Question #92279

Prove that c2=a2+b2- 2abcosy is equal to a= bcosy+√c2- b2×sin2y

Expert's answer

We have quadratic equation


c2=a2+b22abcos(y)c^2=a^2+b^2-2abcos(y)

a2a(2bcos(y))+b2c2=0a^2-a(2bcos(y))+b^2-c^2=0

Using quadratic formula


a1,2=bcos(y)±b2cos2(y)(b2c2)a_{1,2}=bcos(y)±\sqrt{b^2cos^2(y)-(b^2-c^2)}

a1,2=bcos(y)±b2cos2(y)b2+c2a_{1,2}=bcos(y)±\sqrt{b^2cos^2(y)-b^2+c^2}

a1,2=bcos(y)±b2(1cos2(y))+c2a_{1,2}=bcos(y)±\sqrt{-b^2(1-cos^2(y))+c^2}

a1,2=bcos(y)±b2sin2(y)+c2a_{1,2}=bcos(y)±\sqrt{-b^2sin^2(y)+c^2}


a1,2=bcos(y)±c2b2sin2(y)a_{1,2}=bcos(y)±\sqrt{c^2-b^2sin^2(y)}

So this is not strict equivalence:

if\ a=bcos(y)+\sqrt{c^2-b^2sin^2(y)},\

then a is the root of the equation

c2=a2+b22abcos(y)c^2=a^2+b^2-2abcos(y)


if a is the root of the equation


c2=a2+b22abcos(y)c^2=a^2+b^2-2abcos(y)

then


a=bcos(y)+c2b2sin2(y)a=bcos(y)+\sqrt{c^2-b^2sin^2(y)}

or


a=bcos(y)c2b2sin2(y).a=bcos(y)-\sqrt{c^2-b^2sin^2(y)}.


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