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Question Number 93368 by Rio Michael last updated on 12/May/20

given that f(r)= sin (1 + 2r)θ show that f(r)−f(r−1) = 2 cos 2r θ sin θ hence show that Σ_(r=1) ^n cos 2r θ sin θ = cos (n +1)θ sin nθ

$$\mathrm{given}\:\mathrm{that}\:{f}\left({r}\right)=\:\:\mathrm{sin}\:\left(\mathrm{1}\:+\:\mathrm{2}{r}\right)\theta \\ $$$$\mathrm{show}\:\mathrm{that}\:{f}\left({r}\right)−{f}\left({r}−\mathrm{1}\right)\:=\:\mathrm{2}\:\mathrm{cos}\:\mathrm{2}{r}\:\theta\:\mathrm{sin}\:\theta \\ $$$$\mathrm{hence}\:\mathrm{show}\:\mathrm{that}\: \\ $$$$\:\:\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\:\mathrm{cos}\:\mathrm{2}{r}\:\theta\:\mathrm{sin}\:\theta\:\:=\:\mathrm{cos}\:\left({n}\:+\mathrm{1}\right)\theta\:\mathrm{sin}\:{n}\theta \\ $$

Answered by mr W last updated on 12/May/20

f(r)=sin (1+2r)θ f(r−1)=sin (2r−1)θ=−sin (1−2r)θ f(r)−f(r−1)=sin (1+2r)θ+sin (1−2r)θ =sin θ cos 2rθ+cos θ sin 2rθ+sin θ cos 2rθ−cos θ sin 2rθ =2 cos 2rθ sin θ Σ_(r=1) ^n cos 2rθ sin θ=Σ_(r=1) ^n (1/2){f(r)−f(r−1)} =(1/2){f(n)−f(0)} =(1/2){sin (1+2n)θ−sin θ} =(1/2){sin θ cos 2nθ+cos θ sin 2nθ−sin θ} =(1/2){−sin θ 2 sin^2 nθ+2 cos θ sin nθ cos nθ} ={−sin θ sin nθ+cos θ cos nθ }sin nθ =cos (n+1)θ sin nθ

$${f}\left({r}\right)=\mathrm{sin}\:\left(\mathrm{1}+\mathrm{2}{r}\right)\theta \\ $$$${f}\left({r}−\mathrm{1}\right)=\mathrm{sin}\:\left(\mathrm{2}{r}−\mathrm{1}\right)\theta=−\mathrm{sin}\:\left(\mathrm{1}−\mathrm{2}{r}\right)\theta \\ $$$${f}\left({r}\right)−{f}\left({r}−\mathrm{1}\right)=\mathrm{sin}\:\left(\mathrm{1}+\mathrm{2}{r}\right)\theta+\mathrm{sin}\:\left(\mathrm{1}−\mathrm{2}{r}\right)\theta \\ $$$$=\mathrm{sin}\:\theta\:\mathrm{cos}\:\mathrm{2}{r}\theta+\mathrm{cos}\:\theta\:\mathrm{sin}\:\mathrm{2}{r}\theta+\mathrm{sin}\:\theta\:\mathrm{cos}\:\mathrm{2}{r}\theta−\mathrm{cos}\:\theta\:\mathrm{sin}\:\mathrm{2}{r}\theta \\ $$$$=\mathrm{2}\:\mathrm{cos}\:\mathrm{2}{r}\theta\:\mathrm{sin}\:\theta \\ $$$$\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{cos}\:\mathrm{2}{r}\theta\:\mathrm{sin}\:\theta=\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{2}}\left\{{f}\left({r}\right)−{f}\left({r}−\mathrm{1}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{{f}\left({n}\right)−{f}\left(\mathrm{0}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{sin}\:\left(\mathrm{1}+\mathrm{2}{n}\right)\theta−\mathrm{sin}\:\theta\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{sin}\:\theta\:\mathrm{cos}\:\mathrm{2}{n}\theta+\mathrm{cos}\:\theta\:\mathrm{sin}\:\mathrm{2}{n}\theta−\mathrm{sin}\:\theta\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{−\mathrm{sin}\:\theta\:\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:{n}\theta+\mathrm{2}\:\mathrm{cos}\:\theta\:\mathrm{sin}\:{n}\theta\:\mathrm{cos}\:{n}\theta\right\} \\ $$$$=\left\{−\mathrm{sin}\:\theta\:\mathrm{sin}\:{n}\theta+\mathrm{cos}\:\theta\:\mathrm{cos}\:{n}\theta\:\right\}\mathrm{sin}\:{n}\theta \\ $$$$=\mathrm{cos}\:\left({n}+\mathrm{1}\right)\theta\:\mathrm{sin}\:{n}\theta \\ $$

Commented by Rio Michael last updated on 12/May/20

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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