By Katharina JüRges (P2657) on Saturday, June 24, 2000 - 01:49 pm:
Dear Nrich team,
My name's Katharina and I'm 17 years old. I go to school in France
and the equivalent of the class I attend after the summer holidays
is the upper sixth form class, if this is the last class before
A-levels.
I've already encountered trigonometry at school, even quite often
but I've never come across a method to calculate the exact value of
an angle when I have the sine, cosine or tangent, or vice versa:
the sine, cosine or tangent with the value of the angle.
Does such a method exist? Please send me any idea you have about
this subject.
Thank you in advance.
Katharina
By Simon Munday (Sjm78) on Saturday, June 24, 2000 - 05:46 pm:
I don't think there is any general way to
evaluate the trigonometric functions, or their inverses, exactly.
In fact, there's no reason to suppose that any particular value of
a trigonometric function is expressible exactly in terms of things
like rational numbers or roots, or any other types of real number
that you can write down exactly.
However, there are some special cases that you can deduce by
looking at special sorts of triangles. For example, take an
equilateral triangle and divide it into two congruent right-angled
triangles. You know the interior angles of these triangles and
their side lengths, so you can work out the sines and cosines of
the angles. The other thing you can do is use trigonometric
identities such as
sin(A+B) = sin A cos B + cos A sin B
to extend such results.
The most important exact values associated with these, and indeed
any, functions are their zeros, ie the values x for which f(x) = 0.
Make sure you know what these are for all the trig functions.
I can't see any reason why you would want to evaluate trig
functions in general. If you were a mathematician, you would be
perfectly happy to leave a number as, say, sin p/6 (if indeed you actually used a specific number
like that, which is unlikely). On the other hand, if you wanted to
know the value of a trig function for some practical purpose (as an
engineer might), the first few decimal places of the value would be
much more useful to you that some complicated exact
expression.
In summary, trig functions can be evaluated exactly in a few
special cases, but no one ever wants their exact numerical value.
If I were you, I wouldn't worry about it.
By Sean Hartnoll (Sah40) on Saturday, June 24, 2000 - 06:03 pm:
For non-standard cases (i.e. 30,60,90
degrees etc.), one would usually use a calculator.
To get an approximate values, you could use Taylor expansions for
the trigonometric functions and inverse trigonometric functions
(this is probably what the calculator is doing):
sin(x) = x - x3/3! + x5/5! + ...
close to x=0 (x in radians here), there are similar expansions for
cos, tan etc. I can write about Taylor expansions if you
want.
As has been said above, the standard cases above can be done using
simple triangles that have some symmetry in them. E.g. use an
equilateral triangle to find sin(60).
Sean
By Katharina JüRges (P2657) on Monday, June 26, 2000 - 07:28 pm:
Thank you both, Simon and Sean,
I agree that numbers like sin p/6, as
you said Simon, are really sufficient as a final result. I was just
curious and I couldn't ask a teacher because I already have summer
holidays.
I am very happy that you answered me so quickly. I'd like to know
more about Taylor expansions. I've already heard the name but I
don't know what it means, so please write me about it.
Katharina
By Katharina JüRges (P2657) on Monday, June 26, 2000 - 08:16 pm:
I would like to add something.
I stumbled over the notation tan(x)²:
Is this the tangent of x² or the square of the tangent of
x?
I'm not sure because I don't know the English notations very
well.
Thank you in advance.
Katharina
By David Loeffler (P865) on Monday, June 26, 2000 - 10:55 pm:
Well, I suppose that in general one would assume that it meant
(tan(x))2, but this would be normally written as
tan2(x). I can't help thinking that the trig functions
of squares of angles are so overwhelmingly weird that whenever
they're used it'd be made very clear!
Incidentally, what use ARE these functions such as
sin(x2)? I heard somewhere that they were used in
optics, and I once came across them constructing a curve of
linearly increasing curvature, but is there any other application
for them?
David Loeffler
By Simon Munday (Sjm78) on Wednesday, June 28, 2000 - 03:00 pm:
Katharina,
You have to be careful when writing squares of functions. There are
various ambiguities. For example, tan2 x could be
interpreted as tan (tan x), when in fact it means the same thing as
(tan x)2. In fact the former notation is almost always
used for the square (or other powers) of trig functions. Also, tan
x2 should be taken to mean tan (x2) rather
than (tan x)2, although it is common in this case to
make things clearer by writing it as tan (x2). I hope
that clears that up.
