Question Number 86041 by M±th+et£s last updated on 26/Mar/20
$$\left.\mathrm{1}\right){if}\: \\ $$$${sin}\left(\theta−{x}\right)={k}\:{sin}\left(\theta+\alpha\right) \\ $$$${find}\:{tan}\left(\theta\right)\:{and}\:{k} \\ $$$$ \\ $$$${then}\:{find}\:\theta\:{in}\left[\mathrm{0},\mathrm{2}\pi\right]\:\:{when}\:{k}=\frac{\mathrm{1}}{\mathrm{2}}\:{and}\:\alpha=\pi \\ $$$$ \\ $$$$\left.\mathrm{2}\right){if}\:{x}={sin}\left({t}\right)\:\:{and}\:\:{y}={cos}\left(\mathrm{2}{t}\right) \\ $$$${show}\:{that} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+\mathrm{4}=\mathrm{0} \\ $$
Answered by Rio Michael last updated on 26/Mar/20
$$\left.\mathrm{2}\right)\:\:{x}\:=\:\mathrm{sin}\:{t}\:\:\mathrm{and}\:\:{y}=\:\mathrm{cos}\:\left(\mathrm{2}{t}\right) \\ $$$$\:\Rightarrow\:\frac{{dx}}{{dt}}\:=\:\mathrm{cos}\:{t}\:\:\mathrm{and}\:\frac{{dy}}{{dt}}\:=\:−\mathrm{2sin}\:\left(\mathrm{2}{t}\right) \\ $$$$\:\:\Rightarrow\:\:\frac{{dy}}{{dx}}\:=\:\frac{{dy}}{{dt}}\:.\:\frac{{dt}}{{dx}}\:=\:\frac{−\mathrm{2sin}\left(\mathrm{2}{t}\right)}{\mathrm{cos}\:{t}}\:=\:\frac{−\mathrm{2}\left(\mathrm{2sin}\:{t}\:\mathrm{cos}\:{t}\right)}{\mathrm{cos}\:{t}}\:=\:−\mathrm{4sin}\:{t} \\ $$$$\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:=\:\frac{{d}}{{dx}}\left(\frac{{dy}}{{dx}}\right)\:\frac{{dt}}{{dx}}\:=\:−\mathrm{4cos}\:{t}×\:\frac{\mathrm{1}}{\mathrm{cos}\:{t}}\:=\:−\mathrm{4} \\ $$$$\Rightarrow\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} \:\:}\:=\:−\mathrm{4}\:\:\mathrm{hence}\:\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:+\:\mathrm{4}\:=\:\mathrm{0} \\ $$
Commented by M±th+et£s last updated on 26/Mar/20
$${god}\:{bless}\:{you} \\ $$
Answered by mind is power last updated on 27/Mar/20
$${sin}\left(\theta−\alpha\right)={ksin}\left(\theta+\alpha\right)..\:{Quation}\:\mathrm{1}? \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by M±th+et£s last updated on 27/Mar/20
$${yes}\:{sir}\: \\ $$