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Question Number 161668 by mnjuly1970 last updated on 21/Dec/21

Answered by mr W last updated on 21/Dec/21

$$\frac{{\sin }^{2}x}{1-{\sin }^{2}x}=1+{\sin }^{2}x$$$${\sin }^{4}x+{\sin }^{2}x-1=0$$$${\sin }^{2}x=\frac{-1+\sqrt{5}}{2}=\varphi$$$$4{\sin }^2\frac{x}{2}(1-{\sin }^2\frac{x}{2})=\varphi$$$${\sin }^4\frac{x}{2}-{\sin }^2\frac{x}{2}+\frac{\varphi }{4}=0$$$${\sin }^2\frac{x}{2}=\frac{1\pm \sqrt{1-\varphi }}{2}$$$${\left({\sin }^2\frac{x}{2}\right)}_{max}=\frac{1+\sqrt{1-\varphi }}{2}$$$$=\frac{1+\sqrt{\frac{3-\sqrt{5}}{2}}}{2}=\frac{1}{2}+\frac{\sqrt{6-2\sqrt{5}}}{4}=\frac{1}{2}+\frac{\sqrt{5}-1}{4}=\frac{\sqrt{5}+1}{2}$$$$$$$$\frac{\sqrt{6-2\sqrt{5}}}{4}=\frac{\sqrt{6-\sqrt{20}}}{4}$$$$\Rightarrow answer\left(3\right)iscorrect:$$

Commented by mnjuly1970 last updated on 21/Dec/21

$$geatefulsirW...veryexcellent$$

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