Question Number 48526 by Pk1167156@gmail.com last updated on 25/Nov/18
$$\mathrm{The}\:\mathrm{maximum}\:\mathrm{and}\:\mathrm{minimum}\:\mathrm{values} \\ $$$$\mathrm{of}\:\:\:{a}\:\mathrm{cos}\:\mathrm{2}\theta+\:{b}\:\mathrm{sin}\:\mathrm{2}\theta\:\:\mathrm{are} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 25/Nov/18
$${a}={rsin}\alpha\:\:\:{b}={rcos}\alpha \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={r}^{\mathrm{2}} \:\:{so}\:{r}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\: \\ $$$${S}={acos}\mathrm{2}\theta+{bsin}\mathrm{2}\theta \\ $$$$\:\:={rsin}\alpha{cos}\mathrm{2}\theta+{rcos}\alpha{sin}\mathrm{2}\theta \\ $$$$={rsin}\left(\alpha+\mathrm{2}\theta\right) \\ $$$${max}\:{value}\:{of}\:{sin}\left(\alpha+\mathrm{2}\theta\right)=\mathrm{1} \\ $$$${min}\:{value}\:{of}\:{sin}\left(\alpha+\mathrm{2}\theta\right)=−\mathrm{1} \\ $$$${so}\:{max}\:{value}\:{of}\:{acos}\mathrm{2}\theta+{bsin}\mathrm{2}\theta\:\:{is}\: \\ $$$$={r}×\mathrm{1} \\ $$$$=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:\: \\ $$$${min}\:{value}\:{of}\:{acos}\mathrm{2}\theta+{bsin}\mathrm{2}\theta \\ $$$$={r}×−\mathrm{1} \\ $$$$=−\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\: \\ $$$$ \\ $$
Answered by ajfour last updated on 25/Nov/18
$$\:{a}\mathrm{cos}\:\mathrm{2}\theta+{b}\mathrm{sin}\:\mathrm{2}\theta\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\mathrm{sin}\:\left(\mathrm{tan}^{−\mathrm{1}} \frac{{a}}{{b}}+\mathrm{2}\theta\right) \\ $$$${so}\:{maximum}\:{value}\:=\:\sqrt{\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} } \\ $$$${and}\:{minimum}\:{value}\:=\:−\sqrt{\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} }\:. \\ $$