# A calculus of the absurd

#### 14.7 Cool stuff with trigonometry

##### 14.7.1 Proving identities
• Example 14.7.1 Show that

$\cos (a + b) = \cos (a)\cos (b) - \sin (a)\sin (b)$

Solution:

We can write $$\cos (a + b)$$ as the real part of $$e^{(a+b)i}$$. This because $$\cos (a+b)$$ is equal to the real part of $$\cos (a+b) + i\sin (a+b)$$, which is equal to $$e^{(a+b)i}$$.

• Example 14.7.2 Express $$\sin (3x)$$ in terms of $$\sin (x)$$.

Solution: Firstly, we can write $$\sin (3x)$$ as the equation

$$\sin (3x) = \Im (\cos (3x) + i\sin (3x))$$

We can then apply De Moivre’s theorem106106 $$\cos (nx) + i\sin (nx) = (\cos (x) + i\sin (x))^n$$ to rewrite the expression in terms of $$\sin (x)$$ and $$\cos (x)$$

$$\Im (\cos (3x) + i\sin (3x)) = \Im ((\cos (x) + i\sin (x))^3)$$

We can now expand the binomial obtained, which leads to the result that

$$\Im ((\cos (x) + i\sin (x))^3) = \Im ((\cos ^3(x) + 3\cos ^2(x)i\sin (x) + 3\cos (x)i^2\sin ^2(x) + i^3\sin ^3(x)))$$

Then, we can tidy this up a bit, leading to the expression

$$\Im ((\cos ^3(x) + 3\cos ^2(x)\sin (x)i - 3\cos (x)\sin ^2(x) - i\sin ^3(x)))$$

We are only interested in the imaginary parts of the expansion, so it is therefore equal to just

$$3\cos ^2(x)\sin (x) - \sin ^3(x)$$

We want $$\sin (3x)$$ in terms of $$\sin (x)$$, however! There’s a rogue gatecrasher 107107 A handy way to remember whether $$\cos (x)$$ or $$\sin (x)$$ shows a certain property is (as previously mentioned, TODO: mention) that $$\sin (x)$$ generally behaves "better" than $$\cos (x)$$. in the previous expression - the $$\cos (x)$$! Fortunately we can remove the $$\cos ^2(x)$$ without too much difficulty using the Pythagorean identity.

\begin{align} 3(1-\sin ^2(x))\sin (x) - \sin ^3(x) &= 3\sin (x)-3\sin ^3(x) - \sin ^3(x) \\ &= 3\sin (x)-4\sin ^3(x) \end{align}

Thus, we can say that

$$\sin (3x) = 3\sin (x)-4\sin ^3(x)$$