# Trig identity question

Show that $\sin(2nx)=\sin((2n+1)x)\cos(x)-\cos((2n+1)x)\sin(x)$.

I have the mark scheme in front of me, but it doesn't make sense to me...

$$\sin((2n+1)x)\cos(x)-\cos((2n+1)x)\sin(x)=\sin((2n+1)x)-x=\sin(2nx)$$

Initially, I would think that it would be a double angle identity so $\sin2\theta=2\sin\theta \cos\theta$, where $\theta$ here is $nx$ but that doesn't seem to be it. How does the cosine disappear from the solution in the mark scheme? If someone could explain every step, I'd really appreciate it.

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Hint: $(2n+1)x = 2nx + x$ –  John Joy Jul 9 '14 at 13:40

Using angle sum and difference identity: $\sin (A-B) = \sin A \cos B - \cos A \sin B$, the RHS can be rewritten as \begin{align} \sin (2n+1)x \cos x - \cos (2n+1)x \sin x&=\sin((2n+1)x-x)\\ &=\sin(2nx+x-x)\\ &=\sin 2nx. \end{align} Thus, RHS = LHS.
Use the identity $$\sin (a-b) = \sin a \cos b + \cos a \sin b$$ Where $a$ = $(2n+1)x$ and $b=x$