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


Question and Answers Forum

All Questions Topic List
Others Questions
Previous in All Question Next in All Question
Previous in Others Next in Others
Question Number 93368 by Rio Michael last updated on 12/May/20

$given that f (r) = \sin (1 + 2 r) θ$ $show that f (r) - f (r - 1) = 2 \cos 2 r θ \sin θ$ $hence show that$ $\underset{r = 1}{\sum^{n}} \cos 2 r θ \sin θ = \cos (n + 1) θ \sin n θ$
	Answered by mr W last updated on 12/May/20
	$f(r)=sin (1+2r)θ f(r−1)=sin (2r−1)θ=−sin (1−2r)θ f(r)−f(r−1)=sin (1+2r)θ+sin (1−2r)θ =sin θ cos 2rθ+cos θ sin 2rθ+sin θ cos 2rθ−cos θ sin 2rθ =2 cos 2rθ sin θ Σ_(r=1) ^n cos 2rθ sin θ=Σ_(r=1) ^n (1/2){f(r)−f(r−1)} =(1/2){f(n)−f(0)} =(1/2){sin (1+2n)θ−sin θ} =(1/2){sin θ cos 2nθ+cos θ sin 2nθ−sin θ} =(1/2){−sin θ 2 sin^2 nθ+2 cos θ sin nθ cos nθ} ={−sin θ sin nθ+cos θ cos nθ }sin nθ =cos (n+1)θ sin nθ$
	$f (r) = \sin (1 + 2 r) θ$ $f (r - 1) = \sin (2 r - 1) θ = - \sin (1 - 2 r) θ$ $f (r) - f (r - 1) = \sin (1 + 2 r) θ + \sin (1 - 2 r) θ$ $= \sin θ \cos 2 r θ + \cos θ \sin 2 r θ + \sin θ \cos 2 r θ - \cos θ \sin 2 r θ$ $= 2 \cos 2 r θ \sin θ$ $\underset{r = 1}{\sum^{n}} \cos 2 r θ \sin θ = \underset{r = 1}{\sum^{n}} \frac{1}{2} {f (r) - f (r - 1)}$ $= \frac{1}{2} {f (n) - f (0)}$ $= \frac{1}{2} {\sin (1 + 2 n) θ - \sin θ}$ $= \frac{1}{2} {\sin θ \cos 2 n θ + \cos θ \sin 2 n θ - \sin θ}$ $= \frac{1}{2} {- \sin θ 2 \sin^{2} n θ + 2 \cos θ \sin n θ \cos n θ}$ $= {- \sin θ \sin n θ + \cos θ \cos n θ} \sin n θ$ $= \cos (n + 1) θ \sin n θ$
	Commented by Rio Michael last updated on 12/May/20

	$thank you sir$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum