# $(\cos \alpha, \sin \alpha)$ - possible value pairs

We introduced the complex numbers as elements of $\mathrm{Mat}(2\times 2, \mathbb{R})$ with

$$\mathbb{C} \ni x = \left(\begin{array}{cc} a & -b \\ b & a \\ \end{array}\right) = \frac{1}{\sqrt{a^2+b^2}} \left(\begin{array}{cc} \frac{a}{\sqrt{a^2+b^2}} & \frac{-b}{\sqrt{a^2+b^2}} \\ \frac{b}{\sqrt{a^2+b^2}} & \frac{a}{\sqrt{a^2+b^2}} \\ \end{array}\right)$$

Then we concluded that $0 \leq \frac{a}{\sqrt{a^2+b^2}} \leq 1$ and $0 \leq \frac{b}{\sqrt{a^2+b^2}} \leq 1$ and therefore we could find an $\alpha$ so that $\cos \alpha = \frac{a}{\sqrt{a^2+b^2}}$ and $\sin \alpha = \frac{b}{\sqrt{a^2+b^2}}$. Then we can write the matrix with $\cos$ and $\sin$ and can write it as the Euler form as well. So far so good.

My question ist about the following:

Why is it that we cand find an $\alpha$ so that $\cos \alpha = \frac{a}{\sqrt{a^2+b^2}}$ and $\sin \alpha = \frac{b}{\sqrt{a^2+b^2}}$ for every possible value combination of $a$ and $b$? Can't be there a combination of $a$ and $b$ where we can't find one and the same angle $\alpha$ so that the identites are true?

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There are two key factors here: one is the one noted, that each of these quantities is between $0$ and $1$.

But the other, which is very important, is that the sum of their squares is equal to $1$: $$\left(\frac{a}{\sqrt{a^2+b^2}}\right)^2 + \left(\frac{b}{\sqrt{a^2+b^2}}\right)^2 = 1.$$ Because the sum of their squares is equal to $1$, the point $$\left(\frac{a}{\sqrt{a^2+b^2}}, \frac{b}{\sqrt{a^2+b^2}}\right)$$ satisfies the equation $$x^2+y^2=1.$$ But this is the equation of the circle of radius $1$ with center at the origin. Every point on that circle is of the form $(\cos\alpha,\sin\alpha)$, where $\alpha$ is the angle between the positive $x$-axis and the line that goes from the origin to the point. So that means that $\alpha$ exists.

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Find an $\alpha$ so that $\cos(\alpha)= \frac{a}{\sqrt{a^2+b^2}}$. This is possible because the fraction is between $-1$ and $1$.

Then note that

$$\sin^2(\alpha)= 1- \cos^2(\alpha)=1- \frac{a^2}{a^2+b^2}=\frac{b^2}{a^2+b^2} \,.$$

Thus either $\sin(\alpha)=\frac{b}{\sqrt{a^2+b^2}}$ or $\sin(\alpha)=\frac{-b}{\sqrt{a^2+b^2}}$.

In the first situation you are done, while if $\sin(\alpha)=\frac{-b}{\sqrt{a^2+b^2}}$ then

$$\cos(-\alpha)= \frac{a}{\sqrt{a^2+b^2}} \,;\, \sin(-\alpha)=\frac{b}{\sqrt{a^2+b^2}}$$

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I suppose you could simply plot the point $(a,b)$ in the $x$-$y$ plane and let $\alpha$ be the angle formed by the positive $x$-axis and the ray joining the origin with $(a,b)$ (measured counterclockwise starting from the positive $x$-axis to the ray).

Then the distance from $(a,b)$ to the origin is $\sqrt{a^2+b^2}$ and $\cos\alpha ={a\over\sqrt{a^2+b^2}}$ and $\sin\alpha ={b\over\sqrt{a^2+b^2}}$, essentially by definition. (Use similar triangles and the unit circle definition of the trigonometric functions if you like.)

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