19 ( 6 3 ) − 2 ( 5 2 ) 2 6 + 18 ( 6 3 ) = 380 − 20 64 + 360 = 360 424 = 45 53 S(3)=\dfrac{19\displaystyle \binom{6}{3}-2\displaystyle \binom{5}{2}}{2^6+18\displaystyle \binom{6}{3}}=\dfrac{380-20}{64+360}=\dfrac{360}{424}=\boxed{\dfrac{45}{53}}

8 Helpful 0 Interesting 1 Brilliant 0 Confused

It's been over a day! Post proof! Oh, did you just made a FLT reference?

Pi Han Goh - 6 years ago

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No, I have a proof, and I also made a FLT reference :D But I have been very very busy, I hope I can post it soon.

Alan Enrique Ontiveros Salazar - 6 years ago

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WOW, such marvelous demonstration of your proposition!

It's more interesting to note that for all positive integer n n , the value of b n − a n b_n - a_n for the lowest form of S n S_n is always a perfect power of 2 2 . In this case, n = 3 , b 3 − a 3 = 2 3 n = 3, b_3 - a_3 = 2^3 . Can you explain why?

Pi Han Goh - 6 years ago

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@Pi Han Goh – Well, let b n k = 18 ( 2 n n ) + 2 2 n b_nk=18\binom{2n}{n}+2^{2n} and a n k = 19 ( 2 n n ) − 2 ( 2 n − 1 n − 1 ) a_nk=19\binom{2n}{n}-2\binom{2n-1}{n-1} for some positive integer k k . Now,

b n k − a n k = 18 ( 2 n n ) + 2 2 n − ( 19 ( 2 n n ) − 2 ( 2 n − 1 n − 1 ) ) = 2 2 n − ( 2 n n ) + 2 ( 2 n − 1 n − 1 ) b_nk-a_nk=18\binom{2n}{n}+2^{2n}-(19\binom{2n}{n}-2\binom{2n-1}{n-1})=2^{2n}-\binom{2n}{n}+2\binom{2n-1}{n-1}

k ( b n − a n ) = 2 2 n k(b_n-a_n)=2^{2n}

But I don't know how to determine k k , it is the gcd \text{gcd} of 19 ( 2 n n ) − 2 ( 2 n − 1 n − 1 ) 19\binom{2n}{n}-2\binom{2n-1}{n-1} and 18 ( 2 n n ) + 2 2 n 18\binom{2n}{n}+2^{2n} , do you know how to?

Also, did you mean n = 3 n=3 ?

Alan Enrique Ontiveros Salazar - 6 years ago

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@Alan Enrique Ontiveros Salazar – Oh fixed! No, that's where I'm stuck, I'm pretty surprised that it will always leave a power of 2, but I can't determine the exact power of 2. I got the same S n S_n as you (maybe not, some error in my working). Just with the numerator and denominator differ by exactly 1 1 , the resultant b n − a n b_n - a_n is a power of 2 2 . Isn't that weird?

Pi Han Goh - 6 years ago

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@Pi Han Goh – Yes, the first values for b n − a n b_n-a_n for n n from 1 1 to 15 15 are: 2 0 , 2 2 , 2 3 , 2 6 , 2 7 , 2 9 , 2 10 , 2 14 , 2 15 , 2 17 , 2 18 , 2 21 , 2 22 , 2 24 , 2 25 2^0,2^2,2^3,2^6,2^7,2^9,2^{10},2^{14},2^{15},2^{17},2^{18},2^{21},2^{22},2^{24},2^{25} . I don't see any pattern there.

Alan Enrique Ontiveros Salazar - 6 years ago

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@Alan Enrique Ontiveros Salazar – No, I mean, isn't it weird that they are always a power of 2 2 ?

Pi Han Goh - 6 years ago

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@Pi Han Goh – Yes, it is very strange. But there must be a reason.

Alan Enrique Ontiveros Salazar - 6 years ago

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@Alan Enrique Ontiveros Salazar – Ahah, come to think of it, you (me too) did the long way, from a glance, we can see that from the original (generalized) expression, the numerator is one less than the denominator, since you found that the numerator equals to 1 2 2 n ⋅ ( A ) \frac{1}{2^{2n}} \cdot (A) , then S ( n ) = 1 2 2 n ⋅ ( A ) 1 2 2 n ⋅ ( A ) + 1 S(n) =\frac{ \frac{1}{2^{2n}} \cdot (A)}{\frac{1}{2^{2n}} \cdot (A) + 1} , multiply top and bottom by 2 2 n 2^{2n} gives the possibly unsimplified values of b n , a n b_n, a_n , either way, when you find their differences, we are left with 2 2 n 2^{2n} , so it's ALWAYS a power of 2 2 . A better way to phrase my bonus question is to ask the reader to prove that b n − a n b_n - a_n has at most 2 n − 1 2n-1 proper factors.

