7A=180 THEN PROVE THAT COS A COS 2A COS 3A =1/8

given: 7A = 180.........(1)
LHS of the given equation is:
$$\begin{gathered} \cos A\cos 2A\cos 3A=\cos A\cos 2A\cos (180-4A) [\sin ce 3A=180-4A] \\ =\cos A\cos 2A*[-\cos 4A] \\ =-\cos A\cos 2A\cos 4A \\ =-\frac{1}{2\sin A}\left(2\sin A\cos A\right)\cos 2A\cos 4A [multiplying and dividing by 2\sin A \\ =-\frac{1}{2\sin A}*\sin 2A\cos 2A\cos 4A \\ =-\frac{1}{4\sin A}(2\sin 2A\cos 2A)\cos 4A \\ =-\frac{1}{4\sin A}\sin 4A\cos 4A \\ =-\frac{1}{8\sin A}(2\sin 4Aco4A) \end{gathered}$$
$$\begin{gathered} =-\frac{1}{8\sin A}*\sin 8A \\ =-\frac{1}{8\sin A}*\sin (7A+A) \\ =-\frac{1}{8\sin A}*\sin (180+A) \\ =-\frac{1}{8\sin A}*(-\sin A) \\ =\frac{1}{8} \end{gathered}$$
= RHS

hope this helps you