Trigonometry! 6

Geometry Level 4

Simplify ∑ r = 1 n − 1 cos ⁡ 2 ( r π n ) . \displaystyle \sum_{r=1}^{n-1} \cos^{2} \left(\frac {r \pi}{n}\right).

This problem is part of the set Trigonometry .

n 2 \frac {n}{2} n + 1 2 \frac {n+1}{2} n − 1 2 \frac {n-1}{2} n − 2 2 \frac {n-2}{2}

Omkar Kulkarni by Omkar Kulkarni , 22, India

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2 solutions

Mas Mus
Mar 13, 2015

Let S n − 1 = ∑ r = 1 n − 1 cos ⁡ 2 ( r π n ) S_{n-1}=\displaystyle \sum_{r=1}^{n-1} \cos^{2} \left(\frac {r \pi}{n}\right) , n = 2 , 3 , 4 , … n=2, 3, 4, \dots .

S 1 = cos ⁡ 2 π 2 = 0 S_{1}=\cos^{2} \frac { \pi}{2}=0

S 2 = cos ⁡ 2 π 3 + cos ⁡ 2 2 π 3 = 1 4 + 1 4 = 1 2 S_{2}=\cos^{2} \frac { \pi}{3}+\cos^{2} \frac {2 \pi}{3}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}

S 3 = cos ⁡ 2 π 4 + cos ⁡ 2 2 π 4 + cos ⁡ 2 3 π 4 = 1 2 + 0 + 1 2 = 1 S_{3}=\cos^{2} \frac { \pi}{4}+\cos^{2} \frac {2 \pi}{4}+\cos^{2} \frac {3 \pi}{4}=\frac{1}{2}+0+\frac{1}{2}=1

S 4 = cos ⁡ 2 π 5 + cos ⁡ 2 2 π 5 + cos ⁡ 2 3 π 5 + cos ⁡ 2 4 π 5 = 2 cos ⁡ 2 π 5 + 2 cos ⁡ 2 2 π 5 = ( 1 + cos ⁡ 2 π 5 ) + ( 1 + cos ⁡ 4 π 5 ) = 2 + 2 cos ⁡ 3 π 5 cos ⁡ π 5 = 2 + cos ⁡ 3 π 5 sin ⁡ 2 π 5 sin ⁡ π 5 = 2 + sin ⁡ 5 π 5 − sin ⁡ π 5 2 sin ⁡ π 5 = 2 − 1 2 = 3 2 \begin{aligned}S_{4} &=\cos^{2} \frac { \pi}{5}+\cos^{2} \frac {2 \pi}{5}+\cos^{2} \frac {3 \pi}{5}+\cos^{2} \frac {4 \pi}{5}\\ &=2\cos^{2} \frac { \pi}{5}+2\cos^{2} \frac {2 \pi}{5}\\ &=(1+\cos\frac{2\pi}{5})+(1+\cos\frac{4\pi}{5})=2+2\cos\frac{3\pi}{5}\cos\frac{\pi}{5}\\ &=2+\frac{\cos\frac{3\pi}{5} \sin\frac{2\pi}{5}}{\sin\frac{\pi}{5}}=2+\frac{\sin\frac{5\pi}{5}-\sin\frac{\pi}{5}}{2\sin\frac{\pi}{5}}\\ &=2-\frac{1}{2}=\frac{3}{2}\end{aligned}

S 5 = cos ⁡ 2 π 6 + cos ⁡ 2 2 π 6 + cos ⁡ 2 3 π 6 + cos ⁡ 2 4 π 6 + cos ⁡ 2 5 π 6 = 3 4 + 0 + 1 4 + 0 + 1 4 + 3 4 = 2 \begin{aligned}S_{5} &=\cos^{2} \frac { \pi}{6}+\cos^{2} \frac {2 \pi}{6}+\cos^{2} \frac {3 \pi}{6}+\cos^{2} \frac {4 \pi}{6}+\cos^{2} \frac {5 \pi}{6}\\ &=\frac{3}{4}+0+\frac{1}{4}+0+\frac{1}{4}+\frac{3}{4}=2\end{aligned}

Now, we have S 1 , S 2 , S 3 , S 4 , S 5 , … , S n − 1 = 0 , 1 2 , 1 , 3 2 , 2 , … , S n − 1 S_1, S_2, S_3, S_4, S_5, \dots, S_{n-1}=0, \frac{1}{2}, 1, \frac{3}{2}, 2, \dots, S_{n-1} , an arithmetic progression which has initial term 0 and common difference 1 2 \frac{1}{2} .

So, S n − 1 = 0 + ( ( n − 1 ) − 1 ) × 1 2 = n − 2 2 S_{n-1}=0+((n-1)-1) \times \frac{1}{2}=\boxed{\frac{n-2}{2}} .

5 Helpful 0 Interesting 0 Brilliant 0 Confused

∑ r = 1 n − 1 c o s 2 ( r π n ) \sum_{r=1}^{n-1} cos^2(\frac{r\pi}{n})

= ∑ r = 1 n − 1 1 + c o s 2 π r n 2 = \sum_{r=1}^{n-1} \frac{1+cos\frac{2\pi r }{n}}{2}

= n − 1 2 + 1 2 s i n ( ( n − 1 ) 2 π n ) c o s ( 2 π ( 1 + n − 1 ) 2 n ) s i n ( 2 π n ) = \frac{n-1}{2} + \frac{1}{2} \frac{sin(\frac{(n-1)2\pi}{n}) cos(\frac{2\pi(1+n-1)}{2n})}{sin(\frac{2\pi}{n})}

= n − 1 − 1 2 = \frac{n-1-1}{2}

= n − 2 2 = \frac{n-2}{2}

Anurag Pandey - 4 years, 5 months ago

An extremely nice solution. Your approach was easy to understand. Also, I love the way you calculated S4. Keep up the good work :D

Pranav Saxena - 4 years, 11 months ago
Yogesh Shivran
Jan 4, 2015

I did it by hit and trial. Just put n=2, we match it with option. But i cann't solve it by generalisation.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

∑ r = 1 n − 1 1 + cos ⁡ 2 r π 2 \displaystyle \sum_{r=1}^{n-1} \dfrac{1 + \cos2r\pi}{2}

∑ r = 1 n − 1 1 2 + 1 2 ∑ r = 1 n − 1 cos ⁡ 2 r π n \displaystyle \sum_{r=1}^{n-1} \dfrac{1}{2} + \dfrac{1}{2} \displaystyle \sum_{r=1}^{n-1} \cos\dfrac{2r\pi}{n}

= n − 1 2 + 1 4 sin ⁡ r π ∑ r = 1 n − 1 2 cos ⁡ 2 r π n sin ⁡ r π n = \dfrac{n - 1}{2} + \dfrac{1}{4 \sin r\pi} \displaystyle \sum_{r=1}^{n-1} 2\cos\dfrac{2r\pi}{n}\sin\dfrac{r\pi}{n}

= n − 1 2 + 1 4 sin ⁡ r π ∑ r = 1 n − 1 sin ⁡ 3 r π n − s i n r π n = \dfrac{n - 1}{2} + \dfrac{1}{4 \sin r\pi} \displaystyle \sum_{r=1}^{n-1} \sin \dfrac{3r\pi}{n} - sin\dfrac{r\pi}{n}

= n − 1 2 − 1 2 = \dfrac{n - 1}{2} - \dfrac{1}{2}

U Z - 6 years, 4 months ago

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Can you please explain the third last step ? Thanks.

Abhijeet Verma - 5 years, 11 months ago

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