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Question

For the function $$\sin\pi x$$ centred at $$a=0.5$$.using taylor series expansion,find approximate value of $$\sin\left(\dfrac{\pi}{2} + \dfrac{\pi}{10} \right)$$


A
0.9511
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B
0.9633
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C
0.8962
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D
0.2134
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Solution

The correct option is A $$0.9511$$
The taylor series is given by $$\sum_{k=0}^{\infty}\dfrac{f^{(k)}(a)}{k!}(x-a)^k$$
$$f(x)\approx\sum_{k=0}^{n}\dfrac{f^{(k)}(a)}{k!}(x-a)^k=\sum_{k=0}^{2}\dfrac{f^{(k)}(a)}{k!}(x-a)^k$$
Finally after simplifying, we get
$$f(x)\approx\dfrac{1}{0!}\left (x-\dfrac{1}{2}\right)^0+\dfrac{0}{1!}\left (x-\dfrac{1}{2}\right)^1+\dfrac{-(\pi)^2}{2!}(x-\dfrac{1}{2})^2$$ 
$$f(x)=1-\dfrac{(\pi)^2}{2}(x-\dfrac{1}{2})^2$$
$$\sin(\pi x)=1-\dfrac{\pi^2}{2}\left (x-\dfrac{1}{2}\right)^2$$
$$\sin\left (\pi \left (\dfrac{1}{2}+\dfrac{1}{10}\right)\right)=1-\dfrac{\pi^2}{2}\left (\dfrac{1}{2}+\dfrac{1}{10}-\dfrac{1}{2}\right)^2=0.95065\approx0.951$$

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