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Question Number 105 by mr.guddukr@gmail.com last updated on 25/Jan/15

sin5θ= ?

$${sin}\mathrm{5}\theta=\:? \\ $$

Answered by mreddy last updated on 03/Dec/14

sin 2θ=2sin θcos θ cos 2θ=1−2sin^2 θ sin 3θ=sin (2θ+θ)=2sin θcos^2 θ+(1−2sin^2 θ)sin θ =3sin θ−4sin^3 θ cos 3θ=cos θ(1−4sin^2 θ) sin 5θ=sin (3θ+2θ) =sin 3θcos 2θ+cos 3θsin 2θ =(3sin θ−4sin^3 θ)(1−2sin^2 θ)+cos θ(1−4sin^2 θ) 2sin θcos θ =(3sin θ−4sin^3 θ)(1−2sin^2 θ)+(1−4sin^2 θ)2sinθcos^2 θ =(3sin θ−4sin^3 θ)(1−2sin^2 θ)+(1−4sin^2 θ)(2sin θ−2sin^3 θ) =5sin θ−20sin^3 θ+16sin^5 θ

$$\mathrm{sin}\:\mathrm{2}\theta=\mathrm{2sin}\:\theta\mathrm{cos}\:\theta \\ $$$$\mathrm{cos}\:\mathrm{2}\theta=\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \theta \\ $$$$\mathrm{sin}\:\mathrm{3}\theta=\mathrm{sin}\:\left(\mathrm{2}\theta+\theta\right)=\mathrm{2sin}\:\theta\mathrm{cos}^{\mathrm{2}} \theta+\left(\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \theta\right)\mathrm{sin}\:\theta \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{3sin}\:\theta−\mathrm{4sin}^{\mathrm{3}} \theta \\ $$$$\mathrm{cos}\:\mathrm{3}\theta=\mathrm{cos}\:\theta\left(\mathrm{1}−\mathrm{4sin}^{\mathrm{2}} \theta\right) \\ $$$$ \\ $$$$\mathrm{sin}\:\mathrm{5}\theta=\mathrm{sin}\:\left(\mathrm{3}\theta+\mathrm{2}\theta\right) \\ $$$$=\mathrm{sin}\:\mathrm{3}\theta\mathrm{cos}\:\mathrm{2}\theta+\mathrm{cos}\:\mathrm{3}\theta\mathrm{sin}\:\mathrm{2}\theta \\ $$$$=\left(\mathrm{3sin}\:\theta−\mathrm{4sin}^{\mathrm{3}} \theta\right)\left(\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \theta\right)+\mathrm{cos}\:\theta\left(\mathrm{1}−\mathrm{4sin}^{\mathrm{2}} \theta\right)\:\mathrm{2sin}\:\theta\mathrm{cos}\:\theta \\ $$$$=\left(\mathrm{3sin}\:\theta−\mathrm{4sin}^{\mathrm{3}} \theta\right)\left(\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \theta\right)+\left(\mathrm{1}−\mathrm{4sin}^{\mathrm{2}} \theta\right)\mathrm{2sin}\theta\mathrm{cos}^{\mathrm{2}} \theta \\ $$$$=\left(\mathrm{3sin}\:\theta−\mathrm{4sin}^{\mathrm{3}} \theta\right)\left(\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \theta\right)+\left(\mathrm{1}−\mathrm{4sin}^{\mathrm{2}} \theta\right)\left(\mathrm{2sin}\:\theta−\mathrm{2sin}^{\mathrm{3}} \theta\right) \\ $$$$=\mathrm{5sin}\:\theta−\mathrm{20sin}^{\mathrm{3}} \theta+\mathrm{16sin}^{\mathrm{5}} \theta \\ $$

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