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May 28th, 2008, 04:50 AM   #1
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Why sin(x) expansion needs x to be mentioned in radians?

I understand that in sin(x) we can express the angle x in any manner, say radians or degrees. for example sin(90) = sin(PI/2) = 1, because in both cases 90 or PI/2 mean right angle.

Similarly, sin(180 degrees) = sin(PI) = 0, because in both cases we mean to say that it is sine of straight angle.

But, when we see the Taylor series expansion of sin(x), i.e.

x - x^3 / 3! + x^5 / 5! - ....

we find that only the radian values provide the correct values.

So, just to convince myself, I wanted to replace x in the above series with PI and wanted to get the sum to be 0. But I could not do it.

Could somoone please help me to prove that

x - x^3 / 3! + x^5 / 5! - .... = 0 when x = PI ?

Also, is there any concept involved during Taylor series expansion that clearly tells us that this series would yield correct values only for angles mentioned in radians?
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May 28th, 2008, 06:02 AM   #2
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Essentially all computational formulas use angles in radians. Degrees and gradients are just conveniences.
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May 28th, 2008, 07:43 AM   #3
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The value of sine and cosine come from an analysis of the unit circle, so r=1 unit. Of course, c=2π in this case. "radians" as a sort of quasi-unit come out of this. They aren't actually a unit, 1 radian = 1.
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June 1st, 2008, 07:03 AM   #4
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Re: Why sin(x) expansion needs x to be mentioned in radians?

Quote:
Originally Posted by conjecture
Could somoone please help me to prove that

x - x^3 / 3! + x^5 / 5! - .... = 0 when x = PI ?

Also, is there any concept involved during Taylor series expansion that clearly tells us that this series would yield correct values only for angles mentioned in radians?
The Taylor series (and other formulas involving trigonometric functions) work with radians instead of degrees because they were formulated to work with radians. Radians are simply a more mathematically simple way of describing angles; values in radians lend themselves to much simpler identities, etc.

As for your Taylor series, I think it's natural to start from how the series is derived in the first place to show you why it works.

Taylor's theorem says that you can approximate any function f(x) linearly around a point a as follows:

f(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)^2/2! + f'''(a)(x-a)^3/3! + ...

To approximate sin(x), we let f(x) = sin(x). Let's do it around a = 0. (Doing it around another a will get you a similar power series, but shifted. We do it around a = 0 for simplicity.)

we get sin(x) = sin(0) + cos(0)(x - 0) - sin(0)(x - 0)^2/2! - cos(0)(x - 0)^3/3! ...

Every term with a sin or -sin in it drops out, because sin(0) = 0. We're left with:

x - x^3/3! + x^5/5! - x^7/7! + ...

But remember that this is an infinite series, so the result is better approximated the more terms we have. With a finite number of terms, you're going to get something that resembles the true value of the function, but it's not going to be exact. This is reflected in the statement of Taylor's theorem, which states that a function is equal to a certain sum (the Taylor polynomial) plus a remainder. The remainder is your error.

A better way to see that your approximation of sin(pi) really gives you the answer you want would be to calculate it for increasingly more accurate (longer) Taylor polynomials and note that the result approaches sin(pi).
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June 1st, 2008, 08:02 AM   #5
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Re: Why sin(x) expansion needs x to be mentioned in radians?

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Originally Posted by karasuman
Taylor's theorem says that you can approximate any function f(x) linearly around a point a as follows:

f(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)^2/2! + f'''(a)(x-a)^3/3! + ...

To approximate sin(x), we let f(x) = sin(x). Let's do it around a = 0. (Doing it around another a will get you a similar power series, but shifted. We do it around a = 0 for simplicity.)

we get sin(x) = sin(0) + cos(0)(x - 0) - sin(0)(x - 0)^2/2! - cos(0)(x - 0)^3/3! ...

Every term with a sin or -sin in it drops out, because sin(0) = 0. We're left with:

x - x^3/3! + x^5/5! - x^7/7! + ...
What actually goes wrong here if the angle x is measured in degrees is that in this case the derivative of sin(x) is not cos but Pi/180*cos(x) (1/degree). Similarly d/dx[cos(x)]=-Pi/180*sin(x) (1/degree). (See http://au.answers.yahoo.com/question...9223349AAi6win)

One could also think about the unit of sin(x). If x is not measured in radians, the expression sin(x)=x-x^3/3!+x^5/5!... doesn't make much sense. (What does it mean to add different powers of the unit degree? Not much.) If one uses the right forms of the derivatives, which have the unit 1/degree, then the degrees disappear in the formula, and sin is dimensionless.
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June 1st, 2008, 09:02 AM   #6
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Right... My explanation of the Taylor series wasn't meant to address radians/degrees, but rather why he wasn't getting the correct value for sin(pi) using the series.
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June 1st, 2008, 09:41 AM   #7
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Quote:
Originally Posted by karasuman
Right... My explanation of the Taylor series wasn't meant to address radians/degrees, but rather why he wasn't getting the correct value for sin(pi) using the series.
And my explanation of the units wasn't meant to correct you, but to explain the thing with units. You just did good job with the taylor series so I copied it.

E: By the way: I wonder if it's possible to derive the taylor series without using the Taylor's theorem. (For example by somehow geometrically approximating "the opposite side".) That could be interesting
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June 2nd, 2008, 05:17 AM   #8
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If you use just six terms of the series and evaluate it for x = pi, you will get 0.000, correct to three decimal places.
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January 3rd, 2018, 05:17 AM   #9
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The series converges for |x|< pi/2.
For other angles, it should be transformed to make |x|< pi/2, using trigonometry
Reference: Computer oriented numerical methods by Rajaraman

Last edited by skipjack; January 3rd, 2018 at 09:20 PM.
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January 3rd, 2018, 05:21 AM   #10
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Quote:
Originally Posted by RajeshRjy View Post
The series converges for |x|< pi/2.
For other angles, it should be transformed to make |x|< pi/2, using trigonometry
Reference: Computer oriented numerical methods by Rajaraman
The series converges for all real numbers, not just the ones with $|x|<\pi/2$.

Last edited by skipjack; January 3rd, 2018 at 09:19 PM.
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