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Question Number 109337 by ZiYangLee last updated on 22/Aug/20

$$x=\cos \theta ,\mathrm{where}\frac{3\pi }{2}<\theta <2\pi ,\mathrm{and}\mathrm{that}2\cos \theta -\sin \theta =2,$$ $$\mathrm{show}\mathrm{that}\sqrt{1-x^2}=2(1-x).$$ $$\mathrm{Hence}\mathrm{or}\mathrm{otherwise},\mathrm{find}x\mathrm{and}\mathrm{deduce}\mathrm{that}\tan 2\theta =\frac{24}{7}$$

Answered by Aziztisffola last updated on 22/Aug/20

$$\sqrt{1-x^2}=\sqrt{1-{\cos }^2\theta }=-\sin \theta$$ $$=2-2\cos \theta =2(1-\cos \theta )=2(1-x)$$ $$\sqrt{1-x^2}=2(1-x)\iff 1-x^2=4(x^2-2x+1)$$ $$5x^2-8x+3=0$$ $$△=4\Rightarrow x_1=1\wedge x_2=\frac{3}{5}$$ $$\theta \ne 2\pi \Rightarrow x\ne 1\Rightarrow x=\frac{3}{5}$$ $$\tan 2\theta =\frac{2\tan \left(\theta \right)}{1-{\tan }^2\left(\theta \right)}$$ $$\tan \theta =\frac{-\frac{4}{5}}{\frac{3}{5}}=-\frac{4}{3}$$ $$\tan 2\theta =\frac{2\times \frac{-4}{3}}{1-{\left(\frac{-4}{3}\right)}^2}=\frac{\frac{-8}{3}}{1-\frac{16}{9}}$$ $$=\frac{8\times 3}{7}=\frac{24}{7}$$

Commented byZiYangLee last updated on 23/Aug/20

$$\mathrm{NICE}!$$

Answered by Don08q last updated on 22/Aug/20

$$x=\cos \theta$$ $$x^2={\cos }^2\theta$$ $$1-x^2={\sin }^2\theta$$ $$\pm \sqrt{1-x^2}=\sin \theta$$ $$Butfortheinterval,\frac{3\pi }{2}<\theta <2\pi \mathrm{sine}$$ $$hasanegativevalue.$$ $$\Rightarrow -\sqrt{1-x^2}=\sin \theta$$ $$so\sqrt{1-x^2}=-\sin \theta ............\left(1\right)$$ $$$$ $$But,2\cos \theta -\sin \theta =2$$ $$\Rightarrow 2-2x=-\sin \theta$$ $$2(1-x)=-\sin \theta ...........\left(2\right)$$ $$from\left(1\right)and\left(2\right),$$ $$\sqrt{1-x^2}=2(1-x)qed$$ $$$$ $$Itfollowsthat$$ $$1-x^2=4(1-2x+x^2)$$ $$5x^2-8x+3=0$$ $$x=1orx=\frac{3}{5}$$ $$x\mathrm{cannot}\mathrm{be}1,\mathrm{since}\theta <2\pi$$ $$\therefore x=\frac{3}{5}$$ $$\tan \theta =\frac{\sin \theta }{\cos \theta }$$ $$\tan \theta =\frac{-2(1-x)}{x}$$ $$\tan \theta =\frac{2(x-1)}{x}$$ $$\tan \theta =\frac{2(\frac{3}{5}-1)}{\frac{3}{5}}$$ $$\tan \theta =-\frac{4}{3}$$ $$But\tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }$$ $$\tan 2\theta =\frac{2(-\frac{4}{3})}{1-{(-\frac{4}{3})}^2}$$ $$\therefore \tan 2\theta =\frac{24}{7}qed$$ $$$$ $$$$ $$$$ $$$$ $$$$ $$$$ $$$$ $$$$ $$$$

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