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Question Number 179550 by cortano1 last updated on 30/Oct/22
16cos^5 θ−cos 5θ = (5/2) 0<θ<2π θ =?
$$\:\:\:\mathrm{16cos}\:^{\mathrm{5}} \theta−\mathrm{cos}\:\mathrm{5}\theta\:=\:\frac{\mathrm{5}}{\mathrm{2}}\: \\ $$$$\:\:\:\mathrm{0}<\theta<\mathrm{2}\pi \\ $$$$\:\:\:\theta\:=? \\ $$
Answered by greougoury555 last updated on 30/Oct/22
{ ((cos θ= c)),((sin θ = s)) :}⇒cos 5θ=(4c^3 −3c)(2c^2 −1)−2cs(3s−4s^3 ) ⇒cos 5θ=c(4c^2 −3)(2c^2 −1)−(3−4(1−c^2 ))(2cs^2 ) ⇒ =c(4c^2 −3)(2c^2 −1)−(4c^2 −1)(2c)(1−c^2 ) = c(16c^4 −20c^2 +5)=16c^5 −20c^3 +5c ∵ 16cos^5 θ−cos 5θ = 16c^5 −(16c^5 −20c^3 +5c) = 20c^3 −5c ⇔ 16cos^5 θ−cos 5θ=(5/2) ⇔ 20c^3 −5c=(5/2) ⇔ 4c^3 −c = (1/2) ⇔ 8c^3 −2c−1=0 ⇔ (2c−1)(4c^2 +2c+1)=0 ⇒ c=(1/2) ⇒ θ = (π/3) and θ = ((5π)/3)
$$\:\:\begin{cases}{\mathrm{cos}\:\theta=\:{c}}\\{\mathrm{sin}\:\theta\:=\:{s}}\end{cases}\Rightarrow\mathrm{cos}\:\mathrm{5}\theta=\left(\mathrm{4}{c}^{\mathrm{3}} −\mathrm{3}{c}\right)\left(\mathrm{2}{c}^{\mathrm{2}} −\mathrm{1}\right)−\mathrm{2}{cs}\left(\mathrm{3}{s}−\mathrm{4}{s}^{\mathrm{3}} \right) \\ $$$$\:\Rightarrow\mathrm{cos}\:\mathrm{5}\theta={c}\left(\mathrm{4}{c}^{\mathrm{2}} −\mathrm{3}\right)\left(\mathrm{2}{c}^{\mathrm{2}} −\mathrm{1}\right)−\left(\mathrm{3}−\mathrm{4}\left(\mathrm{1}−{c}^{\mathrm{2}} \right)\right)\left(\mathrm{2}{cs}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\:\:\:\:\:\:\:\:\:\:\:\:\:\:={c}\left(\mathrm{4}{c}^{\mathrm{2}} −\mathrm{3}\right)\left(\mathrm{2}{c}^{\mathrm{2}} −\mathrm{1}\right)−\left(\mathrm{4}{c}^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{2}{c}\right)\left(\mathrm{1}−{c}^{\mathrm{2}} \right) \\ $$$$\:\:\:=\:{c}\left(\mathrm{16}{c}^{\mathrm{4}} −\mathrm{20}{c}^{\mathrm{2}} +\mathrm{5}\right)=\mathrm{16}{c}^{\mathrm{5}} −\mathrm{20}{c}^{\mathrm{3}} +\mathrm{5}{c} \\ $$$$\:\because\:\mathrm{16cos}\:^{\mathrm{5}} \theta−\mathrm{cos}\:\mathrm{5}\theta\:=\:\mathrm{16}{c}^{\mathrm{5}} −\left(\mathrm{16}{c}^{\mathrm{5}} −\mathrm{20}{c}^{\mathrm{3}} +\mathrm{5}{c}\right) \\ $$$$\:\:\:\:\:\:=\:\mathrm{20}{c}^{\mathrm{3}} −\mathrm{5}{c}\: \\ $$$$\Leftrightarrow\:\mathrm{16cos}\:^{\mathrm{5}} \theta−\mathrm{cos}\:\mathrm{5}\theta=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\Leftrightarrow\:\mathrm{20}{c}^{\mathrm{3}} −\mathrm{5}{c}=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\Leftrightarrow\:\mathrm{4}{c}^{\mathrm{3}} −{c}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Leftrightarrow\:\mathrm{8}{c}^{\mathrm{3}} −\mathrm{2}{c}−\mathrm{1}=\mathrm{0} \\ $$$$\Leftrightarrow\:\left(\mathrm{2}{c}−\mathrm{1}\right)\left(\mathrm{4}{c}^{\mathrm{2}} +\mathrm{2}{c}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\:{c}=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:\theta\:=\:\frac{\pi}{\mathrm{3}}\:{and}\:\theta\:=\:\frac{\mathrm{5}\pi}{\mathrm{3}} \\ $$
Commented by Frix last updated on 30/Oct/22
(2c−1)(4c^2 +2c+1)=8c^3 −1
$$\left(\mathrm{2}{c}−\mathrm{1}\right)\left(\mathrm{4}{c}^{\mathrm{2}} +\mathrm{2}{c}+\mathrm{1}\right)=\mathrm{8}{c}^{\mathrm{3}} −\mathrm{1} \\ $$
Answered by Frix last updated on 30/Oct/22
cos^5 θ =((cos 5θ +5cos 3θ +10cos θ)/(16)) 5cos 3θ +10cos θ =(5/2) cos^3 θ −((cos θ)/4)−(1/8)=0 I don′t think it makes sense to use the exact solution cos θ =(((1/(16))−((√(69))/(144))))^(1/3) +(((1/(16))+((√(69))/(144))))^(1/3) ⇒ θ≈.846833∨θ≈5.43635
$$\mathrm{cos}^{\mathrm{5}} \:\theta\:=\frac{\mathrm{cos}\:\mathrm{5}\theta\:+\mathrm{5cos}\:\mathrm{3}\theta\:+\mathrm{10cos}\:\theta}{\mathrm{16}} \\ $$$$\mathrm{5cos}\:\mathrm{3}\theta\:+\mathrm{10cos}\:\theta\:=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\mathrm{cos}^{\mathrm{3}} \:\theta\:−\frac{\mathrm{cos}\:\theta}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{8}}=\mathrm{0} \\ $$$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{it}\:\mathrm{makes}\:\mathrm{sense}\:\mathrm{to}\:\mathrm{use}\:\mathrm{the}\:\mathrm{exact} \\ $$$$\mathrm{solution}\:\mathrm{cos}\:\theta\:=\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{16}}−\frac{\sqrt{\mathrm{69}}}{\mathrm{144}}}+\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{16}}+\frac{\sqrt{\mathrm{69}}}{\mathrm{144}}} \\ $$$$\Rightarrow \\ $$$$\theta\approx.\mathrm{846833}\vee\theta\approx\mathrm{5}.\mathrm{43635} \\ $$

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