There is the well-known trigonometric identity: $\sin^2(x)+\cos^2(x) = 1$. Without giving a second thought I took use of it to convert from cosine to sine. Especially in physics involving reflections: $\frac{n_2}{n_1}\,\cos(\alpha_2) = \frac{n_2}{n_1}\,\sqrt{1-\sin^2(\alpha_2)}$. This helped many times. Just now I realized that plotting both sides of the equation actually lead to two different graphs. Even plotting $\pm\sqrt{1-\sin^2(\alpha_2)}$ does not equal the simple cosine. It seems like there has to be a piecewise function defined giving the positive square for positive $\alpha_2$ and vice versa. But why the first substitution without any case differentiation works in physical context?

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    $\begingroup$ You are restricting the domain where $\alpha_2$ is defined. That formula is true if the angle is between $0$ and $90^\circ$. $\endgroup$
    – Andrei
    Aug 17, 2021 at 16:50
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    $\begingroup$ Oh right, that's always the provided domain in reflection-physics. $\endgroup$
    – Leon
    Aug 17, 2021 at 16:54
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    $\begingroup$ Perhaps $\sin(\pi/2-x)=\cos(x)$, which holds for all x, will interest you, as it graphs nicely $\endgroup$
    – FShrike
    Aug 17, 2021 at 17:11

2 Answers 2

  1. $$f(\alpha_2)=\cos(\alpha_2)$$ is a function because each input $\alpha_2$ returns a certain corresponding output $f(\alpha_2).$ Likewise for the function $$g(\alpha_2)=\sqrt{1-\sin^2(\alpha_2)}.$$ As $\alpha_2$ varies over the real numbers (let's work in radians), the values of $f(\alpha_2)$ and $g(\alpha_2)$ coincide precisely when $\alpha_2$ lies in the $1$st or $4$th quadrant (e.g., when $\alpha_2$ is acute). In this case, ($\cos\alpha_2$ is non-negative and) $$\cos(\alpha_2)=\sqrt{1-\sin^2(\alpha_2)}\tag1$$ instead of the more general $$\cos(\alpha_2)=\pm\sqrt{1-\sin^2(\alpha_2)}\tag2.$$
  2. On the other hand, $$h(\alpha_2)=\pm\sqrt{1-\sin^2(\alpha_2)}$$ is not a function: its graph is symmetrical in the $x-$axis because each input value of $\alpha_2$ corresponds to two possible outputs.
  3. The equality $$\cos^2(\alpha_2)=1-\sin^2(\alpha_2)$$ is called an identity because it is true for every possible value of $\alpha_2$. On the other hand, $(1)$ and $(2)$ are equations that are true for particular values of $\alpha_2$ (and which we are using to solve for $\alpha_2,$ for conversion between trigonometric ratios, etc.).

Let's start at the beginning.


$$\sin^2(x)+\cos^2(x) = 1$$


$$ \cos^2(x) - (1 - \sin^2(x)) = 0 $$

$$ \cos^2(x) - \left(\sqrt{1 - \sin^2(x)}\right)^2 = 0 $$

$$ \left(\cos(x) - \sqrt{1 - \sin(x)}\right) \left(\cos(x) + \sqrt{1 - \sin(x)}\right) = 0 $$

So the logic here is as follows. We know that $\sin^2(x)+\cos^2(x) = 1$ is true. It means that for all x, at least one of the following two statements has to be true:

(1) $\cos(x) = \sqrt{1-\sin^2(x)}$

(2) $\cos(x) = - \sqrt{1-\sin^2(x)}$

And it can be a different statement for a different x! And (based on some more other logical conclusions which you could go through on your own) it just so happens that:

  • for all $x \in [-\frac{\pi}{2} + 2\pi k; \frac{\pi}{2} + 2\pi k]$ ($k \in Z $), (1) is true,
  • and for all $x \in [\frac{\pi}{2} + 2\pi k; \frac{3\pi}{2} + 2\pi k]$ ($k \in Z $), (2) is true.

For example, if in some formula you were to say that for some $x \in (\frac{\pi}{2}; \frac{3\pi}{2})$, $\cos(x) = \sqrt{1-\sin^2(x)}$, this would be a mistake which would actually flip the sign of your expression.

So be careful and always check the domain in these kinds of situations.)


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