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Question Number 70597 by Maclaurin Stickker last updated on 06/Oct/19

solve cos^2 β+cos^2 3β=1

$${solve} \\ $$$$\mathrm{cos}\:^{\mathrm{2}} \beta+\mathrm{cos}\:^{\mathrm{2}} \mathrm{3}\beta=\mathrm{1} \\ $$

Answered by Kunal12588 last updated on 08/Oct/19

cos 2α = 2cos^2 α−1⇒cos^2 α=((1+cos 2α)/2) cos^2 β=((1+cos 2β)/2) cos^2 3β=((1+cos 6β)/2) ∴ 1+cos 2β+1+cos 6β=2 ⇒cos 2β+cos 6β=0 ⇒2cos 4β cos 2β =0 ⇒cos 4β=0 and cos 2β=0 ⇒4β=2nπ±(π/2) and 2β=2nπ±(π/2) ⇒β=((nπ)/2)±(π/8), nπ±(π/4)

$${cos}\:\mathrm{2}\alpha\:=\:\mathrm{2}{cos}^{\mathrm{2}} \alpha−\mathrm{1}\Rightarrow{cos}^{\mathrm{2}} \:\alpha=\frac{\mathrm{1}+{cos}\:\mathrm{2}\alpha}{\mathrm{2}} \\ $$$${cos}^{\mathrm{2}} \beta=\frac{\mathrm{1}+{cos}\:\mathrm{2}\beta}{\mathrm{2}} \\ $$$${cos}^{\mathrm{2}} \mathrm{3}\beta=\frac{\mathrm{1}+{cos}\:\mathrm{6}\beta}{\mathrm{2}} \\ $$$$\therefore\:\mathrm{1}+{cos}\:\mathrm{2}\beta+\mathrm{1}+{cos}\:\mathrm{6}\beta=\mathrm{2} \\ $$$$\Rightarrow{cos}\:\mathrm{2}\beta+{cos}\:\mathrm{6}\beta=\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}{cos}\:\mathrm{4}\beta\:{cos}\:\mathrm{2}\beta\:=\mathrm{0} \\ $$$$\Rightarrow{cos}\:\mathrm{4}\beta=\mathrm{0}\:\:\:{and}\:{cos}\:\mathrm{2}\beta=\mathrm{0} \\ $$$$\Rightarrow\mathrm{4}\beta=\mathrm{2}{n}\pi\pm\frac{\pi}{\mathrm{2}}\:\:\:\:{and}\:\mathrm{2}\beta=\mathrm{2}{n}\pi\pm\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow\beta=\frac{{n}\pi}{\mathrm{2}}\pm\frac{\pi}{\mathrm{8}},\:{n}\pi\pm\frac{\pi}{\mathrm{4}} \\ $$

Commented by mr W last updated on 06/Oct/19

nice way! better solution than mine!

$${nice}\:{way}! \\ $$$${better}\:{solution}\:{than}\:{mine}! \\ $$

Commented by mathmax by abdo last updated on 06/Oct/19

error cos^2 α=((1+cos(2α))/2) ..

$${error}\:\:{cos}^{\mathrm{2}} \alpha=\frac{\mathrm{1}+{cos}\left(\mathrm{2}\alpha\right)}{\mathrm{2}}\:.. \\ $$

Commented by Kunal12588 last updated on 08/Oct/19

thanks sir

$${thanks}\:{sir} \\ $$

Answered by mr W last updated on 06/Oct/19

cos 3β=cos β (4 cos^2 β−3) cos^2 β+cos^2 β (4 cos^2 β−3)^2 =1 let t=cos^2 β≥0, ≤1 ⇒t+t(4t−3)^2 =1 ⇒16t^3 −24t^2 +10t−1=0 let s=2t≥0,≤2 ⇒2s^3 −6s^2 +5s−1=0 ⇒(s−1)(2s^2 −4s+1)=0 ⇒s=1 ⇒s=((2±(√2))/2) ⇒t=cos^2 β=(1/2) ⇒cos β=±((√2)/2) ⇒β=nπ±(π/4) ⇒t=cos^2 β=((2+(√2))/4) ⇒cos β=±((√(2+(√2)))/2) ⇒β=nπ±(π/8) ⇒t=cos^2 β=((2−(√2))/4) ⇒cos β=±((√(2−(√2)))/2) ⇒β=nπ±((3π)/8)

$$\mathrm{cos}\:\mathrm{3}\beta=\mathrm{cos}\:\beta\:\left(\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\beta−\mathrm{3}\right) \\ $$$$\mathrm{cos}^{\mathrm{2}} \:\beta+\mathrm{cos}^{\mathrm{2}} \:\beta\:\left(\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\beta−\mathrm{3}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$${let}\:{t}=\mathrm{cos}^{\mathrm{2}} \:\beta\geqslant\mathrm{0},\:\leqslant\mathrm{1} \\ $$$$\Rightarrow{t}+{t}\left(\mathrm{4}{t}−\mathrm{3}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\Rightarrow\mathrm{16}{t}^{\mathrm{3}} −\mathrm{24}{t}^{\mathrm{2}} +\mathrm{10}{t}−\mathrm{1}=\mathrm{0} \\ $$$${let}\:{s}=\mathrm{2}{t}\geqslant\mathrm{0},\leqslant\mathrm{2} \\ $$$$\Rightarrow\mathrm{2}{s}^{\mathrm{3}} −\mathrm{6}{s}^{\mathrm{2}} +\mathrm{5}{s}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\left({s}−\mathrm{1}\right)\left(\mathrm{2}{s}^{\mathrm{2}} −\mathrm{4}{s}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{s}=\mathrm{1} \\ $$$$\Rightarrow{s}=\frac{\mathrm{2}\pm\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\Rightarrow{t}=\mathrm{cos}^{\mathrm{2}} \:\beta=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\mathrm{cos}\:\beta=\pm\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\Rightarrow\beta={n}\pi\pm\frac{\pi}{\mathrm{4}} \\ $$$$\Rightarrow{t}=\mathrm{cos}^{\mathrm{2}} \:\beta=\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{4}}\:\Rightarrow\mathrm{cos}\:\beta=\pm\frac{\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}}{\mathrm{2}}\:\Rightarrow\beta={n}\pi\pm\frac{\pi}{\mathrm{8}} \\ $$$$\Rightarrow{t}=\mathrm{cos}^{\mathrm{2}} \:\beta=\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{4}}\:\Rightarrow\mathrm{cos}\:\beta=\pm\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}}{\mathrm{2}}\:\Rightarrow\beta={n}\pi\pm\frac{\mathrm{3}\pi}{\mathrm{8}} \\ $$

