Prove: if $x^2 + y^2 < \frac{1}{2}$, then $\cos(x+y) > 0$

Question

For an exercise about finding the minimum of a function I have to prove that $cos(x+y) > 0$ for $x^2 + y^2 < \frac{1}{2}$. Which means that $-\frac{\pi}{2}< x + y < \frac{\pi}{2}$. However, I am struggling with drawing any conclusions about $x + y$ from the expression with the squares. Does anyone have tips for that?

only thing I have so far, is this: $-\frac{\pi}{2} < -\frac{1}{\sqrt{2}} < -\sqrt{x^2 + y^2} \leq \sqrt{x^2 + y^2} < \frac{1}{\sqrt{2}} < \frac{\pi}{2}$

Answer 1

We have that

$$x^2 + y^2 < \frac{1}{2} \implies |x| < \frac{\sqrt 2}2 \;\land \; |y| <\frac{\sqrt 2}2$$

then

$$-\frac \pi 2< -\sqrt 2< -|x|-|y|\le x+y \le |x|+|y| < \sqrt 2< \frac \pi 2$$

therefore $\cos (x+y)>0$.

Answer 2

It may be easier drawing the circle. It is easy to see that $f(x, y) = x + y$ is maximized in the disk when $x = y$. This means $2x^2 = \frac{1}{2}$ and $x = y = \frac{1}{2}$. Repeating the process for the minimum, you get that $-1 \leq f(x, y) \leq 1$ and you immediately get the result you want

Answer 3

Use polar coordinate,

$$x^2+y^2<\frac12\Leftrightarrow r<\frac{1}{\sqrt2}$$

hence

$$x+y=r\cos\theta+r\sin\theta=\sqrt2 ~r\sin\left(\theta+\frac{\pi}4\right)$$

we get

$$-\frac{\pi}2<-1<-\sqrt2 ~r\le x+y\le\sqrt2 ~r<1<\frac{\pi}2$$

So

$$\cos(x+y)>0$$

Answer 4

We have $\left(\frac{x+y}{2}\right)^2\le\frac{x^2+y^2}{2}$, so $(x+y)^2\le 2(x^2+y^2)<1$. So $|x+y|<1$. Then $\cos(x+y)=\cos(|x+y|)>\cos 1>\cos\frac{\pi}{2}=0$ because $\cos$ is decreasing in the interval $[0,\pi]$.