If we know that two acute angles have the same value for some trigonometric function, then we know the angles are equal .

Question

This is an excerpt from a book that I'm reading.

If we know that two acute angles have the same value for some trigonometric function (e.g. $\sin \alpha = \sin\beta$), then we know the angles are equal ($\alpha = \beta$). (Can you prove this?)

In order to prove the assertion, I squared both sides of the equation to get

$\sin^2\alpha = \sin^2\beta$

Then I added $\cos^2\alpha$ to both sides of the equation, so that

$\sin^2\alpha$ + $\cos^2\alpha$ = $\sin^2\beta$ + $\cos^2\alpha$

Then since $\sin^2\alpha$ + $\cos^2\alpha$ = $1$, it follows that $\sin^2\beta$ + $\cos^2\alpha$ = $1$.

From the trig identity, $\sin^2\theta$ + $\cos^2\theta$ = $1$, we find that $\theta = \beta = \alpha$

I wanted to know if anything needs to be corrected?

Answer 1

How can you conclude that $\beta=\alpha$ from $\sin^2\beta+\cos^2\alpha=1$, without using the mentioned assertion?

You only need the fact that all the six trigonometric functions are strictly monotonic on $\displaystyle \left(0,\frac{\pi}{2}\right)$.

For example, $\cos x$ is strictly decreasing on $\displaystyle \left(0,\frac{\pi}{2}\right)$. Suppose that $\cos \alpha=\cos\beta$ for some acute angles $\alpha$ and $\beta$. $\alpha$ cannot be smaller than $\beta$, as it would implies that $\cos\alpha>\cos\beta$. Similarly, $\alpha$ cannot be larger than $\beta$, as it would implies that $\cos\alpha<\cos\beta$. The only possibiility left is $\alpha=\beta$.