Question Number 130103 by mohammad17 last updated on 22/Jan/21
$${by}\:{using}\:{demover}\:{theorem}\:{prove}\:{that} \\ $$$$ \\ $$$${sin}\mathrm{2}\theta=\mathrm{2}{sin}\theta{cos}\theta \\ $$
Answered by Olaf last updated on 22/Jan/21
$$\left(\mathrm{cos}\theta+{i}\mathrm{sin}\theta\right)^{{n}} \:=\:\mathrm{cos}\left({n}\theta\right)+{i}\mathrm{sin}\left({n}\theta\right) \\ $$$$\mathrm{For}\:{n}\:=\:\mathrm{2}\:: \\ $$$$\left(\mathrm{cos}\theta+{i}\mathrm{sin}\theta\right)^{\mathrm{2}} \:=\:\mathrm{cos}\left(\mathrm{2}\theta\right)+{i}\mathrm{sin}\left(\mathrm{2}\theta\right) \\ $$$$\mathrm{cos}^{\mathrm{2}} \theta−\mathrm{sin}^{\mathrm{2}} \theta+\mathrm{2}{i}\mathrm{sin}\theta\mathrm{cos}\theta\:=\:\mathrm{cos}\left(\mathrm{2}\theta\right)+{i}\mathrm{sin}\left(\mathrm{2}\theta\right) \\ $$$$\Rightarrow\begin{cases}{\mathrm{cos}^{\mathrm{2}} \theta−\mathrm{sin}^{\mathrm{2}} \theta\:=\:\mathrm{cos}\left(\mathrm{2}\theta\right)}\\{\mathrm{2sin}\theta\mathrm{cos}\theta\:=\:\mathrm{sin}\left(\mathrm{2}\theta\right)}\end{cases} \\ $$
Commented by mohammad17 last updated on 23/Jan/21
$${very}\:{nice}\:{thank}\:{you}\:{sir} \\ $$