Could someone tell me how $\sin(2x)$ arrives?
I know that there is a trig identity that says $2\cos(x)\sin(x) = \sin(2x)$ ... but this isn't isn't the case here... so could someone explain why?
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$\begingroup$Notice that:
$$\sin(a+b)=\sin a\cos b+\sin b\cos a$$
Since $2x = x+x$ we have:
$$\sin(2x)=\sin(x+x)=\sin x\cos x+\sin x\cos x=2\sin x\cos x$$
Now this simplifies to $\sin x\cos x = \sin(2x)/2$ as you want.
$\endgroup$ 1 $\begingroup$Using de Moivre's theorem: $$ e^{nxi}=(\cos{x}+i\sin{x})^n = \cos{2x}+i\sin{2x} $$
We expand the brackets and take the imaginary part (which is always $\sin(nx)i$): $$ e^{2ix}=(\cos{x}+i\sin{x})^2=\cos^2{x}+2i\cos{x}\sin{x}-\sin^2{x} \\ \text{Im}(e^{2ix})=2\cos{x}\sin{x}=\sin{2x} $$
Bonus: If we take the real part, we get an identity for $\cos{2x}$: $$ \text{Re}(e^{2ix}) = \cos{2x} = \cos^2{x}-\sin^2{x} $$
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