It can't be all of them!

Geometry Level 4

M = ∑ k = 1 n − 1 cos ⁡ ( 2 π n k ) \large M= \large \sum_{k=1}^{n-1} \cos\left( \frac{2\pi}n k \right)

Above shows a trigonometric identity. What is the value of M M ?

There is insufficient information 1 1 − 1 -1 2 2 0 0

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Atharva Bagul by Atharva Bagul , 20, India

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2 solutions

Let e 2 ⋅ π ⋅ i n e^{\frac {2 \cdot \pi \cdot i}{n}} be one primitive root of x n − 1 x^n - 1 = 0 = ( x - 1)( x n − 1 + x n − 2 + . . . + x + 1 x^{n-1} + x^{n-2} + ... + x + 1 ). Then e 2 ⋅ ( n − 1 ) ⋅ π ⋅ i n + e 2 ⋅ ( n − 2 ) ⋅ π ⋅ i n + . . . + e 2 ⋅ π ⋅ i n + 1 = 0 e^{\frac {2 \cdot {(n-1)} \cdot \pi \cdot i}{n}} + e^{\frac {2 \cdot {(n-2)} \cdot \pi \cdot i}{n}} + ... + e^{\frac {2 \cdot \pi \cdot i}{n}} + 1 = 0 . Now only have to review the real part of this last identity and it gives the solution. (n > 1)

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Can you please explain the last sentence of your solution?

Atharva Bagul - 5 years, 7 months ago

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Yes, e i x = c o s ( x ) + i s i n ( x ) e^{ix} = cos (x) + i sin (x) and for Moivre's Theorem, ( c o s ( x ) + i s i n ( x ) ) n (cos (x) + i sin(x))^{n} = ( e i x ) n (e^{ix})^{n} = e i n x e^{inx} = cos (nx) + i sin(nx) . Therefore, e 2 ⋅ ( n − 1 ) ⋅ π ⋅ i n + e 2 ⋅ ( n − 2 ) ⋅ π ⋅ i n + . . . + e 2 ⋅ π ⋅ i n + 1 = 0 e^{\frac {2 \cdot {(n-1)} \cdot \pi \cdot i}{n}} + e^{\frac {2 \cdot {(n-2)} \cdot \pi \cdot i}{n}} + ... + e^{\frac {2 \cdot \pi \cdot i}{n}} + 1 = 0 ⇒ \Rightarrow cos 2 ⋅ ( n − 1 ) ⋅ π n \frac {2 \cdot {(n-1)} \cdot \pi}{n} + i sin 2 ⋅ ( n − 1 ) ⋅ π n \frac {2 \cdot {(n-1)} \cdot \pi}{n} + cos 2 ⋅ ( n − 2 ) ⋅ π n \frac {2 \cdot {(n-2)} \cdot \pi}{n} + i sin 2 ⋅ ( n − 2 ) ⋅ π n \frac {2 \cdot {(n-2)} \cdot \pi}{n} + ... + cos 2 ⋅ π n \frac {2 \cdot \pi}{n} + i sin 2 ⋅ π n \frac {2 \cdot \pi}{n} + 1 = 0. Now, take the real part of this identity and it will give you the result. You also can take the imaginary part and it gives another beautiful result. e 2 ⋅ ( n − 1 ) ⋅ π ⋅ i n + e 2 ⋅ ( n − 2 ) ⋅ π ⋅ i n + . . . + e 2 ⋅ π ⋅ i n + 1 = 0 e^{\frac {2 \cdot {(n-1)} \cdot \pi \cdot i}{n}} + e^{\frac {2 \cdot {(n-2)} \cdot \pi \cdot i}{n}} + ... + e^{\frac {2 \cdot \pi \cdot i}{n}} + 1 = 0 because ( e 2 ⋅ i ⋅ π n − 1 e^{\frac {2 \cdot i \cdot \pi}{n}} - 1 ) ≠ \neq 0 (n > 1)

Guillermo Templado - 5 years, 7 months ago
Rishi Sharma
Nov 12, 2015

J E E s t y l e : − s i n c e t h i s i s a n i d e n t i t y i n n l e t n = 2 M = c o s ( 2 π 2 ( 2 − 1 ) ) M = c o s ( π ) = − 1 P S : − i t i s n o t t h e r i g h t w a y y o u s h o u l d u s e g u i l l e r m o ′ s w a y . JEE\quad style:-\\ since\quad this\quad is\quad an\quad identity\quad in\quad n\quad let\quad n=2\\ M=cos(\frac { 2\pi }{ 2 } (2-1))\\ M=cos(\pi )=-1\\ PS:-\quad it\quad is\quad not\quad the\quad right\quad way\quad you\quad should\quad use\\ guillermo's\quad way.

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