PROVE THAT SIN^2(18) AND COS^2(36) ARE THE ROOTS OF THE QUADRATIC EQUATION 16X^2-12X+1=0

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$$\begin{gathered} \mathrm{We} \mathrm{have}: \\ 16{\mathrm{x}}^2-12\mathrm{x}+1=0 \\ \mathrm{Using} \mathrm{Shridharacharya}\text{'}\mathrm{s} \mathrm{formula} \mathrm{we} \mathrm{get} \\ \mathrm{x}=\frac{-\left(-12\right)\pm \sqrt{{\left(-12\right)}^2-4\times 16\times 1}}{2\times 16} \\ =\frac{12\pm \sqrt{144-64}}{32} \\ =\frac{12\pm \sqrt{80}}{32} \\ =\frac{12\pm 4\sqrt{5}}{32} \\ =2\left(\frac{6\pm 2\sqrt{5}}{32}\right) \\ =\frac{6\pm 2\sqrt{5}}{16} \\ =\frac{5+1\pm 2\sqrt{5}}{16} \\ \mathrm{x}=\frac{\left(\sqrt{5}\right)+1^2\pm 2\sqrt{5}}{4^2} \\ \Rightarrow \mathrm{x}=\frac{{\left(\sqrt{5}\pm 1\right)}^2}{4^2} \\ \Rightarrow \mathrm{x}={\left(\frac{\sqrt{5}+1}{4}\right)}^2 \mathrm{or} \mathrm{x}={\left(\frac{\sqrt{5}-1}{4}\right)}^2....\left(\mathrm{i}\right) \mathrm{are} \mathrm{roots} \mathrm{of} \mathrm{above} \mathrm{equation}. \\ \mathrm{We}\mathrm{will}\mathrm{first}\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\sin 18° \\ \mathrm{Suppose}\mathrm{that}\mathrm{A}=18°\mathrm{whcih}\mathrm{gives} \\ \cos 36°=\cos 2\mathrm{A} \\ =1-2{\sin }^2\mathrm{A} \\ =1-2{\sin }^{2}18°......\left(1\right) \\ \mathrm{It}\mathrm{can}\mathrm{be}\mathrm{noted}\mathrm{that}5\mathrm{A}=90\mathrm{so}\mathrm{that} \\ 2\mathrm{A}+3\mathrm{A}=90° \\ 2\mathrm{A}=90°-3\mathrm{A} \\ \mathrm{This}\mathrm{implies}\mathrm{that}, \\ \sin 2\mathrm{A}=\sin \left(90°-3\mathrm{A}\right) \\ \sin 2\mathrm{A}=\cos 3\mathrm{A}\left[\mathrm{since}\sin \left(90°-\mathrm{\theta }\right)=\mathrm{cos\theta }\right] \\ 2\mathrm{sinA}\mathrm{cosA}=4{\cos }^3\mathrm{A}-3\mathrm{cosA} \\ \mathrm{Take}\mathrm{all}\mathrm{the}\mathrm{terms}\mathrm{on}\mathrm{LHS}\mathrm{to}\mathrm{get}, \\ 2\mathrm{sinA}\mathrm{cosA}-4{\cos }^3\mathrm{A}+3\mathrm{cosA}=0 \\ \mathrm{cosA}\left(2\mathrm{sinA}-4{\cos }^2\mathrm{A}+3\right)=0 \\ \mathrm{Here}\cos \mathrm{A}=\cos 18\ne 0\mathrm{which}\mathrm{further}\mathrm{implies}\mathrm{that} \\ 2\mathrm{sinA}-4{\cos }^2\mathrm{A}+3=0 \\ 2\mathrm{sinA}-4\left(1-{\sin }^2\mathrm{A}\right)+3=0 \\ 4{\sin }^2\mathrm{A}+2\mathrm{sinA}-1=0 \\ \mathrm{Use}\mathrm{the}\mathrm{quadratic}\mathrm{formula}\mathrm{to}\mathrm{get}, \\ \mathrm{sinA}=\frac{-2\pm \sqrt{2^2-4\left(4\right)\left(-1\right)}}{2\left(4\right)} \\ =\frac{-2\pm \sqrt{20}}{8} \\ =\frac{-2\pm 2\sqrt{5}}{8} \\ =\frac{-1\pm \sqrt{5}}{4} \\ \mathrm{This}\mathrm{gives}\mathrm{sinA}=\sin 18°=\frac{\sqrt{5}-1}{4}\left(\mathrm{since}\mathrm{A}\mathrm{lies}\mathrm{in}\mathrm{Quadrant}\mathrm{I}\right) \\ \mathrm{Using}\left(1\right),\mathrm{the}\mathrm{value}\mathrm{of}\cos 36°\mathrm{can}\mathrm{be}\mathrm{found}\mathrm{as}, \\ \cos 36°=1-2{\sin }^2\mathrm{A} \\ =1-2{\left(\frac{-1+\sqrt{5}}{4}\right)}^2 \\ =1-2\left(\frac{1+5-2\sqrt{5}}{16}\right) \\ =1-\left(\frac{6-2\sqrt{5}}{8}\right) \\ =\frac{8-\left(6-2\sqrt{5}\right)}{8} \\ =\frac{2+2\sqrt{5}}{8} \\ \cos 36°=\frac{1+\sqrt{5}}{4} \\ \mathrm{Hence} \sin 18°=\frac{\sqrt{5}-1}{4} \mathrm{and} \cos 36°=\frac{\sqrt{5}+1}{4} \\ \mathrm{Put} \mathrm{this} \mathrm{in} \left(\mathrm{i}\right) \\ \mathbf{x}\mathbf{=}{\mathbf{cos}}^{\mathbf{2}}\mathbf{36}\mathbf{°}\mathbf{ }\mathbf{or}\mathbf{ }{\mathbf{sin}}^{\mathbf{2}}\mathbf{18}\mathbf{°}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{Hence}\mathbf{ }\mathbf{proved} \end{gathered}$$

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