Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

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) )) .

$$\mathrm{cos}\:\mathrm{5}\alpha\:=\mathrm{cos}\:\mathrm{3}\alpha\:\mathrm{cos}\:\mathrm{2}\alpha−\mathrm{sin}\:\mathrm{3}\alpha\:\mathrm{sin}\:\mathrm{2}\alpha \\ $$$$=\left(\mathrm{4cos}\:^{\mathrm{3}} \alpha−\mathrm{3cos}\:\alpha\right)\left(\mathrm{2cos}\:^{\mathrm{2}} \alpha−\mathrm{1}\right) \\ $$$$\:\:\:−\left(\mathrm{3sin}\:\alpha−\mathrm{4sin}\:^{\mathrm{3}} \alpha\right)\left(\mathrm{2sin}\:\alpha\mathrm{cos}\:\alpha\right) \\ $$$$=\mathrm{cos}\:\alpha\left(\mathrm{4cos}\:^{\mathrm{2}} \alpha−\mathrm{3}\right)\left(\mathrm{2cos}\:^{\mathrm{2}} \alpha−\mathrm{1}\right) \\ $$$$\:\:\:\:−\mathrm{2cos}\:\alpha\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \alpha\right)\left(\mathrm{4cos}\:^{\mathrm{2}} \alpha−\mathrm{1}\right) \\ $$$$=\mathrm{cos}\:\alpha\left(\mathrm{8cos}\:^{\mathrm{4}} \alpha−\mathrm{10cos}\:^{\mathrm{2}} \alpha+\mathrm{3}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{8cos}\:^{\mathrm{4}} \alpha−\mathrm{10cos}\:^{\mathrm{2}} \alpha+\mathrm{2}\right) \\ $$$$=\mathrm{cos}\:\alpha\left(\mathrm{16cos}\:^{\mathrm{4}} \alpha−\mathrm{20cos}\:^{\mathrm{2}} \alpha+\mathrm{5}\right) \\ $$$${let}\:\alpha=\frac{\pi}{\mathrm{2}}\:\:\:\:{and}\:\:\mathrm{cos}\:^{\mathrm{2}} \alpha={t} \\ $$$$\Rightarrow\:\mathrm{16}{t}^{\mathrm{2}} −\mathrm{20}{t}+\mathrm{5}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\mathrm{16}\left({t}−\frac{\mathrm{5}}{\mathrm{8}}\right)^{\mathrm{2}} =\frac{\mathrm{25}}{\mathrm{4}}−\mathrm{5} \\ $$$$\:\:\:{t}=\mathrm{cos}\:^{\mathrm{2}} \alpha=\frac{\mathrm{5}}{\mathrm{8}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{8}} \\ $$$$\Rightarrow\mathrm{cos}\:\alpha\:=\:\mathrm{cos}\:\frac{\pi}{\mathrm{10}}\:=\sqrt{\frac{\mathrm{5}+\sqrt{\mathrm{5}}}{\mathrm{8}}\:}\:. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com