sin-3A-cos-3A-1-2-

Question Number 119229 by benjo_mathlover last updated on 23/Oct/20

$$\:\:\:\mathrm{sin}\:\mathrm{3}{A}+\mathrm{cos}\:\mathrm{3}{A}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Answered by bemath last updated on 23/Oct/20

let 3A = x ; cos x = X and sin x = Y ⇒X+Y = (1/2) ; X^2 +2XY +Y^2 = (1/4) ⇔ 2XY = −(3/4) ; sin 2x = −(3/4) ⇒ sin 6A = −(3/4) ⇒ 6A = arc sin (−(3/4)) ⇒ A = ((arc sin (−(3/4))+2nπ)/6) ⇒A = ((π−arc sin (−(3/4))+2nπ)/6) sin^(−1) (−(3/4)) −0.848062

$${let}\:\mathrm{3}{A}\:=\:{x}\:;\:\mathrm{cos}\:{x}\:=\:{X}\:{and}\:\mathrm{sin}\:{x}\:=\:{Y} \\ $$$$\Rightarrow{X}+{Y}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:;\:{X}^{\mathrm{2}} +\mathrm{2}{XY}\:+{Y}^{\mathrm{2}} \:=\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Leftrightarrow\:\mathrm{2}{XY}\:=\:−\frac{\mathrm{3}}{\mathrm{4}}\:;\:\mathrm{sin}\:\mathrm{2}{x}\:=\:−\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow\:\mathrm{sin}\:\mathrm{6}{A}\:=\:−\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow\:\mathrm{6}{A}\:=\:\mathrm{arc}\:\mathrm{sin}\:\left(−\frac{\mathrm{3}}{\mathrm{4}}\right)\: \\ $$$$\Rightarrow\:{A}\:=\:\frac{\mathrm{arc}\:\mathrm{sin}\:\left(−\frac{\mathrm{3}}{\mathrm{4}}\right)+\mathrm{2}{n}\pi}{\mathrm{6}} \\ $$$$\Rightarrow{A}\:=\:\frac{\pi−\mathrm{arc}\:\mathrm{sin}\:\left(−\frac{\mathrm{3}}{\mathrm{4}}\right)+\mathrm{2}{n}\pi}{\mathrm{6}} \\ $$$$\mathrm{sin}^{−\mathrm{1}} \left(−\frac{\mathrm{3}}{\mathrm{4}}\right)\: \\ $$$$−\mathrm{0}.\mathrm{848062} \\ $$

Answered by 1549442205PVT last updated on 23/Oct/20

sin 3A+cos 3A = (1/2) ⇔(√2)(((√2)/2)sin3A+((√2)/2)cos3A)=(1/2) ⇔sin3Acos45+cos3Asin45=((√2)/4) ⇔sin(3A+45°)=((√2)/4)⇔3A+45°= sin^(−1) (((√2)/4))+k.360°≈20°42′17′′+k360° ⇔A≈−8°5′54′′+k.120°(k∈Z)

$$\:\:\:\mathrm{sin}\:\mathrm{3}{A}+\mathrm{cos}\:\mathrm{3}{A}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Leftrightarrow\sqrt{\mathrm{2}}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{sin3A}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{cos3A}\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Leftrightarrow\mathrm{sin3Acos45}+\mathrm{cos3Asin45}=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$$$\Leftrightarrow\mathrm{sin}\left(\mathrm{3A}+\mathrm{45}°\right)=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\Leftrightarrow\mathrm{3A}+\mathrm{45}°= \\ $$$$\mathrm{sin}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\right)+\mathrm{k}.\mathrm{360}°\approx\mathrm{20}°\mathrm{42}'\mathrm{17}''+\mathrm{k360}° \\ $$$$\Leftrightarrow\mathrm{A}\approx−\mathrm{8}°\mathrm{5}'\mathrm{54}''+\mathrm{k}.\mathrm{120}°\left(\mathrm{k}\in\mathrm{Z}\right) \\ $$

Answered by mathmax by abdo last updated on 24/Oct/20

sin(3a)+cos(3a)=(1/2) ⇒(√2)(cos((π/4))cos(3a)+sin((π/4))sin(3a))=(1/2) ⇒(√2)cos(3a−(π/4))=(1/2) ⇒cos(3a−(π/4))=(1/(2(√2))) ∃α_0 /cos(α_0 )=((√2)/4) so e⇒cos(3a−(π/4))=cosα_o ⇒ 3a−(π/4) =α_0 +2kπ or 3a−(π/4) =−α_0 +2kπ ⇒ 3a =α_0 +(π/4) +2kπ or 3a=(π/4)−α_0 +2kπ ⇒ a =(α_0 /3) +(π/(12)) +((2kπ)/3) or a =(π/(12))−(α_0 /3) +((2kπ)/3) (k ∈Z) and α_0 =arcos(((√2)/4))

$$\mathrm{sin}\left(\mathrm{3a}\right)+\mathrm{cos}\left(\mathrm{3a}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\sqrt{\mathrm{2}}\left(\mathrm{cos}\left(\frac{\pi}{\mathrm{4}}\right)\mathrm{cos}\left(\mathrm{3a}\right)+\mathrm{sin}\left(\frac{\pi}{\mathrm{4}}\right)\mathrm{sin}\left(\mathrm{3a}\right)\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\sqrt{\mathrm{2}}\mathrm{cos}\left(\mathrm{3a}−\frac{\pi}{\mathrm{4}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\mathrm{cos}\left(\mathrm{3a}−\frac{\pi}{\mathrm{4}}\right)=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$$$\exists\alpha_{\mathrm{0}} \:/\mathrm{cos}\left(\alpha_{\mathrm{0}} \right)=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\:\:\mathrm{so}\:\mathrm{e}\Rightarrow\mathrm{cos}\left(\mathrm{3a}−\frac{\pi}{\mathrm{4}}\right)=\mathrm{cos}\alpha_{\mathrm{o}} \:\Rightarrow \\ $$$$\mathrm{3a}−\frac{\pi}{\mathrm{4}}\:=\alpha_{\mathrm{0}} \:+\mathrm{2k}\pi\:\:\mathrm{or}\:\mathrm{3a}−\frac{\pi}{\mathrm{4}}\:=−\alpha_{\mathrm{0}} \:+\mathrm{2k}\pi\:\Rightarrow \\ $$$$\mathrm{3a}\:=\alpha_{\mathrm{0}} +\frac{\pi}{\mathrm{4}}\:+\mathrm{2k}\pi\:\mathrm{or}\:\mathrm{3a}=\frac{\pi}{\mathrm{4}}−\alpha_{\mathrm{0}} \:+\mathrm{2k}\pi\:\Rightarrow \\ $$$$\mathrm{a}\:=\frac{\alpha_{\mathrm{0}} }{\mathrm{3}}\:+\frac{\pi}{\mathrm{12}}\:+\frac{\mathrm{2k}\pi}{\mathrm{3}}\:\mathrm{or}\:\mathrm{a}\:=\frac{\pi}{\mathrm{12}}−\frac{\alpha_{\mathrm{0}} }{\mathrm{3}}\:+\frac{\mathrm{2k}\pi}{\mathrm{3}}\:\:\:\left(\mathrm{k}\:\in\mathrm{Z}\right)\:\:\mathrm{and}\:\alpha_{\mathrm{0}} =\mathrm{arcos}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\right) \\ $$

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