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 October 26th, 2017, 11:35 AM #1 Newbie   Joined: Oct 2017 From: Wales Posts: 3 Thanks: 0 Taylor series for sin(x) Hiya all, I've been told you can use the Taylor Series to compute functions of sin(x) without a calculator. I have managed to do so for x=61, by using x=61, a=60; however I've had some difficulty doing similarly with x=31, a=30. I would appreciate any help or suggestions!! Thank you Taylor Series I've been using: f(x) = f(a) + (f'(a)(x-a))/1! + (f''(a)(x-a)^2)/2! + ....
October 26th, 2017, 12:06 PM   #2
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 Originally Posted by Yratomanhygoel Hiya all, I've been told you can use the Taylor Series to compute functions of sin(x) without a calculator. I have managed to do so for x=61, by using x=61, a=60; however I've had some difficulty doing similarly with x=31, a=30. I would appreciate any help or suggestions!! Thank you Taylor Series I've been using: f(x) = f(a) + (f'(a)(x-a))/1! + (f''(a)(x-a)^2)/2! + ....
$\displaystyle sin(31) \approx sin(30) + \frac{cos(30)}{1!} \cdot (1) - \frac{sin(30)}{2!} \cdot 1^2 - \frac{cos(30)}{3!} \cdot (1^3)$

$\displaystyle sin(31) \approx \left ( \frac{1}{2} \right ) + \left ( \frac{\sqrt{3}}{2}\right ) - \left ( \frac{1}{2} \cdot \frac{1}{2} \right ) - \left ( \frac{1}{6} \cdot \frac{\sqrt{3}}{2} \right )$

Does that help?

October 26th, 2017, 01:16 PM   #3
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 Originally Posted by topsquark $\displaystyle sin(31) \approx sin(30) + \frac{cos(30)}{1!} \cdot (1) - \frac{sin(30)}{2!} \cdot 1^2 - \frac{cos(30)}{3!} \cdot (1^3)$ $\displaystyle sin(31) \approx \left ( \frac{1}{2} \right ) + \left ( \frac{\sqrt{3}}{2}\right ) - \left ( \frac{1}{2} \cdot \frac{1}{2} \right ) - \left ( \frac{1}{6} \cdot \frac{\sqrt{3}}{2} \right )$ Does that help?
It confirms that what I did was logical; unfortunately calculating it like that yields and answer of about 1.13 - whereas the true value of sin(31) = 0.515. I cannot see any mistake made in using the formula, as it works perfectly for sin(61), any ideas??

Thank you

October 26th, 2017, 01:19 PM   #4
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Quote:
 Originally Posted by Yratomanhygoel It confirms that what I did was logical; unfortunately calculating it like that yields and answer of about 1.13 - whereas the true value of sin(31) = 0.515. I cannot see any mistake made in using the formula, as it works perfectly for sin(61), any ideas?? Thank you
Haven't followed the calculations in detail (or at all) but I wonder if you're talking radians or degrees.

October 26th, 2017, 01:29 PM   #5
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 Originally Posted by Maschke Haven't followed the calculations in detail (or at all) but I wonder if you're talking radians or degrees.
Bingo! Good call.

-Dan

October 26th, 2017, 01:49 PM   #6
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Quote:
 Originally Posted by Maschke Haven't followed the calculations in detail (or at all) but I wonder if you're talking radians or degrees.
It's always the little things isn't it?? I can't believe I overlooked that, thank you - it works now, though I couldn't say why.

Thank you both for your help!!!

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