Question-162589

Question Number 162589 by mkam last updated on 30/Dec/21

Answered by Ar Brandon last updated on 30/Dec/21

T = \cos ϑ \cos 2 ϑ \cos 4 ϑ \dots \cos (2^{n - 1} ϑ)

2 \sin ϑ T = \sin 2 ϑ \cos 2 ϑ \cos 4 ϑ \dots \cos (2^{n - 1} ϑ)

4 \sin ϑ T = \sin 4 ϑ \dots \cos (2^{n - 1} ϑ)

2^{n} \sin ϑ T = \sin (2^{n} ϑ)

T = \frac{1}{2^{n}} \cdot \frac{\sin (2^{n} ϑ)}{\sin ϑ}, ϑ = \frac{π}{2^{n} + 1}

T = \frac{1}{2^{n}} \cdot \frac{\sin (\frac{2^{n}}{2^{n} + 1} π)}{\sin (\frac{1}{2^{n} + 1} π)} = \frac{1}{2^{n}} \cdot \frac{\sin (π - \frac{π}{2^{n} + 1})}{\sin (\frac{π}{2^{n} + 1})}

T = \frac{1}{2^{n}} \cdot \frac{\sin (\frac{π}{2^{n} + 1})}{\sin (\frac{π}{2^{n} + 1})} = \frac{1}{2^{n}}, \sin (π - t) = \sin (t)

Commented by peter frank last updated on 31/Dec/21

thank you