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How to find the approximation to $\sin(1.58)$ ? By using the Newton's method

$$x_{n+1} = x_{n} - \frac{f(x)}{f'(x)}$$

You always will get $0$. Using this method:

$f(x+\Delta x) \approx f(x) + f'(x)dx$

You get $\sin(1.58) \approx 0.5974$ , but that's not very precise.

So, is there a method to precisely approximate $\sin(x)$?

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Which $f$ are you using in Newton's method? Which starting point? –  lhf Aug 8 '11 at 10:51
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Precisely approximate? –  Peter Taylor Aug 8 '11 at 10:53
    
@lhf I start with $x_0 = 0$ –  Cobold Aug 8 '11 at 10:54
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The Newton Method is a method for solving equations approximately. So it is not directly applicable to function evaluation. There are workarounds. For example, you can approximate $\sqrt{2}$ by approximating the positive root of $x^2-2=0$ using the Newton Method. For your second method, you need to start from an $x$ which is near your target value. Since $1.58\approx \pi/2$, and since you know $\sin(\pi/2)$, the tangent line approximation would, in this case, give you the approximation $1$, since $\cos(\pi/2)=0$. That's reasonably good. –  André Nicolas Aug 8 '11 at 11:07
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4 Answers 4

up vote 12 down vote accepted

Notice that 1.58 is very close to $\pi/2 \approx 1.57079633$

So you can use the Taylor expansion,

$$\sin(x+\delta x) \approx \sin(x) + \cos(x)\delta x - \tfrac{1}{2} \sin(x) \delta x^2$$

with $x=\pi/2$ and $\delta x = 1.58 - \pi/2 \approx 0.00920367321$

This gives

$$\sin(1.58) \approx \sin(\pi/2) + \cos(\pi/2) \delta x - \tfrac{1}{2}\sin(\pi/2) \delta x^2 = 0.999957646$$

compared to the exact value

$$\sin(1.58) = 0.999957646\dots$$

so it is in exact agreement, up to the accuracy provided by Google calculator. The actual error will be of the order $\delta x^4$, or around $10^{-8}$.

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You didn't write $\frac{1}{2}$ before the last term so the approximation is the same as the real answer –  Cobold Aug 8 '11 at 11:08
    
Oops, fixed! Thanks. –  Chris Taylor Aug 8 '11 at 12:27
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A question like the one that you asked has to be looked at in context. If you just happen to need the value of $\sin(1.58)$, calculators or computer programs reliably produce an answer of greater than needed accuracy in any real-world application.

If the context is a test or homework problem, the situation is different. There will be a body of ideas and techniques that have been learned within the term, and you are expected to show some command of these.

In a test or homework context, the number $1.58$ is too conveniently close to $\pi/2$ for this to be an accident. Thus, if tangent line (linear) approximation or its generalization to approximation by higher degree polynomials is part of the course, that is likely what you are intended to use.

You quoted almost correctly the linear approximation formula $$f(x+\Delta x)\approx f(x)+f'(x) \Delta x.$$

It can be useful to write this as $f(a+\Delta x) \approx f(a)+f'(a) \Delta x$ to emphasize that $a$ is, at least temporarily, a constant, known. So the formula says that if we know $f(a)$ and $f'(a)$, we can approximate $f$ "near" $a$ in a simple way. Informally, it says that under high magnification, the graph of $y=f(x)$ near $x=a$ looks very much like a straight line.

The actual calculation is easy. We choose $a=\pi/2$. Then $f(a)=1$ and $f'(a)=0$, so if if $\Delta x$ is close to $0$, then $\sin(\pi/2 +\Delta x)\approx 1$.

The fact that the answer is nearly $1$ should be clear from the shape of the curve $y=f(x)$ near $x=\pi/2$. The curve is nearly flat there, so a little change in $x$ produces nearly no change in $\sin x$.

If the recent background to the question includes polynomial approximations of higher degree than $1$ (Taylor polynomials), then presumably that is what you are intended to use. Ordinarily, when on a test or homework it is intended that a Taylor polynomial be used, the degree is specified.

Comment: The problem of approximating $\sin(1.58)$ has an artificial appearance, since after all we have devices that produce excellent approximations. However, the idea of linear approximation is an essential part of the theory. It is also a key factor in many of the applications of the differential calculus.

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Hint: You can naturally use Taylor series. It consists in a polynomial approximation of any "reasonable" function around a point. Depending on how many terms you consider in the expansion, you can even obtain a numerical bound of the error occuring by this method. I leave you the care to calculate Taylor series of $sin(x)$ considering it can be a good exercise to become familiar with this method.

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$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \ldots$ which converges for all $x$.

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So can you show the steps to approximate $sin(1.58)$ –  Cobold Aug 8 '11 at 10:55
    
@Cobold What is the expression for the remainder term in Taylor's formula? Use that to estimate $\sin(1.58)$. –  user38268 Aug 8 '11 at 12:13
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