Sines of the angles

In a right-angled triangle, the sum of the cosines of the acute angles is equal to $\frac{2\sqrt3}{3}$. Calculate the product of the sines of these angles.

Peter presented the following solution to the class:

(1) Let's sketch a right-triangle:

(2) In a right-triangle, cosine is defined as the ratio of the length of the adjacent side to that of the hypotenuse. From the picture we can see that $$ \begin{gather} \cos{\alpha=\frac{b}{c}},~\cos{\beta=\frac{a}{c}} \end{gather} $$ (3) We know that $$ \cos{\alpha+\cos{\beta=\frac{2\sqrt3}{3}}} $$ i.e. $$ \frac{a+b}{c}=\frac{2\sqrt3}{3} $$

(4) Now, we square both sides of the above equality and use the Pythagorean theorem: $$ \begin{gather} \frac{a^2+2ab+b^2}{c^2}=\frac{4\cdot 3}{9} \cr \frac{c^2+2ab}{c^2}=\frac{4}{3} \end{gather} $$

(5) Finally, we simplify the resulting equality and get: $$ \begin{gather} 1+\frac{2ab}{c^2}=\frac{4}{3}\cr \frac{ab}{c^2}=-\frac{1}{6} \cr \frac{a}{c}\cdot\frac{b}{c}=-\frac{1}{6} \end{gather} $$

(6) Now, just notice that on the left side we have the product of sines that we should have calculated, i.e. $$ \sin{\alpha\cdot\sin{\beta=-\frac{1}{6}}} $$ His classmates commented on his solution as follows:

a) Luisa: Peter's solution is not quite right. He made an error in Step (5).

b) Filip: Peter's solution is quite right.

c) Ann: Petr made an error in Step (4) when squaring the equality. It should have been: $$ \frac{a^2+2ab+b^2}{c}=\frac{4\cdot 3}{3} $$

d) Sara: Peter was wrong in the definition of the cosine function. It should have been $$ \cos{\alpha=\frac{c}{b}}; \cos{\beta=\frac{c}{a}} $$

e) John: The example is not solved correctly because we lost one solution when squaring the equality in Step (4).

f) Erika: The result is wrong. The sine of each angle in a right-triangle is a positive number, so their product must be positive.

Who was right?