Giải thích các bước giải:

B xem lại đề bài câu $j$

$\begin{array}{l}
a)A = \sin 7x - \sin 9x\\
 = 2\cos \dfrac{{7x + 9x}}{2}\sin \dfrac{{7x - 9x}}{2}\\
 = 2\cos 8x\sin \left( { - x} \right)\\
 =  - 2\sin x\cos 8x\\
b)B = \cos 2x + \cos 14x\\
 = 2\cos \dfrac{{2x + 14x}}{2}\cos \dfrac{{2x - 14x}}{2}\\
 = 2\cos 8x\cos \left( { - 6x} \right)\\
 = 2\cos 8x\cos 6x\\
c)C = \tan x - \tan 5x\\
 = \dfrac{{\sin x}}{{\cos x}} - \dfrac{{\sin 5x}}{{\cos 5x}}\\
 = \dfrac{{\sin x\cos 5x - \sin 5x\cos x}}{{\cos x\cos 5x}}\\
 = \dfrac{{\sin \left( {x - 5x} \right)}}{{\cos x\cos 5x}}\\
 = \dfrac{{ - \sin 4x}}{{\cos x\cos 5x}}\\
d)D = \cot 3x + \cot 2x\\
 = \dfrac{{\cos 3x}}{{\sin 3x}} + \dfrac{{\cos 2x}}{{\sin 2x}}\\
 = \dfrac{{\cos 3x\sin 2x + \cos 2x\sin 3x}}{{\sin 3x\sin 2x}}\\
 = \dfrac{{\sin \left( {3x + 2x} \right)}}{{\sin 3x\sin 2x}}\\
 = \dfrac{{\sin 5x}}{{\sin 3x\sin 2x}}\\
e)E = \sin x + \sin 7x + \sin 13x + \sin 19x\\
 = \left( {\sin x + \sin 19x} \right) + \left( {\sin 7x + \sin 13x} \right)\\
 = 2\sin 10x\cos 9x + 2\sin 10x\cos 3x\\
 = 2\sin 10x\left( {\cos 9x + \cos 3x} \right)\\
f)F = \cos 2x - \cos 4x + \cos 6x - \cos 8x\\
 = \left( {\cos 2x - \cos 8x} \right) + \left( {\cos 6x - \cos 4x} \right)\\
 =  - 2\sin 5x\sin \left( { - 3x} \right) - 2\sin 5x\sin x\\
 = 2\sin 5x\left( {\sin 3x - \sin x} \right)\\
 = 2\sin 5x.2\cos 2x\sin x\\
 = 4\sin 5x\cos 2x\sin x\\
g)G = \cos x + \cos 3x + \cos 5x\\
 = \left( {\cos x + \cos 5x} \right) + \cos 3x\\
 = 2\cos 3x\cos 2x + \cos 3x\\
 = \cos 3x\left( {2\cos x + 1} \right)\\
h)H = \sin 7x - \sin 9x + \sin 11x\\
 = \left( {\sin 7x + \sin 11x} \right) - \sin 9x\\
 = 2\sin 9x\cos 2x - \sin 9x\\
 = \sin 9x\left( {2\cos 2x - 1} \right)\\
i)I = 1 + \cos x + \cos 2x + \cos 3x\\
 = 1 + \cos 2x + \left( {\cos x + \cos 3x} \right)\\
 = 1 + \cos 2x + 2\cos 2x\cos x\\
 = 2{\cos ^2}x + 2\cos 2x\cos x\\
 = 2\cos x\left( {\cos x + \cos 2x} \right)
\end{array}$

$\begin{array}{l}
j)J = 1 - \sin x + \sin 2x - \sin 3x\\
 = 1 + \sin 2x - \left( {\sin x + \sin 3x} \right)\\
 = 1 + \sin 2x - 2\sin 2x\cos x\\
 = 1 + \sin 2x - 4\sin x{\cos ^2}x\\
 = 1 + 2\sin x\cos x - 4\sin x{\cos ^2}x
\end{array}$