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

*

*$m\angle B = \theta$


*$x = BC$


*$y = AC$


*$1 = AB$
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 $\sin \theta + \cos \theta > 1$
