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Question Number 29173 by abdo imad last updated on 04/Feb/18
find cos(5α) interms of cosα then find the value of cos((π/(10))).
$${find}\:{cos}\left(\mathrm{5}\alpha\right)\:{interms}\:{of}\:{cos}\alpha\:{then}\:{find}\:{the}\:{value}\:{of} \\ $$$${cos}\left(\frac{\pi}{\mathrm{10}}\right). \\ $$
Answered by ajfour last updated on 05/Feb/18
cos 5α =cos 3α cos 2α−sin 3α sin 2α =(4cos^3 α−3cos α)(2cos^2 α−1) −(3sin α−4sin^3 α)(2sin αcos α) =cos α(4cos^2 α−3)(2cos^2 α−1) −2cos α(1−cos^2 α)(4cos^2 α−1) =cos α(8cos^4 α−10cos^2 α+3 +8cos^4 α−10cos^2 α+2) =cos α(16cos^4 α−20cos^2 α+5) let α=(π/2) and cos^2 α=t ⇒ 16t^2 −20t+5=0 16(t−(5/8))^2 =((25)/4)−5 t=cos^2 α=(5/8)+((√5)/8) ⇒cos α = cos (π/(10)) =(√(((5+(√5))/8) )) .
$$\mathrm{cos}\:\mathrm{5}\alpha\:=\mathrm{cos}\:\mathrm{3}\alpha\:\mathrm{cos}\:\mathrm{2}\alpha−\mathrm{sin}\:\mathrm{3}\alpha\:\mathrm{sin}\:\mathrm{2}\alpha \\ $$$$=\left(\mathrm{4cos}\:^{\mathrm{3}} \alpha−\mathrm{3cos}\:\alpha\right)\left(\mathrm{2cos}\:^{\mathrm{2}} \alpha−\mathrm{1}\right) \\ $$$$\:\:\:−\left(\mathrm{3sin}\:\alpha−\mathrm{4sin}\:^{\mathrm{3}} \alpha\right)\left(\mathrm{2sin}\:\alpha\mathrm{cos}\:\alpha\right) \\ $$$$=\mathrm{cos}\:\alpha\left(\mathrm{4cos}\:^{\mathrm{2}} \alpha−\mathrm{3}\right)\left(\mathrm{2cos}\:^{\mathrm{2}} \alpha−\mathrm{1}\right) \\ $$$$\:\:\:\:−\mathrm{2cos}\:\alpha\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \alpha\right)\left(\mathrm{4cos}\:^{\mathrm{2}} \alpha−\mathrm{1}\right) \\ $$$$=\mathrm{cos}\:\alpha\left(\mathrm{8cos}\:^{\mathrm{4}} \alpha−\mathrm{10cos}\:^{\mathrm{2}} \alpha+\mathrm{3}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{8cos}\:^{\mathrm{4}} \alpha−\mathrm{10cos}\:^{\mathrm{2}} \alpha+\mathrm{2}\right) \\ $$$$=\mathrm{cos}\:\alpha\left(\mathrm{16cos}\:^{\mathrm{4}} \alpha−\mathrm{20cos}\:^{\mathrm{2}} \alpha+\mathrm{5}\right) \\ $$$${let}\:\alpha=\frac{\pi}{\mathrm{2}}\:\:\:\:{and}\:\:\mathrm{cos}\:^{\mathrm{2}} \alpha={t} \\ $$$$\Rightarrow\:\mathrm{16}{t}^{\mathrm{2}} −\mathrm{20}{t}+\mathrm{5}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\mathrm{16}\left({t}−\frac{\mathrm{5}}{\mathrm{8}}\right)^{\mathrm{2}} =\frac{\mathrm{25}}{\mathrm{4}}−\mathrm{5} \\ $$$$\:\:\:{t}=\mathrm{cos}\:^{\mathrm{2}} \alpha=\frac{\mathrm{5}}{\mathrm{8}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{8}} \\ $$$$\Rightarrow\mathrm{cos}\:\alpha\:=\:\mathrm{cos}\:\frac{\pi}{\mathrm{10}}\:=\sqrt{\frac{\mathrm{5}+\sqrt{\mathrm{5}}}{\mathrm{8}}\:}\:. \\ $$

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