Answered by Rasheed.Sindhi last updated on 06/Oct/19

cos^2 β+cos^2 3β=1 cos^2 β+cos^2 3β=sin^2 β+cos^2 β cos^2 3β=sin^2 β cos^2 3β−sin^2 β=0 (cos 3β−sin β)(cos 3β+sin β)=0 cos 3β−sin β=0 ∨ cos 3β+sin β) cos 3β=sin β ∨ cos 3β=−sin β) cos 3β=sin β ∨ cos 3β=sin (−β) 3β+β=π/2 ∨ 3β−β=π/2 4β=(π/2) ∨ 2β=(π/2) β=(π/8) , (π/4)

$$\mathrm{cos}\:^{\mathrm{2}} \beta+\mathrm{cos}\:^{\mathrm{2}} \mathrm{3}\beta=\mathrm{1} \\ $$$$\mathrm{cos}\:^{\mathrm{2}} \beta+\mathrm{cos}\:^{\mathrm{2}} \mathrm{3}\beta=\mathrm{sin}^{\mathrm{2}} \beta+\mathrm{cos}^{\mathrm{2}} \beta\:\: \\ $$$$\mathrm{cos}\:^{\mathrm{2}} \mathrm{3}\beta=\mathrm{sin}^{\mathrm{2}} \beta \\ $$$$\mathrm{cos}\:^{\mathrm{2}} \mathrm{3}\beta−\mathrm{sin}^{\mathrm{2}} \beta=\mathrm{0} \\ $$$$\left(\mathrm{cos}\:\mathrm{3}\beta−\mathrm{sin}\:\beta\right)\left(\mathrm{cos}\:\mathrm{3}\beta+\mathrm{sin}\:\beta\right)=\mathrm{0} \\ $$$$\left.\mathrm{cos}\:\mathrm{3}\beta−\mathrm{sin}\:\beta=\mathrm{0}\:\vee\:\mathrm{cos}\:\mathrm{3}\beta+\mathrm{sin}\:\beta\right) \\ $$$$\left.\mathrm{cos}\:\mathrm{3}\beta=\mathrm{sin}\:\beta\:\vee\:\mathrm{cos}\:\mathrm{3}\beta=−\mathrm{sin}\:\beta\right) \\ $$$$\mathrm{cos}\:\mathrm{3}\beta=\mathrm{sin}\:\beta\:\vee\:\mathrm{cos}\:\mathrm{3}\beta=\mathrm{sin}\:\left(−\beta\right) \\ $$$$\mathrm{3}\beta+\beta=\pi/\mathrm{2}\:\:\vee\:\mathrm{3}\beta−\beta=\pi/\mathrm{2} \\ $$$$\mathrm{4}\beta=\frac{\pi}{\mathrm{2}}\:\:\vee\:\mathrm{2}\beta=\frac{\pi}{\mathrm{2}} \\ $$$$\beta=\frac{\pi}{\mathrm{8}}\:,\:\frac{\pi}{\mathrm{4}} \\ $$

Answered by ajfour last updated on 06/Oct/19

2β=θ cos^2 (θ−β)+cos^2 (θ+β)=1 cos (2θ−2β)+cos (2θ+2β)=0 ⇒ 2θ+2β = (2n+1)π±(2θ−2β) ⇒ 6β=(2n+1)π±2β ⇒ β = (((2n+1)π)/8) , β = (((2n+1)π)/4) ⇒ β = ((kπ)/8) .

$$\mathrm{2}\beta=\theta \\ $$$$\mathrm{cos}\:^{\mathrm{2}} \left(\theta−\beta\right)+\mathrm{cos}\:^{\mathrm{2}} \left(\theta+\beta\right)=\mathrm{1} \\ $$$$\mathrm{cos}\:\left(\mathrm{2}\theta−\mathrm{2}\beta\right)+\mathrm{cos}\:\left(\mathrm{2}\theta+\mathrm{2}\beta\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:\mathrm{2}\theta+\mathrm{2}\beta\:=\:\left(\mathrm{2}{n}+\mathrm{1}\right)\pi\pm\left(\mathrm{2}\theta−\mathrm{2}\beta\right) \\ $$$$\Rightarrow\:\:\mathrm{6}\beta=\left(\mathrm{2}{n}+\mathrm{1}\right)\pi\pm\mathrm{2}\beta \\ $$$$\Rightarrow\:\:\beta\:=\:\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)\pi}{\mathrm{8}}\:,\:\:\beta\:=\:\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)\pi}{\mathrm{4}} \\ $$$$\Rightarrow\:\:\beta\:=\:\frac{{k}\pi}{\mathrm{8}}\:. \\ $$

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