Giải thích các bước giải:
Ta có :
$\cos\dfrac{x(2k+3)}{2}-\cos \dfrac{x}{2}$
$=-2(\sin\dfrac{\dfrac{x(2k+3)}{2}+\dfrac{x}{2}}{2}.\sin\dfrac{\dfrac{x(2k+3)}{2}-\dfrac{x}{2}}{2})$
$=-2.\sin\dfrac{x(k+2)}{2}.\sin\dfrac{x(k+1)}{2}$
$\to \dfrac{-\dfrac 12(\cos\dfrac{x(2k+3)}{2}-\cos \dfrac{x}{2})}{\sin\dfrac x2}=\dfrac{-\dfrac 12(-2(\sin\dfrac{x(k+2)}{2}.\sin\dfrac{x(k+1)}{2})}{\sin \dfrac x2}$
$\to \dfrac{-\dfrac 12(\cos\dfrac{x(2k+3)}{2}-\cos \dfrac{x}{2})}{\sin\dfrac x2}=\dfrac{\sin\dfrac{x(k+2)}{2}.\sin\dfrac{x(k+1)}{2}}{\sin \dfrac x2}$
