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How to calculate

$$\sin(37°)$$

with a Taylor approximation accurate to 3 decimal digits?

I know it is not a difficult question, but I have no answers of my book and so far I have only determined the Taylor approach of the umpteenth order and not approached or calculated anything in any sense.

Xander Henderson
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No. iD
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    HINT: Convert to radians. $30^\circ=\pi/6$ and $7^\circ \approx \pi/24$. Take the taylor expansion of sine around $\pi/6$. – K.defaoite Oct 28 '20 at 12:21
  • @K.defaoite At that point you’re better just using the exact value of $\sin30°$ as an approximation of $\sin37°$ than using a Taylor approximation of $\sin30°$. – gen-ℤ ready to perish Oct 28 '20 at 12:50
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    @gen-ℤ ready to perish, K.defaoite said to approximate around $\pi/6$, not to approximate $\sin (\pi/6)$. – Ennar Oct 28 '20 at 13:11
  • Yeah, and you’d get a closer answer using the exact value of $\sin(π/6)$ – gen-ℤ ready to perish Oct 28 '20 at 13:13
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    @gen-ℤ ready to perish, it looks like you don't understand what the Taylor expansion around a point $a$ means. It means $f(x) = f(a) + f'(a)(x-a) + f''(a)/2, (x-a)^2 + \ldots$, which obviously assumes that you know the exact value of $f^{(n)}(a)$, for all practical purposes. – Ennar Oct 28 '20 at 13:16
  • @Ennar I know how a Taylor series works. We’re obviously on different pages, and I was willing to explain what I was trying to say, but not if you’re going to insult my intelligence. – gen-ℤ ready to perish Oct 28 '20 at 14:06
  • @gen-ℤ ready to perish, no insult insult is intended, especially not to your intelligence. Sincere apologies if that's how you received my last comment. However, we do discuss mathematics here, and it would be much appreciated if you explained what you meant in greater detail, concretely why do you think that the Taylor expansion of $\sin$ around $a= \pi/6$ wouldn't use exact values of $\sin(\pi/6)$ and $\cos(\pi/6)$? – Ennar Oct 29 '20 at 06:34

6 Answers6

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The easiest approach may be to use the Taylor series $\sin x=x-{1\over6}x^3+{1\over120}x^5-\cdots$, since $37^\circ=37\pi/180\approx0.646\lt1$:

$$\sin(37^\circ)=\sin\left(37\pi\over180\right)\approx\left(37\pi\over180\right)-{1\over6}\left(37\pi\over180\right)^3+{1\over120}\left(37\pi\over180\right)^5$$

(noting that the next term in the alternating sum is considerably less than $1/5040\approx0.0002$). I wouldn't want to complete the decimal calculation by hand, but it's relatively straightforward with a calculator, even if you have to use an approximation like $\pi\approx3.1416$ on a pocket calculator that lacks a button for $\pi$ and only does arithmetic.

Barry Cipra
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    I think this is the best way to do it. You get sufficient accuracy. There isn't a need to center the taylor series around another angle. – Ameet Sharma Oct 28 '20 at 13:05
  • @BarryCipra The magnitude of the next term does not provide an error bound, unless you use some extra information. – PierreCarre Oct 28 '20 at 13:34
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    @PierreCarre, for an alternating sum (with decreasing terms) it does. So if $0\le x\le1$, you can use the next term in the Taylor series for $\sin x$ as an upper bound on the error. Is that the extra information you have in mind? – Barry Cipra Oct 28 '20 at 13:44
  • Yes @barryCipra, that was it. We never know how people reading the post in the future may extrapolate the result to other situations! – PierreCarre Oct 28 '20 at 13:46
  • @PierreCarre, good point. Let's hope future readers read the comments! – Barry Cipra Oct 28 '20 at 13:51
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Hint:

  • $ \sin (a+x) = \sin(a) + x \cos(a) + O(x^2) $

  • $ a = 36^\circ = 2 \pi /10 $

  • $ x = 1^\circ = 2 \pi /360 \approx 0.0175 $

  • $ \sin(a) = \frac{\sqrt{10 - 2\sqrt{5}}}{4}$, $\cos(a)=\frac{1+\sqrt{5}}{4}$ (see Wikipedia)

lhf
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For $\alpha=\frac{37\cdot\pi}{180}=0.65477$ we take 3 terms in the Maclaurin series: $$ \sum_{k=0}^2 \frac{(-1)^k 0.64577^{1 + 2 k}}{(1 + 2 k)!}=0.64577-\frac{0.65477^3}{6}+\frac{0.65477^5}{120}=0.601823 $$ This is as close as $10^{-5}$ to the exact value.

am301
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Expand around $\pi/5$ $$\sin x = \sin(\pi/5)+\cos(\pi/5) \left(x-\frac{\pi }{5}\right)-\frac{1}{2} \sin(\pi/5) \left(x-\frac{\pi }{5}\right)^2+\\+\frac{1}{6} \cos(\pi/5) \left(x-\frac{\pi }{5}\right)^3+\frac{1}{24} \sin(\pi/5)\left(x-\frac{\pi }{5}\right)^4+O\left(x^5\right)$$ You get $0.601$ plugging $x=37/180 \pi$

Raffaele
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Using Taylor's poloynomial with degree $p$ around $a=\frac{\pi}{6}$, the error is bounded by $$ |R_p| =\left| \dfrac{f^{(p+1)}(\xi)}{(p+1)!}\left(\frac{37 \pi}{180}-\frac{\pi}{6}\right)^{p+1}\right|\leq \frac{1}{(p+1)!} \left(\frac{7 \pi}{180}\right)^{p+1} $$ so, just select a convenient $p$ and use the corresponding Taylor polynomial to get the approximation.

