Answer
$2 \sin x \cos x= \sin 2x $ (Verified)
Work Step by Step
The Taylor series for $\sin x $ can be defined as: $\sin x= x-\dfrac{x^3}{3!}+\dfrac{ x^5}{5!}-....$
$\sin^2 x=\dfrac{1-\cos 2x }{2}=\dfrac{2x^2}{2!}-\dfrac{2^3 x^4}{4!}+\dfrac{2^5 x^6}{6!}-....$
Now, $\dfrac{d(\sin^2 x)}{dx}=\dfrac{d(\dfrac{2x^2}{2!}-\dfrac{2^3 x^4}{4!}+\dfrac{2^5 x^6}{6!}-....)}{dx}$
or, $2 \sin x \cos x= 2x-\dfrac{(2x)^3}{3!}+\dfrac{(2x)^5}{5!}-....$
or, $2 \sin x \cos x= \sin 2x $
Hence, the result has been verified.
