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

Question

So in trig, say I have an acute angle $X$. And one can intuitively conclude that $\sin x + \cos x \ge 1$, but how does the fact that $$(\sin x + \cos x)^2 = 1 + 2 \sin x \cos x$$ tell me that it is true that $\sin x + \cos x \ge 1$? I don't quite see the connection.

Thanks

Answer 1

If $x$ is acute $\sin x$ and $\cos x$ are both positive, and therefor $$2\sin x\cos x\ge 0\\ 1+2\sin x\cos x\ge 1\\ \sin^2x+\cos^2x+2\sin x\cos x\ge 1\\ \left(\sin x+\cos x\right)^2\ge 1$$ Where both $\sin x$ and $\cos x$ are positive, so is their addition: $$\sin x+\cos x\ge 1$$

Answer 2

For an acute angle $X$, $\sin X \ge 0$ and $\cos X \ge 0$. So $1 + 2 \sin X \cos X \ge 1$...

Answer 3

Let's use $\theta$ instead of $X$.

We are assuming that $0 < \theta < 90^\circ$. If we don't assume this, many of the things I say are not true.

Let $y = \sin \theta$ and $x = \cos \theta$

Now draw right triangle $ABC$ with

1. $m\angle B = \theta$

2. $x = BC$

3. $y = AC$

4. $1 = AB$

There is a theorem that says that the sum of the lengths of any two sides of a triangle exceeds the length of the third side. So you will always have $x+y> 1$. That is to say, $\sin \theta + \cos \theta > 1$ for all $\theta$ in the first quadrant.

So you don't care if $(\sin \theta + \cos \theta)^2 = 1 + 2 \sin \theta \cos \theta$ because, if $\theta$ is an acute angle, then $\sin \theta + \cos \theta > 1$