Thomas' Calculus 13th Edition

$2 \sin x \cos x= \sin 2x$ ; Verified result
The Taylor series can be written as follows: $\sin x= x-\dfrac{x^3}{3!}+\dfrac{ x^5}{5!}-...........$ We need to prove the result. We have: $\sin^2 x=\dfrac{1-\cos 2x }{2}$ or, $=\dfrac{2x^2}{2!}-\dfrac{2^3 x^4}{4!}+\dfrac{2^5 x^6}{6!}-....$ $\implies \dfrac{d}{dx} (\sin^2 x)=\dfrac{d}{dx}[(\dfrac{2x^2}{2!}-\dfrac{2^3 x^4}{4!}+\dfrac{2^5 x^6}{6!}-....)]$ $\implies 2 \sin x \cos x= 2x-\dfrac{(2x)^3}{3!}+\dfrac{(2x)^5}{5!}-....$ $\implies 2 \sin x \cos x= \sin 2x$ So, the result has been proven.