This question is from an exam paper:
Given that $\alpha$ and $\beta$ are acute angles, show that $\alpha+\beta = \pi/2$ if and only if $\cos^2{\alpha} +\cos^2{\beta} = 1$.
I want to do it in one line rather than doing both directions separately. Is this a valid proof?
$\cos^2{\alpha}+\cos^2{\beta} =\cos^2{\alpha} +\sin^2(\pi/2-\beta)=\cos^2{\alpha}+\sin^2{\alpha}=1$
Edit: Thanks, I now realise I've only proven the 'only if' part. Here is my attempt at the 'if' part, which was more complicated than I thought it would be.
$\begin{align}
&&\cos^2{\alpha} +\cos^2{\beta} = \cos^2{\alpha} +\sin^2(\pi/2-\beta)=1\\
\implies&&\sin^2(\pi/2-\beta)=\sin^2{\alpha}\\
\implies&&\pi/2-\beta=\pm\alpha+\pi n\\
\implies&&\alpha+\beta=\pi/2+\pi n\end{align}$
$\alpha$ and $\beta$ are acute so $0<\alpha + \beta<\pi$ so $n=0$.
Best Answer
The condition that $\alpha$ and $\beta$ are acute implies that the cosines are positive, then $\cos^2{\alpha} +\cos^2{\beta} = 1$ implies $\cos{\alpha} +\cos{\beta} \ge 1$.
Hence we can construct a triangle with sides $1,\cos{\alpha},\cos{\beta}$.
Now, by Pythagoras' theorem, either of the statements $\alpha+\beta = \pi/2$ and $\cos^2{\alpha} +\cos^2{\beta} = 1$ is equivalent to the fact that this triangle is right-angled.