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Let $a,b\in\mathbb{R}$ such that $(a,b)\neq(0,0)$. Let $x$ be a real number.
We make a function such that $A(x)= a\cos(x)+b\sin(x)$. I need to show the existence of a number $\alpha\in\mathbb{R}$ such that:
$$\sin\alpha=\frac{b}{\sqrt{a^2+b^2}}; \cos\alpha=\frac{a}{\sqrt{a^2+b^2}}$$ I thought of using the fact that the sum of the square of the two numbers is equal to one is enough to deduce the existence but I feel like I am wrong. Do you guys have any hints?

Thomas Andrews
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CHOSM
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    $$\left(\frac{a}{\sqrt{a^2+b^2}}\right)^2+\left(\frac{b}{\sqrt{a^2+b^2}}\right)^2=1.$$ – Surb Mar 21 '22 at 19:25
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    That and the fact that both numbers are in $[-1,1]$ is enough to finish. – Ian Mar 21 '22 at 19:27
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    appears you've had calculus. Sine and cosine are continuous, also $a/ \sqrt{a^2 + b^2}$ lies between $-1$ and $1$ inclusive. There is a real $t$ with $\cos t$ equal to it. Then either $\pm t$ has the correct sine – Will Jagy Mar 21 '22 at 19:29
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    @Ian: If I'm not mistaken, if $u,v\in \mathbb R$ are such that $u^2+v^2=1$, you implicitly have that $u,v\in [-1,1]$... – Surb Mar 21 '22 at 19:29
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    @Surb Sure, this is just covering the fact that $\frac{a}{\sqrt{a^2+b^2}}$ and $\frac{b}{\sqrt{a^2+b^2}}$ are also real, which is trivial, but technically not immediately given. – Ian Mar 21 '22 at 19:32

2 Answers2

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Note that $\sin(x)$ is continuous in $x$, and that there is an $x'$ such that $\sin(x')=1$, and there is an $y$ such that $\sin(y)=-1$. So for any $c \in [-1,1]$, there is an $x$ such that $\sin(x)=c$. Now, for any such $a,b$, note that $\frac{b}{\sqrt{a^2+b^2}}$ is in $[-1,1]$, so there indeed is an $\alpha'$ such that $\sin(\alpha')= \frac{b}{\sqrt{a^2+b^2}}$.

Now, we are almost done. Indeed, the equation $\sin^2(\alpha')+\cos^2(\alpha')=1$ must hold, so from this it follows that $\cos(\alpha')$ is $\frac{\pm a}{\sqrt{a^2+b^2}}$, where $\alpha'$ is as in the previous pragraph. If on the one hand $\alpha'$ is such that $\cos(\alpha') = \frac{a}{\sqrt{a^2+b^2}}$, thenset $\alpha=\alpha'$ and we are done.

If on the other hand $\alpha'$ is such that $\cos(\alpha') = \frac{-a}{\sqrt{a^2+b^2}}$, then set $\alpha = \frac{\pi}{2}-\alpha$; then $\sin(\alpha)=\sin(\alpha')$ $=\frac{b}{\sqrt{a^2+b^2}}$, and $\cos(\alpha)=-\cos(\alpha') = -\frac{-a}{\sqrt{a^2+b^2}} = \frac{a}{\sqrt{a^2+b^2}}$, and so now we are done here as well.

Mike
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If you are interested to what should be the right angle, take the following one

$\alpha=arc tg(\frac{b}{a})$

if $a\neq 0$, otherwise $\alpha=\frac{\pi}{2}$

To prove your identity you have to resolve the system

$\begin{cases} sin(\alpha)=\frac{b}{a}cos(\alpha)\\ cos^2(\alpha)(1+(\frac{b}{a})^2)=1 \end{cases}$

Federico Fallucca
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  • If you aren't convinced there is an angle $\alpha$ that gives you $\sin(\alpha)=\frac{b}{\sqrt{a^2+b^2}}$ and $\cos(\alpha)=\frac{a}{\sqrt{a^2+b^2}}$, then why would you believe $\arctan(b/a)$ exists [for nonzero $a$ that is]. And then even given such an $\alpha$ you could still be off by a factor of $-1$ for both $\sin(\alpha)$ and $\cos(\alpha)$ anyway. – Mike Mar 21 '22 at 19:57
  • @Mike it always exists because the point (a/sqrt() , b/sqrt()) is on the trigonometric circumference and any point there is identified by an angle. I wanted just to show that there exists an exact solution for the angle – Federico Fallucca Mar 21 '22 at 20:03
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    I believe that is what the OP was trying to show though. – Mike Mar 21 '22 at 20:13