Question Number 209768 by OmoloyeMichael last updated on 20/Jul/24
$$\boldsymbol{{Express}}\:\boldsymbol{{tan}}\left(\mathrm{3}\right)\:\boldsymbol{{in}}\:\boldsymbol{{surd}}\:\boldsymbol{{form}} \\ $$
Answered by Frix last updated on 22/Jul/24
$$\mathrm{tan}\:\left(\alpha−\beta\right)\:=\frac{\mathrm{tan}\:\alpha\:−\mathrm{tan}\:\beta}{\mathrm{1}+\mathrm{tan}\:\alpha\:\mathrm{tan}\:\beta} \\ $$$$\mathrm{Use}\:\alpha=\mathrm{18}°\:\mathrm{and}\:\beta=\mathrm{15}°.\:\mathrm{The}\:\mathrm{only}\:\mathrm{problem}\:\mathrm{is} \\ $$$$\mathrm{to}\:\mathrm{factorize}\:\mathrm{it}… \\ $$$$\mathrm{I}\:\mathrm{get} \\ $$$$\mathrm{tan}\:\mathrm{3}°\:=\frac{\left(−\mathrm{7}+\mathrm{2}\sqrt{\mathrm{15}}−\mathrm{3}\sqrt{\mathrm{5}}+\mathrm{4}\sqrt{\mathrm{3}}\right)\sqrt{\mathrm{5}−\mathrm{2}\sqrt{\mathrm{5}}}+\left(\mathrm{4}−\sqrt{\mathrm{15}}+\mathrm{2}\sqrt{\mathrm{5}}−\mathrm{3}\sqrt{\mathrm{3}}\right)}{\mathrm{2}} \\ $$
Commented by OmoloyeMichael last updated on 22/Jul/24
$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$
Commented by Frix last updated on 22/Jul/24
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