Question Number 11751 by tawa last updated on 31/Mar/17
$$\mathrm{Simplify}:\:\mathrm{sin}\left(\mathrm{6}\theta\right) \\ $$
Answered by mrW1 last updated on 31/Mar/17
![sin (6θ)=sin (4θ+2θ)=sin (4θ)cos (2θ)+sin (2θ)cos (4θ) =2sin (2θ)cos^2 (2θ)+sin (2θ)[1−2sin^2 (2θ)] =2sin (2θ)[1−sin^2 (2θ)]+sin (2θ)−2sin^3 (2θ) =2sin (2θ)−2sin^3 (2θ)+sin (2θ)−2sin^3 (2θ) =3sin (2θ)−4sin^3 (2θ) =3×2sin θcos θ−4×8×sin^3 θcos^3 θ =2sin θcos θ(3−16sin^2 θcos^2 θ)](Q11756.png)
$$\mathrm{sin}\:\left(\mathrm{6}\theta\right)=\mathrm{sin}\:\left(\mathrm{4}\theta+\mathrm{2}\theta\right)=\mathrm{sin}\:\left(\mathrm{4}\theta\right)\mathrm{cos}\:\left(\mathrm{2}\theta\right)+\mathrm{sin}\:\left(\mathrm{2}\theta\right)\mathrm{cos}\:\left(\mathrm{4}\theta\right) \\ $$$$=\mathrm{2sin}\:\left(\mathrm{2}\theta\right)\mathrm{cos}^{\mathrm{2}} \:\left(\mathrm{2}\theta\right)+\mathrm{sin}\:\left(\mathrm{2}\theta\right)\left[\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \:\left(\mathrm{2}\theta\right)\right] \\ $$$$=\mathrm{2sin}\:\left(\mathrm{2}\theta\right)\left[\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:\left(\mathrm{2}\theta\right)\right]+\mathrm{sin}\:\left(\mathrm{2}\theta\right)−\mathrm{2sin}^{\mathrm{3}} \:\left(\mathrm{2}\theta\right) \\ $$$$=\mathrm{2sin}\:\left(\mathrm{2}\theta\right)−\mathrm{2sin}^{\mathrm{3}} \:\left(\mathrm{2}\theta\right)+\mathrm{sin}\:\left(\mathrm{2}\theta\right)−\mathrm{2sin}^{\mathrm{3}} \:\left(\mathrm{2}\theta\right) \\ $$$$=\mathrm{3sin}\:\left(\mathrm{2}\theta\right)−\mathrm{4sin}^{\mathrm{3}} \:\left(\mathrm{2}\theta\right) \\ $$$$=\mathrm{3}×\mathrm{2sin}\:\theta\mathrm{cos}\:\theta−\mathrm{4}×\mathrm{8}×\mathrm{sin}^{\mathrm{3}} \:\theta\mathrm{cos}^{\mathrm{3}} \:\theta \\ $$$$=\mathrm{2sin}\:\theta\mathrm{cos}\:\theta\left(\mathrm{3}−\mathrm{16sin}^{\mathrm{2}} \:\theta\mathrm{cos}^{\mathrm{2}} \:\theta\right) \\ $$
Commented by tawa last updated on 31/Mar/17
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$