Question Number 17866 by Tinkutara last updated on 11/Jul/17
$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{sin}\:\mathrm{5}{A}\:=\:\mathrm{5}\:\mathrm{cos}^{\mathrm{4}} \:{A}\:\mathrm{sin}\:{A}\:− \\ $$$$\mathrm{10}\:\mathrm{cos}^{\mathrm{2}} \:{A}\:\mathrm{sin}^{\mathrm{3}} \:{A}\:+\:\mathrm{sin}^{\mathrm{5}} \:{A} \\ $$
Answered by alex041103 last updated on 13/Jul/17
![sin5A=sin(2A+3A) We use sin(A+B)=sinAcosB+sinBcosA sin2A=2sinAcosA cos2A=cos^2 A−sin^2 A sin3A=3sinA−4sin^3 A cos3A=4cos^3 A−3cosA And we have sin5A=8sinAcos^4 A−3sinAcos^2 A −3sin^3 A−4sin^3 Acos^2 A+4sin^5 A= =5cos^4 AsinA−10cos^2 Asin^3 A+sin^5 A +3sinA(sin^4 A+2sin^2 Acos^2 A+cos^4 A−sin^2 A−cos^2 A) =5cos^4 AsinA−10cos^2 Asin^3 A+sin^5 A +3sinA[(sin^2 A+cos^2 A)^2 −(sin^2 A+cos^2 A)] =5cos^4 AsinA−10cos^2 Asin^3 A+sin^5 A +3sinA(1^2 −1)= =5cos^4 AsinA−10cos^2 Asin^3 A+sin^5 A +3sinA(0)= =5cos^4 AsinA−10cos^2 Asin^3 A+sin^5 A ⇒sin5A=5cos^4 AsinA−10cos^2 Asin^3 A+sin^5 A](https://www.tinkutara.com/question/Q17967.png)
$${sin}\mathrm{5}{A}={sin}\left(\mathrm{2}{A}+\mathrm{3}{A}\right) \\ $$$${We}\:{use} \\ $$$${sin}\left({A}+{B}\right)={sinAcosB}+{sinBcosA} \\ $$$${sin}\mathrm{2}{A}=\mathrm{2}{sinAcosA} \\ $$$${cos}\mathrm{2}{A}={cos}^{\mathrm{2}} {A}−{sin}^{\mathrm{2}} {A} \\ $$$${sin}\mathrm{3}{A}=\mathrm{3}{sinA}−\mathrm{4}{sin}^{\mathrm{3}} {A} \\ $$$${cos}\mathrm{3}{A}=\mathrm{4}{cos}^{\mathrm{3}} {A}−\mathrm{3}{cosA} \\ $$$${And}\:{we}\:{have} \\ $$$${sin}\mathrm{5}{A}=\mathrm{8}{sinAcos}^{\mathrm{4}} {A}−\mathrm{3}{sinAcos}^{\mathrm{2}} {A} \\ $$$$−\mathrm{3}{sin}^{\mathrm{3}} {A}−\mathrm{4}{sin}^{\mathrm{3}} {Acos}^{\mathrm{2}} {A}+\mathrm{4}{sin}^{\mathrm{5}} {A}= \\ $$$$=\mathrm{5}{cos}^{\mathrm{4}} {AsinA}−\mathrm{10}{cos}^{\mathrm{2}} {Asin}^{\mathrm{3}} {A}+{sin}^{\mathrm{5}} {A} \\ $$$$+\mathrm{3}{sinA}\left({sin}^{\mathrm{4}} {A}+\mathrm{2}{sin}^{\mathrm{2}} {Acos}^{\mathrm{2}} {A}+{cos}^{\mathrm{4}} {A}−{sin}^{\mathrm{2}} {A}−{cos}^{\mathrm{2}} {A}\right) \\ $$$$=\mathrm{5}{cos}^{\mathrm{4}} {AsinA}−\mathrm{10}{cos}^{\mathrm{2}} {Asin}^{\mathrm{3}} {A}+{sin}^{\mathrm{5}} {A} \\ $$$$+\mathrm{3}{sinA}\left[\left({sin}^{\mathrm{2}} {A}+{cos}^{\mathrm{2}} {A}\right)^{\mathrm{2}} −\left({sin}^{\mathrm{2}} {A}+{cos}^{\mathrm{2}} {A}\right)\right] \\ $$$$=\mathrm{5}{cos}^{\mathrm{4}} {AsinA}−\mathrm{10}{cos}^{\mathrm{2}} {Asin}^{\mathrm{3}} {A}+{sin}^{\mathrm{5}} {A} \\ $$$$+\mathrm{3}{sinA}\left(\mathrm{1}^{\mathrm{2}} −\mathrm{1}\right)= \\ $$$$=\mathrm{5}{cos}^{\mathrm{4}} {AsinA}−\mathrm{10}{cos}^{\mathrm{2}} {Asin}^{\mathrm{3}} {A}+{sin}^{\mathrm{5}} {A} \\ $$$$+\mathrm{3}{sinA}\left(\mathrm{0}\right)= \\ $$$$=\mathrm{5}{cos}^{\mathrm{4}} {AsinA}−\mathrm{10}{cos}^{\mathrm{2}} {Asin}^{\mathrm{3}} {A}+{sin}^{\mathrm{5}} {A} \\ $$$$\Rightarrow{sin}\mathrm{5}{A}=\mathrm{5}{cos}^{\mathrm{4}} {AsinA}−\mathrm{10}{cos}^{\mathrm{2}} {Asin}^{\mathrm{3}} {A}+{sin}^{\mathrm{5}} {A} \\ $$$$ \\ $$
Commented by Tinkutara last updated on 13/Jul/17
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$