Trigonometry's pattern

Geometry Level 4

sin ⁡ ( n x ) − cos ⁡ ( n x ) = 0 \large \sin(nx)-\cos(nx) = 0

Which of the following can represent the sum of roots of the equation above for 0 < x < 2 π 0 < x < 2\pi and n n is a positive integer?

n 2 π n^2\pi 2 n π 2^n\pi ( n + 1 2 ) π \left(n+\frac{1}{2}\right)\pi 2 n π 2n\pi ( n − 1 2 ) π \left(n-\frac{1}{2}\right)\pi ( 2 n + 1 2 ) π \left(2n+\frac{1}{2}\right)\pi ( 2 n − 1 2 ) π \left(2n-\frac{1}{2}\right)\pi n π n\pi

Tommy Li by Tommy Li , 22, Hong Kong

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1 solution

The general solution of the equation is x = { ( 2 k + 1 4 n ) π ( 2 k + 5 4 n ) π x = \begin{cases} \left(\dfrac {2k + \frac 14}n \right)\pi \\ \left(\dfrac {2k + \frac 54}n \right)\pi \end{cases} , where k = 0 , 1 , 2 , . . . n − 1 k = 0, 1, 2, ... n-1 .

Therefore, the sum of roots is given by:

S n = ∑ k = 0 n − 1 ( 4 k + 3 2 n ) π = ( 4 n ( n − 1 ) 2 + 3 2 n ) π n = ( 2 n − 1 2 ) π \begin{aligned} S_n & = \sum_{k=0}^{n-1} \left( \frac {4k+\frac 32}n\right) \pi \\ & = \left( \frac {4n(n-1)}2 +\frac 32n \right) \frac \pi n \\ & = \boxed{\left( 2n-\dfrac 12 \right) \pi} \end{aligned}

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