We have quadratic equation
c 2 = a 2 + b 2 − 2 a b c o s ( y ) c^2=a^2+b^2-2abcos(y) c 2 = a 2 + b 2 − 2 ab cos ( y )
a 2 − a ( 2 b c o s ( y ) ) + b 2 − c 2 = 0 a^2-a(2bcos(y))+b^2-c^2=0 a 2 − a ( 2 b cos ( y )) + b 2 − c 2 = 0 Using quadratic formula
a 1 , 2 = b c o s ( y ) ± b 2 c o s 2 ( y ) − ( b 2 − c 2 ) a_{1,2}=bcos(y)±\sqrt{b^2cos^2(y)-(b^2-c^2)} a 1 , 2 = b cos ( y ) ± b 2 co s 2 ( y ) − ( b 2 − c 2 )
a 1 , 2 = b c o s ( y ) ± b 2 c o s 2 ( y ) − b 2 + c 2 a_{1,2}=bcos(y)±\sqrt{b^2cos^2(y)-b^2+c^2} a 1 , 2 = b cos ( y ) ± b 2 co s 2 ( y ) − b 2 + c 2
a 1 , 2 = b c o s ( y ) ± − b 2 ( 1 − c o s 2 ( y ) ) + c 2 a_{1,2}=bcos(y)±\sqrt{-b^2(1-cos^2(y))+c^2} a 1 , 2 = b cos ( y ) ± − b 2 ( 1 − co s 2 ( y )) + c 2
a 1 , 2 = b c o s ( y ) ± − b 2 s i n 2 ( y ) + c 2 a_{1,2}=bcos(y)±\sqrt{-b^2sin^2(y)+c^2} a 1 , 2 = b cos ( y ) ± − b 2 s i n 2 ( y ) + c 2
a 1 , 2 = b c o s ( y ) ± c 2 − b 2 s i n 2 ( y ) a_{1,2}=bcos(y)±\sqrt{c^2-b^2sin^2(y)} a 1 , 2 = b cos ( y ) ± c 2 − b 2 s i n 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 − 2 a b c o s ( y ) c^2=a^2+b^2-2abcos(y) c 2 = a 2 + b 2 − 2 ab cos ( y )
if a is the root of the equation
c 2 = a 2 + b 2 − 2 a b c o s ( y ) c^2=a^2+b^2-2abcos(y) c 2 = a 2 + b 2 − 2 ab cos ( y ) then
a = b c o s ( y ) + c 2 − b 2 s i n 2 ( y ) a=bcos(y)+\sqrt{c^2-b^2sin^2(y)} a = b cos ( y ) + c 2 − b 2 s i n 2 ( y ) or
a = b c o s ( y ) − c 2 − b 2 s i n 2 ( y ) . a=bcos(y)-\sqrt{c^2-b^2sin^2(y)}. a = b cos ( y ) − c 2 − b 2 s i n 2 ( y ) .
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