Prove that: \[ \cos(A+B+C)+\cos(A-B+C) \] \[ +\cos(A+B-C)+\cos(-A+B+C) \] \[ =4\cos A\cos B\cos C \]
Solution
L.H.S.
\[ = \cos(A+B+C)+\cos(A-B+C) \] \[ +\cos(A+B-C)+\cos(-A+B+C) \]Group first two terms and last two terms.
Use identity:
\[ \cos C+\cos D = 2\cos\frac{C+D}{2}\cos\frac{C-D}{2} \]
\[
=
2\cos(A+C)\cos B
\]
\[
+
2\cos(B+C)\cos A
\]
Take common factor 2:
\[ = 2\left[\cos(A+C)\cos B+\cos(B+C)\cos A\right] \]Expand using:
\[ \cos(X+Y)=\cos X\cos Y-\sin X\sin Y \]
\[
=
2[
(\cos A\cos C-\sin A\sin C)\cos B
\]
\[
+
(\cos B\cos C-\sin B\sin C)\cos A
]
\]
\[
=
2[
\cos A\cos B\cos C
-\sin A\sin C\cos B
\]
\[
+
\cos A\cos B\cos C
-\sin B\sin C\cos A
]
\]
\[
=
4\cos A\cos B\cos C
\]
Hence Proved.