Pi Han Goh - 6 years ago

thats a great generalization =) hopefully a proof could be post sooner ..thanks

Keil Cerbito - 6 years ago

Note first that sin ⁡ 6 ( x ) + cos ⁡ 6 ( x ) = \sin^{6}(x) + \cos^{6}(x) =

( sin ⁡ 2 ( x ) + cos ⁡ 2 ( x ) ) ( sin ⁡ 4 ( x ) − sin ⁡ 2 ( x ) cos ⁡ 2 ( x ) + cos ⁡ 4 ( x ) ) = (\sin^{2}(x) + \cos^{2}(x))(\sin^{4}(x) - \sin^{2}(x)\cos^{2}(x) + \cos^{4}(x)) =

( sin ⁡ 2 ( x ) + cos ⁡ 2 ( x ) ) 2 − 3 sin ⁡ 2 ( x ) cos ⁡ 2 ( x ) = 1 − 3 4 sin ⁡ 2 ( 2 x ) . (\sin^{2}(x) + \cos^{2}(x))^{2} - 3\sin^{2}(x)\cos^{2}(x) = 1 - \dfrac{3}{4}\sin^{2}(2x).

Now since sin ⁡ 6 ( x ) = sin ⁡ 6 ( 18 0 ∘ − x ) \sin^{6}(x) = \sin^{6}(180^{\circ} - x) and sin ⁡ 6 ( x ) = cos ⁡ 6 ( 9 0 ∘ − x ) \sin^{6}(x) = \cos^{6}(90^{\circ} - x) we can rewrite the numerator as

sin ⁡ 6 ( 9 0 ∘ ) + 2 ∗ ∑ k = 1 4 ( sin ⁡ 6 ( 10 k ∘ ) + cos ⁡ 6 ( 10 k ∘ ) ) = \sin^{6}(90^{\circ}) + 2*\displaystyle\sum_{k=1}^{4} (\sin^{6}(10k^{\circ}) + \cos^{6}(10k^{\circ})) =

9 − 3 2 ( sin ⁡ 2 ( 2 0 ∘ ) + sin ⁡ 2 ( 4 0 ∘ ) + sin ⁡ 2 ( 6 0 ∘ ) + sin ⁡ 2 ( 8 0 ∘ ) ) = 9 - \dfrac{3}{2}(\sin^{2}(20^{\circ}) + \sin^{2}(40^{\circ}) + \sin^{2}(60^{\circ}) + \sin^{2}(80^{\circ})) =

9 − 3 2 sin ⁡ 2 ( 6 0 ∘ ) − 3 4 ( ( 1 − cos ⁡ ( 4 0 ∘ ) ) + ( 1 − cos ⁡ ( 8 0 ∘ ) ) + ( 1 − cos ⁡ ( 16 0 ∘ ) ) = 9 - \dfrac{3}{2}\sin^{2}(60^{\circ}) - \dfrac{3}{4}((1 - \cos(40^{\circ})) + (1 - \cos(80^{\circ})) + (1 - \cos(160^{\circ})) =

9 − 9 8 − 9 4 + 3 4 ( cos ⁡ ( 4 0 ∘ ) + cos ⁡ ( 8 0 ∘ ) − cos ⁡ ( 2 0 ∘ ) ) = 9 − 27 8 = 45 8 9 - \dfrac{9}{8} - \dfrac{9}{4} + \dfrac{3}{4}(\cos(40^{\circ}) + \cos(80^{\circ}) - \cos(20^{\circ})) = 9 - \dfrac{27}{8} = \dfrac{45}{8}

since cos ⁡ ( 4 0 ∘ ) + cos ⁡ ( 8 0 ∘ ) = 2 cos ⁡ ( 8 0 ∘ + 4 0 ∘ 2 ) cos ⁡ ( 8 0 ∘ − 4 0 ∘ 2 ) = \cos(40^{\circ}) + \cos(80^{\circ}) = 2\cos\left(\dfrac{80^{\circ} + 40^{\circ}}{2}\right)\cos\left(\dfrac{80^{\circ} - 40^{\circ}}{2}\right) =

2 cos ⁡ ( 6 0 ∘ ) cos ⁡ ( 2 0 ∘ ) = cos ⁡ ( 2 0 ∘ ) . 2\cos(60^{\circ})\cos(20^{\circ}) = \cos(20^{\circ}).

Now the method for simplifying the denominator is identical except that the denominator will exceed the numerator by

cos ⁡ 6 ( 0 ∘ ) + cos ⁡ 6 ( 18 0 ∘ ) − sin ⁡ 6 ( 9 0 ∘ ) = 1. \cos^{6}(0^{\circ}) + \cos^{6}(180^{\circ}) - \sin^{6}(90^{\circ}) = 1.

So the given expression is equal to 45 8 45 8 + 1 = 45 53 , \dfrac{\dfrac{45}{8}}{\dfrac{45}{8} + 1} = \dfrac{45}{53}, and so b − a = 53 − 45 = 8 . b - a = 53 - 45 = \boxed{8}.

I'll leave the bonus question for someone else to play with. :)

8 Helpful 0 Interesting 1 Brilliant 0 Confused

I was playing around with this equation on the computer and saw that the answer is the same even if the progression is with increments of 20, 30, 40, 50, 70, 80, 100, 110, 130, etc. So 90, 60 and 120 gave different answers.

Oh, and I didn't change the number of terms, 19, in each progression. But anyway, it's an interesting observation. I think that it depends on the ratio of the 2*pi to the incremental factor. I also tried to do the same with increments of 5, 15, etc. and the answer was 1. Again, I kept the number of terms a constant

But I didn't play around with the powers like the bonus question asked.

vishnu c - 6 years ago

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