Deduce projection formula from cosine formula.

Solution

In any triangle, we have to show that
a = b cosC + c cosB
b = c cosA + a cosC
c = a cosB + b cosA
R.H.S. = $straight b space cosC space plus space straight c space cosB space equals space straight b open square brackets fraction numerator straight a squared plus straight b squared minus straight c squared over denominator 2 ab end fraction close square brackets plus straight c open square brackets fraction numerator straight a squared plus straight c squared minus straight b squared over denominator 2 ac end fraction close square brackets$
= $fraction numerator straight a squared plus straight b squared minus straight c squared over denominator 2 straight a end fraction plus fraction numerator straight a squared plus straight c squared minus straight b squared over denominator 2 straight a end fraction space equals space fraction numerator straight a squared plus straight b squared minus straight c squared plus straight a squared plus straight c squared minus straight b squared over denominator 2 straight a end fraction$
= $fraction numerator 2 straight a squared over denominator 2 straight a end fraction equals straight a equals straight L. straight H. straight S.$
Similarly, we can show that
b = c cosA + a cosC
c = a cosB + b cosA

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Principle Of Mathematical Induction

Deduce projection formula from cosine formula.

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