Víťa Šmíd a.k.a. Zephyrus

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Deriving Common Trigonometric Identities

There are quite a few identities concerning trigonometric functions. The most commonly used are the cosine and sine of double angles and the cosine and sine of the sum or difference of two angles. It might be perhaps surprising that these formulae can be easily derived with the help of basic complex analysis and algebra. Personally, I find it much easier to derive most of the formulae whenever I need them instead of remembering them all.

Preparing The Tools

In complex analysis, the complex exponential, complex cosine, and complex sine functions are defined as follows:

<$$ % These will come handy later. \newcommand{\Re}{\mathop{\mathrm{Re}}} \newcommand{\Im}{\mathop{\mathrm{Im}}} \exp z := \sum_{n = 0}^{\infty} \frac{z^n}{n!}, \quad \cos z := \sum_{n = 0}^{\infty} (-1)^n \frac{z^{2n}}{(2n)!}, \quad \sin z := \sum_{n = 0}^{\infty} (-1)^n \frac{z^{2n+1}}{(2n+1)!}, \quad z \in \mathbb{C}. $$>

Using these definitions and the fact that <$i^2 = -1$>, the so-called Euler’s formula can be easily proved:

<$$ e^{ix} = \cos x + i\sin x. $$>

Now all we need is to make a smart choice of <$x$> to obtain the identity we want.

<$ \cos 2\alpha,\; \sin 2\alpha $>

Letting <$x := 2\alpha$>, we have, by the Euler’s formula,

<$$ e^{i2\alpha} = \cos 2\alpha + i\sin 2\alpha. $$>

But it is also true that

<$$ e^{i2\alpha} = \left(e^{i\alpha}\right)^2 = \left(\cos \alpha + i\sin \alpha\right)^2 = \cos^2\alpha -\sin^2\alpha + 2i\sin\alpha\cos\alpha. $$>

Comparing both results, we obtain the identity

<$$ \cos 2\alpha + i\sin 2\alpha = \cos^2\alpha -\sin^2\alpha + 2i\sin\alpha\cos\alpha. $$>

Now, two complex numbers equal if, and only if, both their real and imaginary parts equal each other, or <$ a = b \Leftrightarrow \Re~a = \Re~b \wedge \Im~a = \Im~b $>. Therefore, if we compare either real or imaginary parts of the above identity they must hold as well. And so we have the well-known identities

<$$ \displaylines{ \cos 2\alpha = \cos^2\alpha -\sin^2\alpha, \\ \sin 2\alpha = 2\sin\alpha\cos\alpha. } $$>

<$\cos(\alpha+\beta),\; \sin(\alpha+\beta)$>

This case is not much more complicated. Letting <$x := \alpha+\beta$>, we have

<$$ e^{i(\alpha+\beta)} = \cos(\alpha+\beta) + i\sin(\alpha+\beta). $$>

On the other hand,

<$$ \displaylines{ e^{i(\alpha+\beta)} = e^{i\alpha}e^{i\beta} = \left(\cos\alpha + i\sin\alpha\right)\left(\cos\beta + i\sin\beta\right) = \\ = \cos\alpha\cos\beta - \sin\alpha\sin\beta + i\left(\cos\alpha\sin\beta + \sin\alpha\cos\beta\right). } $$> We have arrived at <$$ \cos(\alpha+\beta) + i\sin(\alpha+\beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta + i\left(\cos\alpha\sin\beta + \sin\alpha\cos\beta\right), $$>

and comparing again the real and imaginary parts, it is clear that

<$$ \displaylines{ \cos(\alpha+\beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta, \\ \sin(\alpha+\beta) = \cos\alpha\sin\beta + \sin\alpha\cos\beta. } $$>

Note that when <$\alpha = \beta$>, the formulae will be the same as those for double angles.

<$\cos(\alpha-\beta),\; \sin(\alpha-\beta)$>

In this case the procedure is pretty much the same as before if we regard <$\alpha-\beta$> as <$\alpha+(-\beta)$>. We just need to be careful about the minus sign. Since cosine is an even function, <$\cos(-\beta) = \cos\beta$>. Sine, on the other hand, is odd, so <$\sin(-\beta) = -\sin\beta$>.

<$$ \displaylines{ e^{i(\alpha-\beta)} = e^{i\alpha}e^{-i\beta} = \left(\cos\alpha + i\sin\alpha\right)\left(\cos\beta - i\sin\beta\right) = \\ = \cos\alpha\cos\beta + \sin\alpha\sin\beta + i\left(\sin\alpha\cos\beta - \cos\alpha\sin\beta\right). } $$> Evidently <$$ \cos(\alpha-\beta) + i\sin(\alpha-\beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta + i\left(\sin\alpha\cos\beta - \cos\alpha\sin\beta\right), $$>

and so

<$$ \displaylines{ \cos(\alpha-\beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta, \\ \sin(\alpha-\beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta. } $$>

Do you also have your favorite tricks for formulae derivation? :-)

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