Question Number 27821 by bmind4860 last updated on 15/Jan/18
$${If}\:\alpha\:{and}\:\beta\:{satisfy}\:{sin}\alpha{cos}\beta=\:−\frac{\mathrm{1}}{\mathrm{2}}\:{then} \\ $$$${the}\:{greatest}\:{value}\:{of}\:\mathrm{2}{cos}\alpha{sin}\beta \\ $$
Answered by ajfour last updated on 15/Jan/18
$$\mathrm{sin}\:\alpha\mathrm{cos}\:\beta=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2sin}\:\alpha\mathrm{cos}\:\beta=−\mathrm{1} \\ $$$$\Rightarrow\mathrm{sin}\:\left(\alpha+\beta\right)+\mathrm{sin}\:\left(\alpha−\beta\right)=−\mathrm{1} \\ $$$${greatest}\:{value}\:{of}\:\mathrm{sin}\:\left(\alpha+\beta\right)=\mathrm{0} \\ $$$${when}\:\mathrm{sin}\:\left(\alpha−\beta\right)=−\mathrm{1} \\ $$$${and}\:\mathrm{2sin}\:\left(\alpha+\beta\right)=\mathrm{2sin}\:\alpha\mathrm{cos}\:\beta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{2cos}\:\alpha\mathrm{sin}\:\beta \\ $$$${so}\:\mathrm{2cos}\:\alpha\mathrm{sin}\:\beta=\mathrm{2sin}\:\left(\alpha+\beta\right)+\mathrm{1} \\ $$$${greatest}\:{value}\:{of}\: \\ $$$$\:\:\:\mathrm{2cos}\:\alpha\mathrm{sin}\:\beta=\:\mathrm{1}\:. \\ $$
Commented by bmind4860 last updated on 15/Jan/18
$${great}!\:{this}\:{is}\:{the}\:{best}\:{and}\:{far}\:{better}\:{solution}\:{from} \\ $$$${that}\:{lenthy}\:{solution}\:{of}\:{differentiation}. \\ $$$$ \\ $$
Commented by bmind4860 last updated on 15/Jan/18
$${really}\:{sometimes}\:{just}\:{analyzing}\:{the}\:{problem} \\ $$$${gives}\:{easiest}\:{solution} \\ $$$${thank}\:{you}!\:{again} \\ $$$$ \\ $$
Commented by ajfour last updated on 15/Jan/18
$${welcome}\:{sir}!\:{thank}\:{you}\:{too}, \\ $$$${and}\:{Prakash}\:{Sir},\:{who}\:{had} \\ $$$${found}\:{error}\:{in}\:{my}\:{initial} \\ $$$${wrong}\:{answer}. \\ $$
Commented by bmind4860 last updated on 15/Jan/18
$${sir}\:{can}\:{you}\:{please}\:{help}\:{me}\:{in}\:{the}\:{new} \\ $$$${questions}\:{posted}\:{by}\:{me} \\ $$