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    Hi guys, struggling with this question. . .

    use the first 3 terms of the Taylor series expansion of a function about pi/6 to approximate to 3 d.p. the value of sin(32 degrees).

    I have the answer but i am not getting it right. Would someone be nice enough to show me step by step?

    Thanks x
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    (Original post by zoe_bartlett)
    Hi guys, struggling with this question. . .

    use the first 3 terms of the Taylor series expansion of a function about pi/6 to approximate to 3 d.p. the value of sin(32 degrees).

    I have the answer but i am not getting it right. Would someone be nice enough to show me step by step?

    Thanks x
    Post your working?
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    (Original post by BabyMaths)
    Post your working?
    My workings are wrong though

    i know the first step should be getting me to .5 + sq(3)/2x - 1/4x^2

    so far I have differentiated sin(x) 4 times but then calculated the value of sin(x) etc by inputing pi/6 as the value of x which I now think is wrong. But I am at a loss how else I should use the formula :-S it al seemed so simple when we first learnt it. Sigh!

    Any help would be appreciated. Even if its just a nudge in the right direction.
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    (Original post by zoe_bartlett)
    .5 + sq(3)/2x - 1/4x^2
    If you obtained that yourself then you've done the hardest part. On the other hand if you want help getting to that point let us know.

    To use it to find sin(32degrees) first convert 32 degree to radians and then sub it in.
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    I am fine with the differentiation, it's just plugging it back in to the formula (what is supposed to be the simple bit!). I am not sure what I should be putting where.

    i have the formula, f(x+h)=f(x)+h*f'(x) + f''(x)*h^2/2

    my confusion is is f(x) just sin(32) and then its derivatives etc and the h just pi/6?

    But then my answer is not in the format A+Bx + Cx^2.

    sorry if this is a stupid question. It seems like it should best simple.
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    You want x = \pi/6, \quad x+h = 32^{o}, \implies h = \pi / 90.

    Note that you must work in radians, or you will end up in a world of pain.
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    Perhaps you're mixing two slightly different ways of doing this..

    I'd be thinking f(x) \approx f(a) + f'(a)(x-a)+\frac{f''(a)(x-a)^2}{2}

    a=\frac{\pi}{6}

    f(x)=\sin x so f(x)=\sin \frac{\pi}{6}=\frac{1}{2}

    f'(x)=\cos x so f(x)=\cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}

    f''(x)=-\sin x so f''(a)=-\sin \frac{\pi}{6}=-\frac{1}{2}

    Sub it all in and use x=32^\circ = \frac{32 \pi}{180} \text{ radians}.
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    Thank you so much for this. Going to redo the question now and hopefully it all makes sense


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    Did the question. I got 0.529 but my answer says 0.034. Can you let me know which is right?

    Thanks


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    sin(32 degrees) is a little more than 0.5
 
 
 
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