Question Number 58025 by Kunal12588 last updated on 16/Apr/19
$${Trace}\:{the}\:{changes}\:{in}\:{the}\:{sign}\:{and}\:{magnitude} \\ $$$${of}\:\:\frac{\mathrm{sin}\:\mathrm{3}\theta}{\mathrm{cos}\:\mathrm{2}\theta}\:{as}\:{the}\:{angle}\:{increases}\:{from}\:\mathrm{0}\:{to}\:\frac{\pi}{\mathrm{2}}. \\ $$$${also}\:{find}\:{its}\:{minimum}\:{and}\:{maximum}\:{values}. \\ $$
Commented by Kunal12588 last updated on 16/Apr/19
Answered by tanmay last updated on 16/Apr/19
$$\frac{\pi}{\mathrm{2}}\geqslant\theta\geqslant\mathrm{0} \\ $$$$\pi\geqslant\mathrm{2}\theta\geqslant\mathrm{0} \\ $$$$\frac{\mathrm{3}\pi}{\mathrm{2}}\geqslant\mathrm{3}\theta\geqslant\mathrm{0} \\ $$$$\:\: \\ $$$$\left.\:{A}\right)\:\:\:\:{sin}\mathrm{3}\theta=+{ve}\:\:\:\frac{\pi}{\mathrm{6}}\geqslant\theta\geqslant\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{cos}\mathrm{2}\theta=+{ve}\: \\ $$$$\:\:\:\:\:\:\:\:{so}\:\:\:\frac{{sin}\mathrm{3}\theta}{{cos}\mathrm{2}\theta}=+{ve}\:\:{when}\:\frac{\pi}{\mathrm{6}}\geqslant\theta\geqslant\mathrm{0} \\ $$$$\left.{B}\right)\:\:\:{sin}\mathrm{3}\theta=+{ve}\:\:\:\:\:\:\left[{when}\:\frac{\pi}{\mathrm{4}}>\:\theta\geqslant\frac{\pi}{\mathrm{6}}\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{cos}\mathrm{2}\theta=+{ve} \\ $$$$\:\:\:\:\:\:\:\:\:\frac{{sin}\mathrm{3}\theta}{{cos}\mathrm{2}\theta}=++{ve} \\ $$$$\left.{c}\right){sin}\mathrm{3}\theta=\:\:\:\:\:+{ve}\:\:\:\:{when}\:\:\frac{\pi}{\mathrm{3}}\geqslant\theta>\frac{\pi}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:{cos}\mathrm{2}\theta=−{ve} \\ $$$$\:\:\:\:\:\frac{{sin}\mathrm{3}\theta}{{cos}\mathrm{2}\theta}=−{ve} \\ $$$$\left.{d}\right){sin}\mathrm{3}\theta=\:\:−{ve}\:\:\:\:\:\frac{\pi}{\mathrm{2}}\geqslant\theta\geqslant\frac{\pi}{\mathrm{3}} \\ $$$$\:\:\:\:\:{cos}\mathrm{2}\theta=−{ve} \\ $$$${so}\:\frac{{sin}\mathrm{3}\theta}{{cos}\mathrm{2}\theta}=+{ve} \\ $$$${pls}\:{check}… \\ $$$$ \\ $$$${f}\left(\theta\right)=\frac{{sin}\mathrm{3}\theta}{{cos}\mathrm{2}\theta} \\ $$$$\frac{{df}}{{d}\theta}=\frac{{cos}\mathrm{2}\theta×\mathrm{3}{cos}\mathrm{3}\theta+\mathrm{2}{sin}\mathrm{3}\theta×{sin}\mathrm{2}\theta}{{cos}^{\mathrm{2}} \mathrm{2}\theta} \\ $$$$\frac{{df}}{{d}\theta}=\frac{{cos}\mathrm{2}\theta×{cos}\mathrm{3}\theta+\mathrm{2}{cos}\left(\mathrm{3}\theta−\mathrm{2}\theta\right)}{{cos}^{\mathrm{2}} \mathrm{2}\theta} \\ $$$$\frac{{df}}{{d}\theta}=\frac{{cos}\mathrm{3}\theta×{cos}\mathrm{2}\theta+\mathrm{2}{cos}\theta}{{cos}^{\mathrm{2}} \mathrm{2}\theta} \\ $$$${for}\:{max}/{min}\:\frac{{df}}{{d}\theta}=\mathrm{0} \\ $$$${cos}\mathrm{3}\theta×{cos}\mathrm{2}\theta+\mathrm{2}{cos}\theta \\ $$$$\left(\mathrm{4}{x}^{\mathrm{3}} −\mathrm{3}{x}\right)\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}\right)+\mathrm{2}{x} \\ $$$$\mathrm{8}{x}^{\mathrm{5}} −\mathrm{4}{x}^{\mathrm{3}} −\mathrm{6}{x}^{\mathrm{3}} +\mathrm{3}{x}+\mathrm{2}{x} \\ $$$$\mathrm{8}{x}^{\mathrm{5}} −\mathrm{10}{x}^{\mathrm{3}} +\mathrm{5}{x} \\ $$$${x}\left(\mathrm{8}{x}^{\mathrm{4}} −\mathrm{10}{x}^{\mathrm{2}} +\mathrm{5}\right) \\ $$$${x}\left\{\mathrm{2}\left(\mathrm{4}{x}^{\mathrm{4}} −\mathrm{5}{x}^{\mathrm{2}} \right)+\mathrm{5}\right\} \\ $$$${x}\left[\mathrm{2}\left\{\left(\mathrm{2}{x}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}×\mathrm{2}{x}^{\mathrm{2}} ×\frac{\mathrm{5}}{\mathrm{4}}+\frac{\mathrm{25}}{\mathrm{16}}−\frac{\mathrm{25}}{\mathrm{16}}\right\}+\mathrm{5}\right] \\ $$$${x}\left[\mathrm{2}\left(\mathrm{2}{x}^{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{4}}\right)^{\mathrm{2}} −\frac{\mathrm{25}}{\mathrm{8}}+\mathrm{5}\right] \\ $$$${x}\left[\mathrm{2}\left(\mathrm{2}{x}^{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{4}}\right)^{\mathrm{2}} +\frac{\mathrm{15}}{\mathrm{8}}\right] \\ $$$${so}\:\left[\mathrm{2}\left(\mathrm{2}{x}^{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{4}}\right)^{\mathrm{2}} +\frac{\mathrm{15}}{\mathrm{8}}\right]\neq\mathrm{0} \\ $$$${hence}\:\mathrm{8}{x}^{\mathrm{5}} −\mathrm{10}{x}^{\mathrm{3}} +\mathrm{5}{x}=\mathrm{0}\:{when}\:{x}=\mathrm{0} \\ $$$${cos}\theta=\mathrm{0}={cos}\frac{\pi}{\mathrm{2}}\:\:\left[\theta=\frac{\pi}{\mathrm{2}}\right] \\ $$$${f}\left(\frac{\pi}{\mathrm{2}}\right)=\frac{{sin}\left(\frac{\mathrm{3}\pi}{\mathrm{2}}\right)}{{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{2}}\right)}=\frac{−\mathrm{1}}{−\mathrm{1}}=\mathrm{1}\left({maximum}\:{value}\right) \\ $$$${f}\left(\mathrm{0}\right)=\frac{{sin}\left(\mathrm{3}×\mathrm{0}\right)}{{cos}\left(\mathrm{2}×\mathrm{0}\right)}=\mathrm{0} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by Kunal12588 last updated on 17/Apr/19
$${thank}\:{you}\:{sir} \\ $$