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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 + 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