T for Trigo

Geometry Level 4

sin ⁡ π 7 sin ⁡ 2 π 7 sin ⁡ 3 π 7 \sin \dfrac{\pi}7\sin \dfrac{2\pi}7\sin \dfrac{3\pi}7

If the value of the expression above is in the form of A B \dfrac{\sqrt A}B , where A A and B B are coprime positive integers, find A + B A+B .


The answer is 15.

Mohit Gupta by Mohit Gupta , 20, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Hassan Abdulla
Mar 4, 2020

I = s i n ( π 7 ) ⋅ s i n ( 2 π 7 ) ⋅ s i n ( 3 π 7 ) I = s i n ( 6 π 7 ) ⋅ s i n ( 5 π 7 ) ⋅ s i n ( 4 π 7 ) sin ⁡ ( π − x ) = s i n ( x ) I 2 = s i n ( π 7 ) ⋅ s i n ( 2 π 7 ) ⋅ s i n ( 3 π 7 ) ⋅ s i n ( 4 π 7 ) ⋅ s i n ( 5 π 7 ) ⋅ s i n ( 6 π 7 ) I 2 = 7 2 6 = 7 64 ∏ k = 1 n − 1 sin ⁡ ( k π n ) = n 2 n − 1 I = 7 8 \begin{aligned} I &= sin(\frac{\pi}{7}) \cdot sin(\frac{2 \pi}{7}) \cdot sin(\frac{3 \pi}{7}) \\ I &= sin(\frac{6 \pi}{7}) \cdot sin(\frac{5 \pi}{7}) \cdot sin(\frac{4 \pi}{7}) & \color{#D61F06} \sin(\pi - x)=sin(x) \\ I^2 &= sin(\frac{\pi}{7}) \cdot sin(\frac{2 \pi}{7}) \cdot sin(\frac{3 \pi}{7}) \cdot sin(\frac{4 \pi}{7}) \cdot sin(\frac{5 \pi}{7}) \cdot sin(\frac{6\pi}{7}) \\ I^2 &= \frac{7}{2^6}=\frac{7}{64} & \color{#D61F06} \prod_{k=1}^{n-1} \sin\left ( \frac{k \pi}{n} \right ) = \frac{n}{2^{n-1}} \\ I &= \frac{\sqrt{7}}{8} \end{aligned}

prove of ∏ k = 1 n − 1 sin ⁡ ( k π n ) = n 2 n − 1 \prod_{k=1}^{n-1} \sin\left ( \frac{k \pi}{n} \right ) = \frac{n}{2^{n-1}}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...