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Question Number 20434 by pp75718276@gmail.com last updated on 26/Aug/17

$$\mathrm{If}\cos x=\tan y,\cos y=\tan z,\cos z=\tan x,$$$$\mathrm{then}\mathrm{the}\mathrm{value}\mathrm{of}\sin x\mathrm{is}$$

Answered by ajfour last updated on 27/Aug/17

$$1+\tan {}^{2}z=\frac{1}{\cos {}^{2}z}$$$$\Rightarrow 1+\cos {}^{2}y=\frac{1}{\tan {}^{2}x}$$$$\Rightarrow 1+\frac{1}{1+\tan {}^{2}y}=\frac{1}{\tan {}^{2}x}$$$$\Rightarrow 1+\frac{1}{1+\cos {}^{2}x}=\frac{1-\sin {}^{2}x}{\sin {}^{2}x}$$$$let\sin {}^{2}x=t$$$$then1+\frac{1}{2-t}=\frac{1-t}{t}$$$$\Rightarrow \frac{3-t}{2-t}=\frac{1-t}{t}$$$$\Rightarrow 3t-t^2=2+t^2-3t$$$$2t^2-6t+2=0$$$$2{(t-\frac{3}{2})}^2=\frac{5}{2}$$$$t=\frac{3}{2}-\frac{\sqrt{5}}{2}$$$$\sin x=\pm \sqrt{\frac{3-\sqrt{5}}{2}}.$$

Commented by myintkhaing last updated on 27/Aug/17

$$\mathrm{Check}\frac{1}{{\tan }^2\mathrm{x}}$$

Answered by Tinkutara last updated on 27/Aug/17

$${\cos }^{2}x={\tan }^{2}y={\sec }^{2}y-1={\cot }^{2}z-1$$$$1+{\cos }^{2}x={\cot }^{2}z=\frac{{\cos }^{2}z}{1-{\cos }^{2}z}=\frac{{\tan }^{2}x}{1-{\tan }^{2}x}$$$$2-{\sin }^{2}x=\frac{{\sin }^{2}x}{{\cos }^{2}x-{\sin }^{2}x}=\frac{{\sin }^{2}x}{1-{2\sin }^{2}x}$$$$(2-{\sin }^{2}x)(1-{2\sin }^{2}x)={\sin }^{2}x$$$${\sin }^{4}x-{3\sin }^{2}x+1=0$$$${\sin }^{2}x=\frac{3\pm \sqrt{5}}{2}$$$$But{\sin }^{2}x⩽1as\sin x\in [-1,1]$$$$So{\sin }^{2}x\in [0,1]$$$${\sin }^{2}x=\frac{1}{2}(3-\sqrt{5})=\frac{1}{2}\left[\frac{1}{2}(1+5-2\sqrt{5})\right]$$$${\sin }^{2}x={\left(\frac{\sqrt{5}-1}{2}\right)}^2\Rightarrow \mid \sin x\mid =\frac{\sqrt{5}-1}{2}$$$$Similarly,\mid \sin y\mid =\mid \sin z\mid =\frac{\sqrt{5}-1}{2}$$

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