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Question
Prove that:
cos (A + B + C) + cos (A − B + C) + cos (A + B − C) + cos (− A + B + C) = 4 cos A cos Bcos C
Solution
Consider LHS:
\[\cos (A + B + C) + \cos (A - B + C) + \cos (A + B - C) + \cos ( - A + B + C)\]
\[ = 2\cos \left( \frac{A + B + C + A - B + C}{2} \right) \cos \left( \frac{A + B + C - A + B - C}{2} \right) + 2\cos \left( \frac{A + B - C - A + B + C}{2} \right) \cos \left( \frac{A + B - C + A - B - C}{2} \right)\]
\[ = 2\cos\left( A + C \right) \cos B + 2\cos B \cos\left( A - C \right)\]
\[ = 2\cos B\left[ \cos \left( A + C \right) + \cos \left( A - C \right) \right]\]
\[ = 2\cos B\left[ 2\cos \left( \frac{A + C + A - C}{2} \right) \cos \left( \frac{A + C - A + C}{2} \right) \right]\]
\[ = 2\cos B\left[ 2\cos A \cos C \right]\]
\[ = 4\cos A \cos B \cos C\]
= RHS
Hence, LHS = RHS.
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