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Question Number 143450 by bramlexs22 last updated on 14/Jun/21

$$whenx+y=\frac{2\pi }{3};x⩾0;y⩾0$$$$themaximumandtheminimum$$$$of\sin x+\sin yis\mathrm{\_}\mathrm{\_}\mathrm{\_}$$

Answered by EDWIN88 last updated on 16/Jun/21

$$\mathrm{y}=\frac{2\pi }{3}-\mathrm{x}\Rightarrow \sin \mathrm{y}=\sin (\frac{2\pi }{3}-\mathrm{x})$$$$\sin \mathrm{y}=\frac{1}{2}\sqrt{3}\cos \mathrm{x}+\frac{1}{2}\sin \mathrm{x}$$$$\mathrm{let}\sin \mathrm{x}+\sin \mathrm{y}=\mathrm{f}\left(\mathrm{x}\right)$$$$\mathrm{f}\left(\mathrm{x}\right)=\sin \mathrm{x}+\frac{1}{2}\sqrt{3}\cos \mathrm{x}+\frac{1}{2}\sin \mathrm{x}$$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{3}{2}\sin \mathrm{x}+\frac{1}{2}\sqrt{3}\cos \mathrm{x}$$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{k}\cos (\mathrm{x}-\mathrm{\varnothing })\to \{\begin{array}{l}\mathrm{k}=\sqrt{\frac{9}{4}+\frac{3}{4}}=\sqrt{3}\\ \mathrm{\varnothing }={\tan }^{-1}\left(\sqrt{3}\right)\to \mathrm{\varnothing }=\frac{\pi }{3}\end{array}$$$$\left(1\right)\mathrm{f}\left(\mathrm{x}\right)=\sqrt{3}\cos (\mathrm{x}-\frac{\pi }{3})\to \max =\sqrt{3}$$$$\mathrm{when}\mathrm{x}=\mathrm{y}=\frac{\pi }{3}$$$$\left(2\right)\mathrm{f}{\left(\mathrm{x}\right)}_{\min }=\frac{\sqrt{3}}{2}\mathrm{when}\mathrm{x}=0\wedge \mathrm{y}=\frac{2\pi }{3}$$$$\mathrm{or}\mathrm{x}=\frac{2\pi }{3}\wedge \mathrm{y}=0$$

Answered by mnjuly1970 last updated on 14/Jun/21

$$y:=\frac{2\pi }{3}-x$$$$M:=sin\left(x\right)+\frac{\sqrt{3}}{2}cosx-\frac{1}{2}sin\left(x\right)$$$$:=\frac{1}{2}sin\left(x\right)+\frac{\sqrt{3}}{2}cos\left(x\right)$$$$max\left(M\right):=\sqrt{\frac{1}{4}+\frac{3}{4}}:=1$$$$min\left(M\right):=-M=-1$$

Commented by bramlexs22 last updated on 15/Jun/21

$$y=120°-x$$$$\sin y=\sin (120°-x)=\frac{1}{2}\sqrt{3}\cos x-(-\frac{1}{2})\sin x$$$$=\frac{1}{2}\sqrt{3}\cos x+\frac{1}{2}\sin x$$

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