Question Number 173946 by mnjuly1970 last updated on 21/Jul/22
$$ \\ $$$$\:\:\:{If}\:\mathrm{3}{sin}\left({x}\right)+\mathrm{4}{cos}\left({y}\right)−\mathrm{3}{sin}\left({y}\right)=\mathrm{8} \\ $$$$\:\:\:{then}\:..\:\:\:\:{sin}\left({x}\right)\:+\:{sin}\left({y}\right)=? \\ $$$$ \\ $$$$ \\ $$
Answered by mr W last updated on 21/Jul/22
$$\mathrm{3}\:\mathrm{sin}\:{x}+\mathrm{4}\:\mathrm{cos}\:{y}−\mathrm{3}\:\mathrm{sin}\:{y}=\mathrm{8} \\ $$$$\mathrm{3}\:\mathrm{sin}\:{x}+\mathrm{5}\left(\frac{\mathrm{4}}{\mathrm{5}}\:\mathrm{cos}\:{y}−\frac{\mathrm{3}}{\mathrm{5}}\:\mathrm{sin}\:{y}\right)=\mathrm{8} \\ $$$$\mathrm{3}\:\mathrm{sin}\:{x}+\mathrm{5}\left(\mathrm{cos}\:\alpha\:\mathrm{cos}\:{y}−\mathrm{sin}\:\alpha\:\mathrm{sin}\:{y}\right)=\mathrm{8} \\ $$$$\mathrm{3}\:\mathrm{sin}\:{x}+\mathrm{5}\:\mathrm{cos}\:\left({y}+\alpha\right)=\mathrm{8} \\ $$$$\Rightarrow\mathrm{sin}\:{x}=\mathrm{1}\:\wedge\:\mathrm{cos}\:\left({y}+\alpha\right)=\mathrm{1} \\ $$$$\Rightarrow{y}+\alpha=\mathrm{2}{k}\pi \\ $$$$\Rightarrow{y}=\mathrm{2}{k}\pi−\alpha \\ $$$$\Rightarrow\mathrm{sin}\:{y}=−\mathrm{sin}\:\alpha=−\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\Rightarrow\mathrm{sin}\:{x}+\mathrm{sin}\:{y}=\mathrm{1}−\frac{\mathrm{3}}{\mathrm{5}}=\frac{\mathrm{2}}{\mathrm{5}}\:\checkmark \\ $$
Commented by mathlove last updated on 22/Jul/22
$${way} \\ $$$$\frac{\mathrm{4}}{\mathrm{5}}{cosy}−\frac{\mathrm{3}}{\mathrm{5}}{siny}={sin}\alpha{cosh}−{cos}\alpha{siny} \\ $$
Commented by infinityaction last updated on 22/Jul/22
$$\mathrm{3},\mathrm{4},\mathrm{5}\:{is}\:{a}\:{right}\:{angle}\:{triangle} \\ $$
Commented by Tawa11 last updated on 22/Jul/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Commented by mr W last updated on 22/Jul/22
$${for}\:{any}\:{values}\:{a}\:{and}\:{b},\:{we}\:{can}\:{find}\:\alpha \\ $$$${such}\:{that} \\ $$$$\mathrm{sin}\:\alpha=\frac{{a}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }},\:\mathrm{cos}\:\alpha=\frac{{b}}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}. \\ $$$${in}\:{our}\:{case}:\:{a}=\mathrm{3},\:{b}=\mathrm{4}. \\ $$
Commented by mnjuly1970 last updated on 22/Jul/22
$${very}\:{nice}\:{sir}\:{W} \\ $$