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

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

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: $$\sin x+\cos x\ge 1$$

• But just because sinX and cosX are greater than 0 doesn't guarantee that their addition is greater than 0?
– Kat
Oct 13, 2013 at 3:50
• @Kat Order is preserved in addition so if $a>0$ and $b>0$ then $a+b>0+0=0$. Now if $c=a+b$ is positive, and $c^2\ge1^2=1$, then $c\ge1$. (If $c$ is negative, then $c\le-1$.) Oct 13, 2013 at 4:01

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