Answer to Question #92279 in Mechanics | Relativity for Gabriel Ayomide

Question

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

Answer 1

We have quadratic equation

$$c^2=a^2+b^2-2abcos(y)$$

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

Using quadratic formula

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

$$a_{1,2}=bcos(y)±\sqrt{b^2cos^2(y)-b^2+c^2}$$

$$a_{1,2}=bcos(y)±\sqrt{-b^2(1-cos^2(y))+c^2}$$

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

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

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

$$c^2=a^2+b^2-2abcos(y)$$

if a is the root of the equation

$$c^2=a^2+b^2-2abcos(y)$$

then

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

or

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