prove that

cosA +cosB +cosC +cos(A+B+C) = 4cos((A+B) / 2)cos((B+C) / 2)cos((C+A) / 2)

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Question

prove that
cosA +cosB +cosC +cos(A+B+C) = 4cos((A+B) / 2)cos((B+C) /
2)cos((C+A) / 2)

Ankush Jain · Accepted Answer

To prove: $\cos A + \cos B + \cos C + \cos (A + B + C) = 4 \cos (\frac{A + B}{2}) \cos (\frac{B + C}{2}) \cos (\frac{C + A}{2})$ Proof: LHS = $\cos A + \cos B + \cos C + \cos (A + B + C)$ $= 2 \cos \frac{(A + B)}{2} \cos \frac{(A - B)}{2} + 2 \cos \frac{(A + B + 2 C)}{2} \cos \frac{(C - (A + B + C))}{2} = 2 \cos \frac{(A + B)}{2} \cos \frac{(A - B)}{2} + 2 \cos \frac{(A + B + 2 C)}{2} \cos \frac{A + B}{2} = 2 \cos \frac{(A + B)}{2} [\cos \frac{(A - B)}{2} + \cos \frac{(A + B + 2 C)}{2}] = 2 \cos \frac{(A + B)}{2} [2 \cos \frac{\frac{(A - B + A + B + 2 C)}{2}}{2} c os \frac{\frac{(A - B - A - B - 2 C)}{2}}{2}] = 2 \cos \frac{(A + B)}{2} [2 \cos \frac{(2 A + 2 C)}{4} c os \frac{(- 2 B - 2 C)}{4}] = 4 \cos \frac{(A + B)}{2} \cos \frac{(A + C)}{2} c os \frac{(B + C)}{2}$ = RHS [Hence Proved]

prove that cosA +cosB +cosC +cos(A+B+C) = 4cos((A+B) / 2)cos((B+C) / 2)cos((C+A) / 2)

prove that

cosA +cosB +cosC +cos(A+B+C) = 4cos((A+B) / 2)cos((B+C) / 2)cos((C+A) / 2)