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Question Number 174327 by aminee last updated on 30/Jul/22

	Answered by Mathspace last updated on 31/Jul/22
	$we have sin(a+b)=sina.cosb+cosasinb sin(a−b)=sinacosb−cosasinb ⇒ 2sina.cosb=sin(a+b)+sin(a−b) ⇒ sina.cosb=(1/2){sin(a+b)+sin(a−b)} ⇒Σ_(n=1) ^∞ sin((1/2^(n+1) ))cos((3/2^(n+1) )) =(1/2)Σ_(n=1) ^∞ sin((4/2^(n+1) ))−(1/2)Σ_(n=1) ^∞ sin((2/2^(n+1) )) =(1/2)Σ_(n=1) ^∞ sin((2/2^n ))−(1/2)Σ_(n=1) ^∞ sin((1/2^n )) =(1/2){Σ_(n=1) ^∞ sin((1/2^(n−1) ))−Σ_(n=1) ^∞ sin((1/2^n ))} =(1/2)lim_(n→+∞) Σ_(k=1) ^n (sin((1/2^(k−1) ))−sin((1/2^k ))} =(1/2)lim_(n→+∞) {sin(1)−sin((1/2)) +sin((1/2))−sin((1/4))+...sin((1/2^(n−1) ))−sin((1/2^n ))} =(1/2)lim_(n→+∞) {sin(1)−sin((1/2^n ))} =(1/2)sin(1)( en rad)$
	$w e h a v e s i n (a + b) = s i n a . c o s b + c o s a s i n b$ $s i n (a - b) = s i n a c o s b - c o s a s i n b \Rightarrow$ $2 s i n a . c o s b = s i n (a + b) + s i n (a - b) \Rightarrow$ $s i n a . c o s b = \frac{1}{2} {s i n (a + b) + s i n (a - b)}$ $\Rightarrow \sum_{n = 1}^{\infty} s i n (\frac{1}{2^{n + 1}}) c o s (\frac{3}{2^{n + 1}})$ $= \frac{1}{2} \sum_{n = 1}^{\infty} s i n (\frac{4}{2^{n + 1}}) - \frac{1}{2} \sum_{n = 1}^{\infty} s i n (\frac{2}{2^{n + 1}})$ $= \frac{1}{2} \sum_{n = 1}^{\infty} s i n (\frac{2}{2^{n}}) - \frac{1}{2} \sum_{n = 1}^{\infty} s i n (\frac{1}{2^{n}})$ $= \frac{1}{2} {\sum_{n = 1}^{\infty} s i n (\frac{1}{2^{n - 1}}) - \sum_{n = 1}^{\infty} s i n (\frac{1}{2^{n}})}$ $= \frac{1}{2} {l i m}_{n \to + \infty} \sum_{k = 1}^{n} (s i n (\frac{1}{2^{k - 1}}) - s i n (\frac{1}{2^{k}})}$ $= \frac{1}{2} {l i m}_{n \to + \infty} {s i n (1) - s i n (\frac{1}{2})$ $+ s i n (\frac{1}{2}) - s i n (\frac{1}{4}) + . . . s i n (\frac{1}{2^{n - 1}}) - s i n (\frac{1}{2^{n}})}$ $= \frac{1}{2} {l i m}_{n \to + \infty} {s i n (1) - s i n (\frac{1}{2^{n}})}$ $= \frac{1}{2} s i n (1) (e n r a d)$
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