For $p=3$ you get na error bound of $ 0.928304 \times 10^{-5}$ and for $p=4$ na error bound of $0.226828\times 10^{-6}$

PierreCarre
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I agree with the other answers that are already provided. However, I think that there is a possible source of confusion that should be resolved.

In Trigonometry/Analytical Geometry, the domain of the sine and cosine functions are angles, which have a unit of measurement, the degree. So $37^{\circ}$ is a dimensioned number in the same way that 1 foot is a dimensioned number.

In Calculus/Real Analysis, this (necessarily) changes. The domain of the sine and cosine functions are dimensionless real numbers (i.e. $\pi/4$ rather than $45^\circ$). The reason for this alteration in the sine and cosine functions is to facilitate (for examples)

  • Using the Taylor series of the sine and cosine functions to attack problems

  • Using the ArcTan function to attack a problem like $\int_0^1 \frac{dx}{1 + x^2}.$

Some of the confusion centers around ambiguity in the connotation of the term radian. That is, does $(\pi/4)$ radians represent a dimensioned number, with 1 radian representing a unit of measurement of an angle similar to what 1 degree represents? This connotation is often useful in allowing the Trigonometry/Analytical Geometry student to transition into Calculus/Real Analysis and immediately begin solving Calculus problems without requiring that the student first acquire a deep understanding of the theory.

The alternative connotation of 1 radian, is that it is used to represent the dimensionless proportion of a specific arc length of a portion of the unit circle. This proportion (i.e. ratio) is taken against the arc length of 1 complete revolution of the unit circle (i.e. the circumference of the unit circle). The circumference of the unit circle is $2\pi.$

Therefore, the term $(\pi/4)$ radians for example, can be interpreted in two totally distinct ways:

  1. equivalent to $45^\circ$.

  2. representing the arc length of 1/8-th of a complete revolution around the unit circle. This interpretation preserves the idea that $(\pi/4)$ radians (for example) is a dimensionless number.

Often, when a transitioning student uses the term radians, they are intending the first connotation above. When someone experienced in Calculus/Real Analysis uses the term radians, they are intending the second connotation above.

For an illustration of the sine and cosine functions having as their domain, dimensionless numbers, see What is the physical meaning of sine, cosine and tangent of an obtuse angle?.

user2661923
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  • @ryang Why? How is the expression $37^\circ$ any different than the expression $1$ inch? Just as an inch has a measurable length, a degree represents a measurable angle. It is true that you can construe a degree as $\displaystyle \frac{1}{360}$ of a complete revolution, but a complete revolution is itself a measurable angle. – user2661923 Oct 28 '21 at 05:14
  • @ryang Also, when attacking the problem of $\displaystyle \int_0^1 \frac{dt}{1 + t^2} = \text{Arctan}(1) - \text{Arctan}(0) = (\pi/4) - 0 = \pi/4$, notice what happens if you instead use short and tall rectangles to estimate $\displaystyle \int_0^1 \frac{dt}{1 + t^2}$. You find that the area converges to the dimensionless value of $(\pi/4)$. Note that if instead, you evaluate $\text{Arctan}(1) - \text{Arctan}(0)$ as $45^\circ - 0^\circ = 45^\circ$, you get a dimensioned (and nonsensical) answer, rather than the dimensionless answer of $(\pi/4)$. – user2661923 Oct 28 '21 at 05:20
  • @ryang As another example where construing a degree as dimensionless leads to trouble, consider that the $\displaystyle \lim_{x \to 0} \frac{\sin(x)}{x} = 1$. This is based on the idea that the domain of the sine function is the dimensionless arc length of the unit circle. If instead, you regard the domain of the sine function as an angle, measured in degrees, where a degree is supposedly dimensionless, then you have that the $\displaystyle \lim_{x \to 0} \frac{\sin(x)}{x} = \frac{\pi}{180}$. – user2661923 Oct 28 '21 at 05:28
  • $270^\circ$ and $4.71\textrm{ rad}$ are specifications (using different units) of the same dimensionless physical quantity whose SI base unit is $1.$ To be clear, I'm merely (pedantically?) distinguishing between "dimensions" and "units": https://math.stackexchange.com/a/4289650/21813 – ryang Oct 28 '21 at 05:59
  • @ryang We are going to have to agree to disagree here. – user2661923 Oct 28 '21 at 06:02
  • Yes my above comments are actually premised on that. -) – ryang Oct 28 '21 at 06:03