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# 4sin-2-18-2sin18-2-5-

Question Number 142414 by mathdanisur last updated on 31/May/21
$$\mathrm{4}{sin}^{\mathrm{2}} \mathrm{18}°\:+\:\mathrm{2}{sin}\mathrm{18}°\:+\:\mathrm{2},\mathrm{5}\:=\:? \\$$
Commented by som(math1967) last updated on 31/May/21
$$\mathrm{3}.\mathrm{5}\: \\$$
Commented by mathdanisur last updated on 31/May/21
$${how}\:{sir},\:{but}\:{answer}:\:\mathrm{3},\mathrm{75} \\$$
Answered by som(math1967) last updated on 31/May/21
$$\mathrm{4}×\frac{\left(\sqrt{\mathrm{5}}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{16}}\:+\mathrm{2}×\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}\:+\mathrm{2}.\mathrm{5} \\$$$$\frac{\mathrm{5}+\mathrm{1}−\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{4}}\:+\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\:+\mathrm{2}.\mathrm{5} \\$$$$\frac{\mathrm{6}−\mathrm{2}\sqrt{\mathrm{5}}+\mathrm{2}\sqrt{\mathrm{5}}−\mathrm{2}}{\mathrm{4}}\:+\mathrm{2}.\mathrm{5} \\$$$$\mathrm{1}+\mathrm{2}.\mathrm{5}=\mathrm{3}.\mathrm{5} \\$$
Commented by mathdanisur last updated on 31/May/21
$${cool}\:{Sir}\:{thanks} \\$$