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Question Number 185107 by emmanuelson123 last updated on 17/Jan/23

Answered by SEKRET last updated on 17/Jan/23

$$2{\mathbf{sin}}^2\left(\frac{\mathit{\pi }}{2048}\right)+\mathbf{cos}\left(\frac{2\mathit{\pi }}{2048}\right)=1$$$$2(1-{\mathbf{cos}}^2\left(\frac{\mathit{\pi }}{2048}\right))+2{\mathbf{cos}}^2\left(\frac{\mathit{\pi }}{2048}\right)-1=1$$$$2-2{\mathbf{cos}}^2\left(\frac{\mathit{\pi }}{2048}\right)+2{\mathbf{cos}}^2\left(\frac{\mathit{\pi }}{2048}\right)-1=1$$$$1=1$$

Answered by som(math1967) last updated on 17/Jan/23

$$let\frac{\pi }{2048}=\theta \therefore \frac{\pi }{1024}=2\theta$$$$\therefore 2{sin}^2\theta +cos2\theta$$$$=1-cos2\theta +cos2\theta$$$$=1$$

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