Question Number 109428 by mathdave last updated on 23/Aug/20
Commented by som(math1967) last updated on 23/Aug/20
$$\:\mathrm{is}\:\mathrm{it}\:\mathrm{2}.\mathrm{5}\:? \\ $$
Commented by mathdave last updated on 23/Aug/20
$${yes} \\ $$
Answered by Dwaipayan Shikari last updated on 23/Aug/20
$$\mathrm{4}\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} +\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}+\mathrm{2}=\frac{\mathrm{6}−\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{4}}+\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}+\mathrm{2}=\mathrm{1}+\mathrm{2}=\mathrm{3} \\ $$
Answered by som(math1967) last updated on 23/Aug/20
$$\mathrm{4}×\frac{\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{16}}\:+\mathrm{2}×\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}+\mathrm{2}.\mathrm{5}\:\bigstar \\ $$$$=\frac{\mathrm{5}+\mathrm{1}−\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{4}}\:+\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\:+\mathrm{2}.\mathrm{5} \\ $$$$=\frac{\mathrm{6}−\mathrm{2}\sqrt{\mathrm{5}}+\mathrm{2}\sqrt{\mathrm{5}}−\mathrm{2}}{\mathrm{4}}\:+\mathrm{2}.\mathrm{5} \\ $$$$=\frac{\mathrm{4}}{\mathrm{4}}\:+\mathrm{2}.\mathrm{5}=\mathrm{3}.\mathrm{5} \\ $$$$\mathrm{ans}\left(\mathrm{b}\right) \\ $$$$\bigstar\mathrm{sin18}=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}} \\ $$