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Question Number 216074 by CrispyXYZ last updated on 27/Jan/25
find the maximum of y=∣sin x∣+∣sin 2x∣.
$$\mathrm{find}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{of}\:{y}=\mid\mathrm{sin}\:{x}\mid+\mid\mathrm{sin}\:\mathrm{2}{x}\mid. \\ $$
Answered by efronzo1 last updated on 27/Jan/25
y = ∣sin x∣ + 2∣sin x∣ (√(1−sin^2 x)) y= ∣sin x∣ (1+2(√(1−sin^2 x)) ) Let ∣sin x∣ =(√(sin^2 x)) = t y = t (1+2(√(1−t^2 )) ) y′ = 1+2(√(1−t^2 )) +t(((−2t)/( (√(1−t^2 )))) )=0 (√(1−t^2 )) + 2(1−t^2 )−2t^2 = 0 4t^2 −2 = (√(1−t^2 )) 16t^4 −16t^2 +4 = 1−t^2 16t^4 −15t^2 +3 = 0 t^2 = ((15+(√(33)))/(32)) or t^2 = ((15−(√(33)))/(32)) sin x =(√((15+(√(33)))/(32))) or sin x=(√((15−(√(33)))/(32))) = ((√(30+2(√(33))))/8) or = ((√(30−2(√(33))))/8)
$$\:\mathrm{y}\:=\:\mid\mathrm{sin}\:\mathrm{x}\mid\:+\:\mathrm{2}\mid\mathrm{sin}\:\mathrm{x}\mid\:\sqrt{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}} \\ $$$$\:\:\mathrm{y}=\:\mid\mathrm{sin}\:\mathrm{x}\mid\:\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}}\:\right) \\ $$$$\:\:\mathrm{Let}\:\mid\mathrm{sin}\:\mathrm{x}\mid\:=\sqrt{\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}}\:=\:\mathrm{t}\: \\ $$$$\:\:\mathrm{y}\:=\:\mathrm{t}\:\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }\:\right) \\ $$$$\:\:\mathrm{y}'\:=\:\mathrm{1}+\mathrm{2}\sqrt{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }\:+\mathrm{t}\left(\frac{−\mathrm{2t}}{\:\sqrt{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }}\:\right)=\mathrm{0} \\ $$$$\:\:\:\sqrt{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }\:+\:\mathrm{2}\left(\mathrm{1}−\mathrm{t}^{\mathrm{2}} \right)−\mathrm{2t}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\:\:\:\mathrm{4t}^{\mathrm{2}} −\mathrm{2}\:=\:\sqrt{\mathrm{1}−\mathrm{t}^{\mathrm{2}} } \\ $$$$\:\:\:\mathrm{16t}^{\mathrm{4}} −\mathrm{16t}^{\mathrm{2}} +\mathrm{4}\:=\:\mathrm{1}−\mathrm{t}^{\mathrm{2}} \\ $$$$\:\:\:\mathrm{16t}^{\mathrm{4}} −\mathrm{15t}^{\mathrm{2}} +\mathrm{3}\:=\:\mathrm{0}\: \\ $$$$\:\:\:\:\mathrm{t}^{\mathrm{2}} \:=\:\frac{\mathrm{15}+\sqrt{\mathrm{33}}}{\mathrm{32}}\:\mathrm{or}\:\mathrm{t}^{\mathrm{2}} =\:\frac{\mathrm{15}−\sqrt{\mathrm{33}}}{\mathrm{32}} \\ $$$$\:\:\:\:\mathrm{sin}\:\mathrm{x}\:=\sqrt{\frac{\mathrm{15}+\sqrt{\mathrm{33}}}{\mathrm{32}}}\:\mathrm{or}\:\mathrm{sin}\:\mathrm{x}=\sqrt{\frac{\mathrm{15}−\sqrt{\mathrm{33}}}{\mathrm{32}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\sqrt{\mathrm{30}+\mathrm{2}\sqrt{\mathrm{33}}}}{\mathrm{8}}\:\:\:\mathrm{or}\:=\:\frac{\sqrt{\mathrm{30}−\mathrm{2}\sqrt{\mathrm{33}}}}{\mathrm{8}} \\ $$$$\:\:\:\:\:\:\:\:\:\: \\ $$
Answered by Frix last updated on 27/Jan/25
y=∣sin x∣+∣sin 2x∣ is periodic and symmetric It′s enough to look at the interval [0, (π/2)] where y=sin x +sin 2x y′=0 ⇒ cos^2 x +((cos x)/4)−(1/2)=0 cos x =−(1/8)±((√(33))/8) 0≤x≤(π/2) ⇒ x=cos^(−1) (−(1/8)+((√(33))/8)) y_(max) =((√(6(69+11(√(33))))/(16))≈1.76017259
$${y}=\mid\mathrm{sin}\:{x}\mid+\mid\mathrm{sin}\:\mathrm{2}{x}\mid\:\mathrm{is}\:\mathrm{periodic}\:\mathrm{and}\:\mathrm{symmetric} \\ $$$$\mathrm{It}'\mathrm{s}\:\mathrm{enough}\:\mathrm{to}\:\mathrm{look}\:\mathrm{at}\:\mathrm{the}\:\mathrm{interval}\:\left[\mathrm{0},\:\frac{\pi}{\mathrm{2}}\right] \\ $$$$\mathrm{where}\:{y}=\mathrm{sin}\:{x}\:+\mathrm{sin}\:\mathrm{2}{x} \\ $$$${y}'=\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{cos}^{\mathrm{2}} \:{x}\:+\frac{\mathrm{cos}\:{x}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$$$\mathrm{cos}\:{x}\:=−\frac{\mathrm{1}}{\mathrm{8}}\pm\frac{\sqrt{\mathrm{33}}}{\mathrm{8}} \\ $$$$\mathrm{0}\leqslant{x}\leqslant\frac{\pi}{\mathrm{2}}\:\Rightarrow\:{x}=\mathrm{cos}^{−\mathrm{1}} \:\left(−\frac{\mathrm{1}}{\mathrm{8}}+\frac{\sqrt{\mathrm{33}}}{\mathrm{8}}\right) \\ $$$${y}_{\mathrm{max}} =\frac{\sqrt{\mathrm{6}\left(\mathrm{69}+\mathrm{11}\sqrt{\mathrm{33}}\right.}}{\mathrm{16}}\approx\mathrm{1}.\mathrm{76017259} \\ $$

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