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Question-176848




Question Number 176848 by Ar Brandon last updated on 27/Sep/22
Answered by mr W last updated on 27/Sep/22
β=(π/2)−α  cos ((π/2)−2β)=sin 2β=sin (π−2α)  =sin 2α=2 sin α cos α  =2x(√(1−x^2 ))
$$\beta=\frac{\pi}{\mathrm{2}}−\alpha \\ $$$$\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}−\mathrm{2}\beta\right)=\mathrm{sin}\:\mathrm{2}\beta=\mathrm{sin}\:\left(\pi−\mathrm{2}\alpha\right) \\ $$$$=\mathrm{sin}\:\mathrm{2}\alpha=\mathrm{2}\:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\alpha \\ $$$$=\mathrm{2}{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$

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