Question Number 10746 by okhema last updated on 24/Feb/17
$$\left.{i}\right){express}\:{the}\:{function}\:{f}\left(\theta\right)={sin}\theta\:+\:{cos}\theta\:{in}\:{the}\:{form}\:{rsin}\left(\theta+\alpha\right),\:{r}>\mathrm{0}\:{and}\:\mathrm{0}\leqslant\theta\leqslant\leqslant\frac{\pi}{\mathrm{2}} \\ $$$$\left.{ii}\right){hence}\:{find}\:{the}\:{maximum}\:{value}\:{of}\:{f}\:{and} \\ $$$${the}\:{smallest}\:{non}−{negative}\:{value}\:{of}\:\theta\:{at}\:{which}\:{it}\:{occurs}. \\ $$
Answered by mrW1 last updated on 24/Feb/17
$$\left.{i}\right) \\ $$$$\left.{f}\left(\theta\right)={sin}\theta\:+\:{cos}\theta=\sqrt{\mathrm{2}}\left({sin}\theta\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:+\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}{cos}\theta\right)=\sqrt{\mathrm{2}\left(\right.}{sin}\theta\mathrm{cos}\:\frac{\pi}{\mathrm{4}}+\:{cos}\theta\mathrm{sin}\:\frac{\pi}{\mathrm{4}}\right) \\ $$$$=\sqrt{\mathrm{2}}\mathrm{sin}\:\left(\theta+\frac{\pi}{\mathrm{4}}\right) \\ $$$$\left.{ii}\right) \\ $$$${max}.{f}\left(\theta\right)=\sqrt{\mathrm{2}}\:{with}\:\theta=\frac{\pi}{\mathrm{4}} \\ $$$${min}.{f}\left(\theta\right)=\mathrm{1}\:{with}\:\theta=\mathrm{0}\:{or}\:\frac{\pi}{\mathrm{2}} \\ $$