Question

find sin 9 and cos 9

Answer 1

$$\begin{gathered} \left(1\right) \\ Let\theta ={18}^0 \\ Then, \\ 2\theta =2\times {18}^0={36}^0={90}^0-{54}^0={90}^0-3\theta \\ Takin\sin onbothside,wehave, \\ \sin 2\theta =\sin \left({90}^0-3\theta \right) \\ \Rightarrow 2\sin \theta \cos \theta =\cos 3\theta \\ \Rightarrow 2\sin \theta \cos \theta =4{\cos }^3\theta -3\cos \theta \\ \Rightarrow 2\sin \theta \cos \theta =\cos \theta \left(4{\cos }^2\theta -3\right) \\ \Rightarrow 2\sin \theta =\left(4{\cos }^2\theta -3\right)\left(\cos \theta \ne 0\right) \\ \Rightarrow 2\sin \theta =\left(4\left(1-{\sin }^2\theta \right)-3\right) \\ \Rightarrow 4{\sin }^2\theta +2\sin \theta -1=0 \\ \Rightarrow \theta =\frac{-2\pm \sqrt{4+16}}{2\times 4}=\frac{-1\pm \sqrt{5}}{4} \\ Since\sin \theta >0soneglecting\sin \theta =\frac{-1-\sqrt{5}}{4} \\ Thus,\mathit{s}\mathit{i}\mathit{n}{\mathbf{18}}^{\mathbf{0}}\mathbf{=}\frac{\mathbf{-}\mathbf{1}\mathbf{+}\sqrt{\mathbf{5}}}{\mathbf{4}} \end{gathered}$$
$$\begin{gathered} \mathrm{We}\mathrm{know}\mathrm{that},{\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }=1 \\ \mathrm{Put}\mathrm{\theta }={18}^0,\mathrm{we}\mathrm{have} \\ {\sin }^2{18}^0+{\cos }^2{18}^0=1 \\ \Rightarrow {\cos }^2{18}^0=1-{\sin }^2{18}^0 \\ \Rightarrow {\cos }^2{18}^0=1-{\left(\frac{\sqrt{5}-1}{4}\right)}^2=1-\left(\frac{6-2\sqrt{5}}{16}\right)=\frac{10+2\sqrt{5}}{16} \\ \Rightarrow \cos {18}^0=\sqrt{\frac{10+2\sqrt{5}}{16}}=\frac{\sqrt{10+2\sqrt{5}}}{4} \\ \Rightarrow \cos {18}^0=\frac{\sqrt{10+2\sqrt{5}}}{4} \\ \mathrm{Now}\mathrm{we}\mathrm{know}\mathrm{that}\cos 2\mathrm{x}=1-2{\sin }^2\mathrm{x}=2{\cos }^2\mathrm{x}-1 \\ \mathrm{So}\mathrm{sinx}=\sqrt{\frac{1-\cos 2\mathrm{x}}{2}} \\ \mathrm{Putting}\mathrm{x}=9\mathrm{degree} \\ \sin 9=\sqrt{\frac{1-\cos 18}{2}}=\sqrt{\frac{1-\frac{\sqrt{10+2\sqrt{5}}}{4}}{2}}=\sqrt{\frac{4-\sqrt{10+2\sqrt{5}}}{8}} \\ \mathrm{and} \\ \mathrm{cosx}=\sqrt{\frac{1+\cos 2\mathrm{x}}{2}}=\sqrt{\frac{1+\frac{\sqrt{10+2\sqrt{5}}}{4}}{2}}=\sqrt{\frac{4+\sqrt{10+2\sqrt{5}}}{8}} \end{gathered}$$