You also asked about Taylor series. Let me give an example. The
Taylor series for sin x (about 0) is
|
It turns out that more or less any function has a Taylor series. The crux of the theory of Taylor series is Taylor's theorem, which you would encounter in the first year of a degree course in mathematics. This says that if a function f(x) is n-times differentiable at a point a, then we can write
|
where f(i)(a) denotes the i-th derivative of f(x) at x = a, and Rn is a remainder term whose form, crucially, is known. In fact, there are two alternative forms of the remainder term, the most useful (I think) being
Rn = f(n)(c)(x-a)n/n!
where c is some number lying between x and a that we don't know. (If you don't know what it means for a function to be "n-times differentiable at a point", which is the same as saying that its nth derivative exists at that point, don't worry too much. Unfortunately there's not enough room to explain that here. Polynomials and trig functions are examples of infinitely differentiable functions, i.e. at least n-times differentiable for all n)
Notice that the form of the remainder is, in a sense, the crucial part of the theorem. If we didn't know what it was, the theorem wouldn't actually tell us anything at all. The form of the remainder above gives us the ability to estimate the error in the approximation to f(x) that is given by the series on the r.h.s. (so long as we have some idea of the size of the n th derivative of f(x) at x=c. If f(x) is infinitely differentiable, then we can take n as large as we like, and so make the remainder term as small as we like (since it is inversely proportional to n!). Effectively, then, we get an infinite series with no remainder term, i.e. an exact expression, like the one I gave for sin x above.
In certain circumstances in pure mathematics, these exact expressions can be very useful, particularly in complex analysis. However, Taylor series are most widely used in applied mathematics. Here they are typically used to simplify problems involving non-linear functions by approximating them by the first one or two terms in the appropriate Taylor series, which gives an expression of the form
f(x) <> f(a) + (x-a)f'(a)
where <> means "roughly equal to". The right hand side is linear in x, which simplifies things immensely. This approximation is valid only when (x-a)2 is small enough to be ignored, so that the remainder term R2 can be ignored. This approach is called linearisation. Taylor series are also absolutely fundamental to all known algorithms for solving differential equations numerically, which is an extremely important problem in real life.
I'm not sure whether you know enough maths to be able to follow all that - it's fairly advanced stuff. But I hope it was reasonably comprehensible and that it satisfies your curiosity to some extent.
David,
I don't know where the sines and cosines of squares of angles are used in physics - I've never come across them in three years of studying mostly applied maths. However, they're not weird in a mathematical sense at all. A more odd, if rather inocuous looking, >function is sin(1/x). Try drawing that on your graphical calculator.... (Or just think about what happens in the vicinity of x = 0)
Simon
By Katharina JüRges (P2657) on Friday, June 30, 2000 - 03:32 pm:
Hello Simon,
What you explained to me was very clear and I've understood almost
everything apart from some notations. But this doesn't matter, I
will certainly get to know them next year, I think the symbol
"sigma" stands for sums and this is part of the syllabus of the
last (for me next) school year according to my maths teacher. I
already had plenty of differentiation and I know the method of
linearisation, so no problem with this.
I've got a little question:
According to the Taylor theorem and the equation it gives, isn't
it
f(x)~ f(a)(x-a)+(x-a)f'(a) instead of
f(x)~ f(a)+(x-a)f'(a) ? (I'm sorry I can't use the correct "roughly
equals to" symbol.)
This was the last equation you wrote, and if it's not right what I
just wrote, please try to explain my mistake to me.
As I mentioned above, I don't know sums yet, so my mistake could
lie there. However, I think they aren't that difficult to
understand, so just talk about sums if need be.
Thank you again, you've already helped me a lot.
By Katharina JüRges (P2657) on Friday, June 30, 2000 - 03:39 pm:
I just want to add
where I found the notation tan(x)²:
In one of the letters on the Plus-Page, its title was "Taking a
STEP".
If someone could look on this page and tell me then what
tan(x)² means in that case, I would be
grateful.
By Sean Hartnoll (Sah40) on Tuesday, August 22, 2000 - 10:01 pm:
Hi,
The expression Simon gave in his last post was the correct
one.
Here is a very unrigurous argument for where the Taylor series
coefficients come from.
Suppose a function near the point a can be written as a
series
f(x) = a0 + a1(x-a) +
a2(x-a)2 + ...
in which in some sense the terms get smaller so that the infinite
series gives a finite result (this is why we have x-a terms, they
will be small if x is close to a).
now at x=a, this gives. f(a) = a0, so we know what
a0 is (because all the other terms vanished).
now differentiate the expression,
f'(x) = a1 + 2a2(x-a) +
3a3(x-a)2 + ...
and put again x=a, we find a1 = f'(a)
We continue differentiating and putting x=a, to find
a2 = f''(a)/2
a3 = f'''(a)/6
etc.
This gives the Taylor series,
f(x) = f(a) + f'(a) (x-a) + f''(a)(x-a)2/2 +
f'''(a)(x-a)3/6 + ...
as a way of evaluating f(x) when x is near to a.
Hope this answers your question,
Sean
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