Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

3 Answers 3

up vote 5 down vote accepted

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.

share|improve this answer

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

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

share|improve this answer

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.)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.