Using the formulas de Moivre to deduce trigonometric identities.

Question

Yesterday I made a test of complex variables, and this contained a question (in which I could not solve) that asked to use the de Moivre formulas to deduce the following trigonometric identities:

$$\sin 2\theta=2\sin\theta\cos\theta\\\cos 2\theta=\cos^2{\theta}-\sin^2\theta$$How could I have them resolved?

Answer 1

Recall that $e^{ix} = \cos(x) + i \sin(x)$. Replacing $x$ by $2x$, we get $$e^{2ix} = \cos(2x) + i \sin(2x) \tag{$\star$}$$ We also have $$e^{2ix} = (e^{ix})^2 = (\cos(x) + i \sin(x))^2 = \cos^2(x) - \sin^2(x) + i (2\sin(x) \cos(x)) \tag{$\dagger$}$$ Comparing the real and imaginary parts of $\star$ and $\dagger$, we get what you want.

Answer 2

We shall prove the general result, i.e., finding identities for ${\rm sin}\space nA$ and ${\rm cos}\space nA$, where $n \in \Bbb N$ and $A \in (-\frac{\pi}{2},\frac{\pi}{2}).$
We let $y$ be any real number and observe that the complex number $1+ iy$ can be written as $(1+y^2)^{1/2}e^{i{\rm tan^{-1}} y}.$
Now, from the binomial theorem, we have $$(1+iy)^{n}=\sum_{r=0}^{n}{n \choose r}(iy)^{r}=\sum_{r=0}^{\lfloor\frac{n}{2}\rfloor}(-1)^{r}{n \choose 2r}y^{2r}+i\sum_{r=0}^{\lfloor\frac{n-1}{2}\rfloor}(-1)^{r}{n \choose 2r+1}y^{2r+1}.$$ But, $$(1+iy)^{n}=\{(1+y^2)^{1/2}e^{itan^{-1}y}\}^{n}=(1+y^2)^{\frac{n}{2}}[cos(ntan^{-1}y)+isin(ntan^{-1}y)].$$ Hence comparing real and imaginary parts, we have, $$cos(ntan^{-1}y)=(\frac{1}{\sqrt{1+y^{2}}})^{n}\sum_{r=0}^{\lfloor\frac{n}{2}\rfloor}(-1)^{r}{n \choose 2r}y^{2r}$$ and $$sin(ntan^{-1}y)=(\frac{1}{\sqrt{1+y^{2}}})^{n}\sum_{r=0}^{\lfloor \frac {n-1}{2}\rfloor}(-1)^{r}{n \choose 2r+1}y^{2r+1}.$$ Now, we had assumed $y$ is any real number and hence we can find $A \in (-\frac{\pi}{2},\frac{\pi}{2})$ such that $tan \space A =y$. Now, with the observation that $\frac{1}{\sqrt{1+y^2}}=cos\space A$, we have, $$cos(nA)=cos^{n}A\sum_{r=0}^{\lfloor\frac{n}{2}\rfloor}(-1)^{r}{n \choose 2r}tan^{2r}A$$ and $$sin(nA)=cos^{n}A\sum_{r=0}^{\lfloor \frac {n-1}{2}\rfloor}(-1)^{r}{n \choose 2r+1}tan^{2r+1}A.$$ Having proved this true for all $A \in (-\frac{\pi}{2},\frac{\pi}{2})$, it is easy to verify that the relations are true for any $A$ for which $tan\space A$ is defined. Your question is the special case with $n=2$.