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What's the importance of the trig angle formulas, like the sum and difference formulas, the double angle formula, and the half angle formula? I understand that they help us calculate some trig ratios without the aid of a calculator, but I guess I don't really understand the point of learning them since they're not the same algorithms used by calculators to calculate trig ratios (most of them use a Taylor series, right?). Wouldn't our time be better spent learning the calculator's algorithms so that we can calculate the ratio at any arbitrary angle? Thanks!

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    $\begingroup$ I know from experience that they assist a lot in integration and summing infinite series. $\endgroup$ – Eric Auld Jul 31 '13 at 13:51
  • $\begingroup$ You may get a lot of heat from asking this, but in a sense, you are right. Still, they are historically and culturally important. They allow you to communicate and read a lot mathematical text that is written using them, for example. $\endgroup$ – OR. Jul 31 '13 at 13:51
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    $\begingroup$ @RGB I disagree, they are useful beyond just a historical setting. Take for instance the inductive proof of De Moivre's formula for integers values of $n$. (en.wikipedia.org/wiki/…) $\endgroup$ – Dan Rust Jul 31 '13 at 14:12
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    $\begingroup$ Calculators don't use Taylor series. $\endgroup$ – André Nicolas Jul 31 '13 at 15:32
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    $\begingroup$ The early scientific calculators mostly used CORDIC. There are a number of references in answers to this question. $\endgroup$ – André Nicolas Jul 31 '13 at 15:56
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-These are the basis of Chebyshev polynomials, a useful set of orthogonal polynomials.

-They also come up when doing things with waves in physics: A standing wave $\sin(kx)\cos(\omega t)$ can be rewritten as the sum of a wave going forward and backwards; see here. In general this ability to compose/decompose waves comes in handy often enough in wave physics. I use these formulas all the time here.

-They also come up in (and/or come from) complex multiplication:

$$\cos(a+b)+i\sin(a+b)=(\cos a+i\sin a)(\cos b+i\sin b)=[\cos a\cos b-\sin a\sin b]+i[\sin a \cos b+\sin b\cos a]$$

(note that $i^2=-1$). In this sense you don't even really have to "memorize them". Once you've gotten used to the relationship between complex numbers and geometry, they're immediate and natural things to keep stumbling across.

A lot of the reasons to learn these has nothing to do with calculating particular angles. It's really there because $\sin$ and $\cos$ come up a lot (like a lot, not just for triangles as your classes might imply), and learning how to rewrite expressions of them into more useful forms is a natural consequence of trying to use them when you find them hiding in a problem.

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The triple angle formula is used to prove the impossibility of trisecting an angle.

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Trigonometric angle formulae are important as they help simplify expressions which are difficult to solve into easier ones. This is notably helpful when working with limits and integrals.

A few concrete examples include:

$\int\dfrac{\cos^2x}{1+\sin x}\space dx, \int\cos^2 x\space dx, \int \sin^5x\cos^2x \space dx,\int\sin3x\sin7x \space dx$

I'll evaluate the last one, so you can see exactly how they make things easier.

$\int \sin3x \sin7x \space dx=-\dfrac{1}{2}\int(\cos10x-\cos(-4x))\space dx=-\dfrac{1}{2}\int\cos10x \space dx +\dfrac{1}{2}\int\cos4x \space dx=-\dfrac{1}{20}\sin10x+\dfrac{1}{8}\sin4x+C.$

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  • $\begingroup$ Not that I don't believe, because I do, but would you mind giving a concrete example? $\endgroup$ – jrc03c Jul 31 '13 at 16:41
  • $\begingroup$ Sure. Give me a moment- I'll edit it into my answer. $\endgroup$ – Sujaan Kunalan Jul 31 '13 at 16:47
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Consider for instance $2\cos(x)\sin(x)=\sin(2x)$. What happens if you want to calculate the derivative of this?

$\frac{d}{dx}2\cos(x)\sin(x) = -2\sin^2(x)+2\cos^2(x)$

$\frac{d}{dx}\sin(2x) = 2\cos(2x)$

Now suppose that you're doing physics and an expression like this appears. Depending on which version you use, you will come up with potentially different physical interpretations.

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It is clear that trigonometric functions are used. And that, I think, is the main reason why they are important. If you don't know about them then you cannot communicate with people that do use them. A lot of mathematical text is comfortably written using these functions. So, history forces you to use that language.

On the other hand, strictly speaking, you could completely ignore them. The reason is that we have

$$\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$$ and $$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i},$$

where $i=\sqrt{-1}$. This means that trigonometric functions can be replaced everywhere by using instead the exponential $e^{y}$ function. Yo do need to use complex numbers, and since complex numbers were invented later in history, that is the reason why we still talk about trigonometric functions.

To give you an example of how things would work let us see some of those trigonometric identities. Notice first that in the formulas above we have $e^{ix}$ and $e^{-ix}$, which is just $1/e^{ix}$. For convenience typing I will write $z$ instead of $e^{ix}$. So the formulas above read as $$\cos(x)=\frac{z+z^{-1}}{2}$$ and $$\sin(x)=\frac{z-z^{-1}}{2i}.$$

Suppose we want to re-write the formula for the cosine of the double angle. Notice that $e^{i2x}=(e^{ix})^2=z^2$. Then we have $$\cos(2x)=\frac{z^2+z^{-2}}{2}=(\frac{z+z^{-1}}{2})^2-(\frac{z-z^{-1}}{2i})^2=\cos^2(x)-\sin^2(x).$$

As you see, checking the equality in the middle only requires basic simplification of rational functions of $z$. This can be done without thinking. We just need to remember that $i^2=-1$ but the rest is work like we would do to prove $(z-1)(z+1)=z^2-1$, mechanically. So, at least for proving these trigonometric identities, getting rid of the trigonometric functions does represent an advantage. This means that, by this re-writing, most of the trigonometric identities used in school are just rather trivial identities between polynomials. I think students would be very happy if instead of having to prove trigonometric identities they all had to do is to play with polynomials in $z$.

Why is it not done, then. As you see, the work requires complex numbers. Complex numbers were invented much later than trigonometric functions. At the beginning, they were immersed in a mysterious aura. People were using $\sqrt{-1}$, in an ad hoc manner but without an interpretation that would give them confidence about their nature. That is probably the reason why the names imaginary numbers or complex number were used. Later they were given, for example, a geometric interpretation. They are just the points in the plane in which you define a way to multiply them. Nothing more strange that when you define how to multiply points in a line, which you call the real line. Trigonometric function were introduced first as geometric notions. In this sense, trigonometric functions are no more than complex numbers.

So, trigonometric functions are useful, but (I will dare to say 'only', although I am not strongly positive about it) because history made them part of our mathematical language first; and therefore they are useful because you are forced to use them to communicate to each other.

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