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

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.

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

Thank you both, Simon and Sean,

I agree that numbers like $\sin \pi /6$, 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

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

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

You have to be careful when writing squares of functions. There are various ambiguities. For example, ${\tan }^{2}x$ 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, $\tan x^2$ should be taken to mean $\tan (x^2)$ rather than $(\tan x{)}^2$, although it is common in this case to make things clearer by writing it as $\tan (x^2)$. 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

${\displaystyle \sum _{n=0}^{\infty }\frac{(-1{)}^n{x}^{2n+1}}{(2n+1)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots }$

This is an infinite power series that converges for any value of $x$ to the corresponding value of $\sin x$.

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

${\displaystyle f(x)=\sum _{i=0}^{n-1}\frac{f^{(i)}(a)(x-a{)}^i}{i!}+R_n}$

where $f^{(i)}(a)$ denotes the $i$-th derivative of $f(x)$ at $x=a$, and $R_n$ 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

$R_n=f^{(n)}(c)(x-a{)}^n/n!$

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 $n$th 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

$f(x)<>f(a)+(x-a)f\text{'}(a)$

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 $R_2$ 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 innocuous 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

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.

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.

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