How to prove the following? $$2\sin x\cos x = \sin 2x$$
I was trying:
$$2\sin x \cos x = \sin x \sqrt{1-\cos^2 x}\sqrt{1-\sin^2 x}$$ $$2\sin x \cos x = \sin x \sqrt{1-\cos^2 x +1-\sin^2 x}$$ $$2\sin x \cos x = \sin x \sqrt{2-1}$$
but this did not work out
Update: how to prove the double angle formula?
On
$$\require{cancel} 2\sin x \cos x = \cancel{2}\overbrace{\left(\frac{e^{ix}-e^{-ix}}{2i}\right)}^{\sin x}\overbrace{\left(\frac{e^{ix}+e^{-ix}}{\cancel{2}}\right)}^{\cos x}$$ $$=\frac{(e^{ix})^2 - (e^{-ix})^2}{2i}$$ $$=\frac{e^{i2x}-e^{-i2x}}{2i}$$ $$=\sin 2x$$
Addendum: Proof that $e^{ix}=\cos x +i\sin x$:
$$(e^{-ix}(\cos x + i\sin x))' = (e^{-ix})'(\cos x + i\sin x)+ e^{-ix}(\cos x + i\sin x)'$$ $$=-ie^{-ix}(\cos x + i\sin x) + e^{-ix}(-\sin x + i\cos x)$$ $$= 0$$ So the expression $e^{-ix}(\cos x + i\sin x)$ is a constant, independent of $x$. In particular, taking $x=0$, this expression is $1$, which means $$e^{-ix}(\cos x + i\sin x) = 1$$ $$e^{ix}\cdot e^{-ix}(\cos x + i\sin x) = e^{ix}\cdot1$$ $$\cos x + i\sin x = e^{ix}$$
On
$\sin(x + y) = \sin(x)\cos(y) + \cos(x)\sin(y)$
$\Rightarrow \sin(2x) = \sin(x + x) = \sin(x)\cos(x) + \cos(x)\sin(x)$
$\Rightarrow \sin(2x) = 2\sin(x)\cos(x)$
Hence, the above-mentioned trigonometric identity is verified.
On
A (direct) angle of rays in the euclidean vector plane $\mathbb R^2$ is an ordered couple $(\Delta_1, \Delta_2)$ of rays of $\mathbb R^2$, where a ray is a set of the type $\;\mathbb R_+\cdot v=\{\lambda v\;:\;\lambda\geq0\}$, with $\;v\neq0$. Two angles of rays $(\Delta_1, \Delta_2)$ and $(\Delta'_1, \Delta'_2)$ are equal if, by definition, there is a rotation $u\in \mathbb {SO}(2, \mathbb R)$ such that
$$ u(\Delta_1)=\Delta_2 \quad\text{and}\quad u(\Delta'_1)=\Delta'_2\,. $$
The set $\;\mathcal U\;$ of equivalence classes of equal angles of rays is then in a bijection $\;\rho\;$with the group $\mathbb{SO}(2,\mathbb R)$, hence it is itself a group, called the group of (direct) angles, commonly denoted additively. The bijection $$\;\rho\;:\;\mathcal U \;\longrightarrow \;\mathbb{SO}(2,\mathbb R)$$ is therefore an isomorphism of groups. In this context, an angle is a rotation. Note that this notion is independent of any orientation of the plane.
If, however, we assume the plane is oriented, we can proceed to find an especially interesting form of the matrix representation $\rm{Mat}(u)$ of $u=\rho(\theta)$.
Let $(e_1, e_2)$ be an orthonormal basis of $\mathbb R^2$, and call $\Psi$ the orientation form we have chosen, and let suppose that $\Psi(e_1, e_2)=1$, i.e. the basis is direct with respect to the given orientation.
Recall that the representation matrix of a rotation $u$ with respect to a direct orthonormal basis is of the type
$$ \rm{Mat}(u)=\rm{Mat}(\rho(\theta))= \begin{bmatrix}\alpha &-\beta\\[1ex]\beta &\alpha\end{bmatrix} $$
with $\;\det(u)=\alpha^2+\beta^2=1$.
To calculate $\alpha$ and $\beta$, we observe that $ u(e_1)=\alpha e_1+\beta e_2$, so $$\big(e_1\mid u(e_1)\big)=\big(e_1\mid \alpha e_1+\beta e_2\big)= \alpha\big(e_1\mid e_1)+\beta\big(e_1\mid e_2\big)=\alpha $$ which implies that, by definition, $\;\alpha=\cos(\rho^{-1}(u))=\cos(\theta)$.
Similarly,
$$ \Psi\big(e_1, u(e_1)\big)=\Psi\big(e_1,\alpha e_1+\beta e_2\big)=\alpha\Psi(e_1, e_1\big)+\beta\Psi\big(e_1, e_2\big)=\beta,$$
which means $\;\beta=\sin(\theta)$.
The matrix of $u=\rho(\theta)$ is then
$$ \rm{Mat}(\rho(\theta))= \begin{bmatrix} \cos(\theta) & -\sin(\theta) \\[1ex] \sin(\theta) & \cos(\theta) \end{bmatrix}. $$
Finally, from the group properties of $\;\mathcal U \approx \mathbb{SO}(2,\mathbb R)$, we obtain
$$\rm{Mat}(\rho(\theta+\theta')) = \rm{Mat}\big(\rho(\theta)\circ\rho(\theta')\big) = \rm{Mat}(\rho(\theta))\cdot\rm{Mat}(\rho(\theta')) $$ so
$$\begin{bmatrix} \cos(\theta+\theta') & -\sin(\theta+\theta') \\[1ex]\sin(\theta+\theta') & \cos(\theta+\theta') \end{bmatrix} = \hskip45ex$$ \begin{align} &=\begin{bmatrix} \cos(\theta) & -\sin(\theta) \\[1ex] \sin(\theta) & \cos(\theta)\end{bmatrix}\cdot\begin{bmatrix} \cos(\theta') & -\sin(\theta') \\[1ex] \sin(\theta') & \cos(\theta')\end{bmatrix} =\\[2ex] &=\begin{bmatrix} \cos(\theta)\cos(\theta')-\sin(\theta)\sin(\theta') & -\cos(\theta)\sin(\theta') -\sin(\theta)\cos(\theta') \\[1.5ex] \sin(\theta)\cos(\theta') + \cos(\theta)\sin(\theta') & -\sin(\theta)\sin(\theta') + \cos(\theta)\cos(\theta') \end{bmatrix} \end{align}
from which, by comparison, the desired addition formulas.
There are many proof, from basic arguments to anothers which use very strong machinery. The first proof that people learns is the purely geometric one. I'll give you a very simple proof just using complex numbers. Let $a,b\in\mathbb{R}$, then $$\cos (a+b)+i\sin (a+b)=e^{i(a+b)}=e^{ia}e^{ib}=(\cos a+i\sin a)(\cos b+i\sin b)=(\cos a\cos b-\sin a\sin b)+i(\sin a\cos b+\cos a\sin b),$$ now equalizing the imaginary parts we get that $$\sin (a+b)=\sin a\cos b+\sin b\cos a.$$ Also you'll get the formula for the cosine of sum with this proof. Now just put $a=b$.