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Question Number 216074 by CrispyXYZ last updated on 27/Jan/25

$$\mathrm{find}\mathrm{the}\mathrm{maximum}\mathrm{of}y=\mid \sin x\mid +\mid \sin 2x\mid .$$

Answered by efronzo1 last updated on 27/Jan/25

$$\mathrm{y}=\mid \sin \mathrm{x}\mid +2\mid \sin \mathrm{x}\mid \sqrt{1-\sin {}^2\mathrm{x}}$$$$\mathrm{y}=\mid \sin \mathrm{x}\mid (1+2\sqrt{1-\sin {}^2\mathrm{x}})$$$$\mathrm{Let}\mid \sin \mathrm{x}\mid =\sqrt{\sin {}^2\mathrm{x}}=\mathrm{t}$$$$\mathrm{y}=\mathrm{t}(1+2\sqrt{1-{\mathrm{t}}^2})$$$${\mathrm{y}}^{\prime }=1+2\sqrt{1-{\mathrm{t}}^2}+\mathrm{t}\left(\frac{-2\mathrm{t}}{\sqrt{1-{\mathrm{t}}^2}}\right)=0$$$$\sqrt{1-{\mathrm{t}}^2}+2(1-{\mathrm{t}}^2)-{2\mathrm{t}}^2=0$$$${4\mathrm{t}}^2-2=\sqrt{1-{\mathrm{t}}^2}$$$${16\mathrm{t}}^4-{16\mathrm{t}}^2+4=1-{\mathrm{t}}^2$$$${16\mathrm{t}}^4-{15\mathrm{t}}^2+3=0$$$${\mathrm{t}}^2=\frac{15+\sqrt{33}}{32}\mathrm{or}{\mathrm{t}}^2=\frac{15-\sqrt{33}}{32}$$$$\sin \mathrm{x}=\sqrt{\frac{15+\sqrt{33}}{32}}\mathrm{or}\sin \mathrm{x}=\sqrt{\frac{15-\sqrt{33}}{32}}$$$$=\frac{\sqrt{30+2\sqrt{33}}}{8}\mathrm{or}=\frac{\sqrt{30-2\sqrt{33}}}{8}$$$$$$

Answered by Frix last updated on 27/Jan/25

$$y=\mid \sin x\mid +\mid \sin 2x\mid \mathrm{is}\mathrm{periodic}\mathrm{and}\mathrm{symmetric}$$$${\mathrm{It}}^{\prime }\mathrm{s}\mathrm{enough}\mathrm{to}\mathrm{look}\mathrm{at}\mathrm{the}\mathrm{interval}[0,\frac{\pi }{2}]$$$$\mathrm{where}y=\sin x+\sin 2x$$$$y^{\prime }=0\Rightarrow$$$${\cos }^{2}x+\frac{\cos x}{4}-\frac{1}{2}=0$$$$\cos x=-\frac{1}{8}\pm \frac{\sqrt{33}}{8}$$$$0⩽x⩽\frac{\pi }{2}\Rightarrow x={\cos }^{-1}(-\frac{1}{8}+\frac{\sqrt{33}}{8})$$$$y_{\max }=\frac{\sqrt{6(69+11\sqrt{33}}}{16}\approx 1.76017259$$

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