Question Number 27958 by bmind4860 last updated on 17/Jan/18
$${For}\:\alpha\in{R},\:{cos}\alpha{cosx}+{siny}\geqslant{sinx},\:\forall{x}\in{R}, \\ $$$${then}\:{find}\:{the}\:{sum}\:{of}\:{the}\:{possible}\:{values} \\ $$$${of}\:{sin}\alpha+{siny}. \\ $$
Answered by ajfour last updated on 17/Jan/18
![sin y ≥ sin x−cos αcos x ⇒ ≥ (√(1+cos^2 α)) [(1/(√(1+cos^2 α)))sin x−((cos α)/(√(1+cos^2 α)))cos x] sin y≥ (√(1+cos^2 α)) sin [x−tan^(−1) (cos α)] since x can be any value, and sin y ≤ 1 ⇒ cos α=0 ⇒ sin α =±1 then this implies sin y ≥ sin x and again for this to be always true sin y =1 hence sin y+sin α=0 or 2 .](Q27962.png)
$$\:\:\:\:\mathrm{sin}\:{y}\:\geqslant\:\mathrm{sin}\:{x}−\mathrm{cos}\:\alpha\mathrm{cos}\:{x} \\ $$$$\Rightarrow\:\:\:\:\geqslant\:\sqrt{\mathrm{1}+\mathrm{cos}\:^{\mathrm{2}} \alpha}\:\left[\frac{\mathrm{1}}{\sqrt{\mathrm{1}+\mathrm{cos}\:^{\mathrm{2}} \alpha}}\mathrm{sin}\:{x}−\frac{\mathrm{cos}\:\alpha}{\sqrt{\mathrm{1}+\mathrm{cos}\:^{\mathrm{2}} \alpha}}\mathrm{cos}\:{x}\right] \\ $$$$\mathrm{sin}\:{y}\geqslant\:\sqrt{\mathrm{1}+\mathrm{cos}\:^{\mathrm{2}} \alpha}\:\mathrm{sin}\:\left[{x}−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{cos}\:\alpha\right)\right] \\ $$$${since}\:{x}\:{can}\:{be}\:{any}\:{value}, \\ $$$${and}\:\mathrm{sin}\:{y}\:\leqslant\:\mathrm{1}\:\Rightarrow\:\:\mathrm{cos}\:\alpha=\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{sin}\:\alpha\:=\pm\mathrm{1} \\ $$$${then}\:{this}\:{implies} \\ $$$$\:\:\:\:\:\mathrm{sin}\:\boldsymbol{{y}}\:\geqslant\:\mathrm{sin}\:\boldsymbol{{x}} \\ $$$${and}\:{again}\:{for}\:{this}\:{to}\:{be}\:{always} \\ $$$${true}\:\mathrm{sin}\:{y}\:=\mathrm{1} \\ $$$${hence}\:\mathrm{sin}\:{y}+\mathrm{sin}\:\alpha=\mathrm{0}\:{or}\:\mathrm{2}\:. \\ $$
Commented by bmind4860 last updated on 17/Jan/18
$${thank}\:{you}!\:{would}\:{you}\:{suggest}\:{what}\:{do} \\ $$$${you}\:{think}\:{about}\:{the}\:{following}\:{question}, \\ $$$${whether}\:{it}\:{is}\:{right}\:{or}\:{wrong} \\ $$$${find}\:{general}\:{solution},\:{sinx}+{sin}\mathrm{3}{x}+{sin}\sqrt{{x}}=\mathrm{0}. \\ $$
Commented by ajfour last updated on 17/Jan/18
$${cant}\:{say},\:{i}\:{couldn}'{t}\:{find}\:{a}\:{way}\:. \\ $$
Commented by bmind4860 last updated on 17/Jan/18
$${me}\:{too}\:{that}'{s}\:{why}\:{i}\:{think}\:{it}\:{is}\:{wrong} \\ $$$${and}\:{thank}\:{you}\:{for}\:{suggestion} \\ $$