find the maclaurin series expension for the function f(x) = sin^2 x ; x


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Question Number 158469 by zakirullah last updated on 04/Nov/21

$f i n d t h e m a c l a u r i n s e r i e s e x p e n s i o n$ $f o r t h e f u n c t i o n f (x) = {s i n}^{2} x; x_{o} = 0$
	Answered by TheSupreme last updated on 04/Nov/21

	$f (x) = {s i n}^{2} (x)$ $f^{'} (x) = 2 s i n (x) c o s (x) = s i n (2 x)$ $f (x) = \underset{i = 1}{\sum^{\infty}} D^{i} (s i n (2 x)) \frac{x^{i + 1}}{i + 1!}$ $D^{i} (s i n (2 x)) = . . .$ $i = 2 n$ $D^{2 n} {(s i n (2 x))}_{x = 0} = 0$ $i = 2 n + 1$ $D^{2 n + 1} (s i n (2 x)) = {(- 1)}^{n} 2^{n + 1} c o s (2 x) = 2 {(- 2)}^{n}$ $f (x) = Σ \frac{{(- 2)}^{n} 2 x^{2 n + 2}}{(2 n + 2)!}$
	Commented by zakirullah last updated on 04/Nov/21

	$G r e a t s i r G$ $S i r h o w 2 s i n (x) c o s (x) = s i n (2 x)$
	Commented by MJS_new last updated on 05/Nov/21

	$2 \sin x \cos x = 2 \times \frac{e^{i x} - e^{- i x}}{2 i} \times \frac{e^{i x} + e^{- i x}}{2} =$ $[(a - b) (a + b) = a^{2} + b^{2}]$ $= \frac{e^{2 i x} - e^{- 2 i x}}{2 i} = \sin 2 x$
	Commented by zakirullah last updated on 05/Nov/21

	$o k s i r {y o u}^{'} r t h e g r e a t$
	Commented by MJS_new last updated on 05/Nov/21

	$I^{'} m just a tiny snip$
	Answered by mr W last updated on 04/Nov/21

	$w e k n o w \cos x = \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n} x^{2 n}}{(2 n)!} .$ $f (x) = \sin^{2} x = \frac{1 - \cos (2 x)}{2}$ $f (x) = \frac{1}{2} - \frac{1}{2} \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n} {(2 x)}^{2 n}}{(2 n)!}$ $f (x) = \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n + 1} 2^{2 n - 1} x^{2 n}}{(2 n)!}$
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