Product of This Year?? (Part 3)

Algebra Level 5

Given 2 2014 ∏ k = 0 2014 sin ⁡ ( x + k π 2015 ) = sin ⁡ ( p x q ) . 2^{2014}\prod_{k=0}^{2014}\sin\left(x+\frac{k \pi}{2015}\right) = \sin\left(\frac{px}{q}\right). Find p + q p+q .


The answer is 2016.

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Tunk-Fey Ariawan by Tunk-Fey Ariawan , 35, Indonesia

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1 solution

Chew-Seong Cheong
Oct 14, 2014

This can be solved by induction. Let us consider the first few cases of k = 0 , 1 , 2 , 3 k = 0,1,2,3 of:

P ( n ) = 2 n ∏ k = 0 n sin ⁡ ( x + k π n + 1 ) P(n) = 2^ n \prod _{k=0} ^n \sin { \left( x + \dfrac {k\pi} {n+1} \right) }

P ( 0 ) = sin ⁡ x P(0) = \sin {x}

P ( 1 ) = 2 sin ⁡ x sin ⁡ ( x + 1 2 π ) = 2 sin ⁡ x [ ( 0 ) sin ⁡ x + ( 1 ) cos ⁡ x ] P(1) = 2 \sin {x} \sin {(x + \frac {1}{2}\pi) } = 2 \sin {x} [(0) \sin {x }+ (1) \cos {x} ]

= 2 sin ⁡ x cos ⁡ x = sin ⁡ 2 x \quad \quad = 2 \sin {x} \cos{x} = \sin {2x}

P ( 2 ) = 4 sin ⁡ x sin ⁡ ( x + 1 3 π ) sin ⁡ ( x + 2 3 π ) P(2) = 4 \sin {x} \sin {(x + \frac {1}{3}\pi) } \sin {(x + \frac {2}{3}\pi) }

= 4 sin ⁡ x ( 1 2 sin ⁡ x + 3 2 cos ⁡ x ) ( − 1 2 sin ⁡ x + 3 2 cos ⁡ x ) \quad \quad = 4 \sin {x} ( \frac {1}{2} \sin {x }+ \frac {\sqrt{3}}{2} \cos {x} ) ( -\frac {1}{2} \sin {x }+ \frac {\sqrt{3}}{2} \cos {x} )

= 4 sin ⁡ x ( 3 4 cos ⁡ 2 x − 1 4 sin ⁡ 2 x ) = sin ⁡ x ( 3 cos ⁡ 2 x − sin ⁡ 2 x ) \quad \quad = 4 \sin {x} ( \frac {3}{4} \cos^2 {x } - \frac {1}{4} \sin^2 {x} ) = \sin {x} (3 \cos^2 {x } - \sin^2 {x} )

= sin ⁡ x ( 2 cos ⁡ 2 x + cos ⁡ 2 x − sin ⁡ 2 x ) = sin ⁡ 2 x cos ⁡ x + cos ⁡ 2 x sin ⁡ x = sin ⁡ 3 x \quad \quad = \sin {x} (2\cos^2 {x } + \cos^2{x}- \sin^2 {x} ) = \sin {2x} \cos {x} + \cos {2x} \sin {x} = \sin {3x}

P ( 3 ) = 8 sin ⁡ x sin ⁡ ( x + 1 4 π ) sin ⁡ ( x + 1 2 π ) sin ⁡ ( x + 3 4 π ) P(3) = 8 \sin {x} \sin {(x + \frac {1}{4}\pi) } \sin {(x + \frac {1}{2}\pi) } \sin {(x + \frac {3}{4}\pi) }

= 8 sin ⁡ x ( 1 2 sin ⁡ x + 1 2 cos ⁡ x ) ( cos ⁡ x ) ( − 1 2 sin ⁡ x + 1 2 cos ⁡ x ) \quad \quad = 8 \sin {x} ( \frac {1}{\sqrt{2}} \sin {x }+ \frac {1}{\sqrt{2}} \cos {x} ) (\cos {x}) ( -\frac {1}{\sqrt{2}} \sin {x }+ \frac {1}{\sqrt{2}} \cos {x} )

= 8 sin ⁡ x cos ⁡ x ( 1 2 cos ⁡ 2 x − 1 2 sin ⁡ 2 x ) = 2 sin ⁡ 2 x cos ⁡ 2 x = sin ⁡ 4 x \quad \quad = 8 \sin {x} \cos{x} ( \frac {\sqrt{1}}{2} \cos^2 {x} -\frac {1}{2} \sin^2 {x }) = 2 \sin{2x} \cos {2x} = \sin {4x}

From the above, we can induce that p = n + 1 p = n+1 and q = 1 q = 1 , and when n = 2014 ⇒ p + q = 2016 n = 2014 \Rightarrow p + q = \boxed{2016} .

0 Helpful 0 Interesting 0 Brilliant 0 Confused

You have shown the first three steps, but can you prove the induction step in general? From the examples you give it is not quite clear how to generalize them.

(Namely, if the equation is valid for P ( 0 ) , … , P ( n − 1 ) P(0), \dots, P(n-1) , that it is also valid for P ( n ) P(n) .)

Arjen Vreugdenhil - 5 years, 5 months ago

@Chew-Seong Cheong sir while finding p(2) u could have written sin(x+120 ) AS sin(x-60 ) and then applied the identity sinx.sin(x-60)sin(x+60) =1/4 . sin3x only to simplify things a bit . other than this ur solution is great :)

avi solanki - 4 years, 4 months ago

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