Solving the trig inequality $|\sin{x} + \cos{x}| > 1$ $|\sin{x} + \cos{x} |> 1$
How to solve this kind of question? Is there any websites to learn trigonometry inequalities? My teacher only taught us the simple question but not the complicated one. Thank you.
 A: Hint:
$$\sin x + \cos x = \sqrt{2}\sin \left(x + \frac{\pi}{4}\right)$$
So for your expression to be greater than $1$, is suffices to check in which intervals $$\left|\sin y \right| > \frac{1}{\sqrt{2}}$$
where $y = x + \frac{\pi}{4}$. This is not hard to do by looking at a graph of $\sin y$, remembering that $\sin y = \frac{1}{\sqrt{2}} \iff y = 2\pi k + \frac{\pi}{4}$ or $y = 2\pi k + \frac{3\pi}{4}$. 
A: $$|\sin{x} + \cos{x} |> 1$$ Square this expression then you'll get $$\sin^2x +2 \sin(x) \cos(x) +\cos^2x >1 $$ Simplify $$1+\sin2x>1$$ or $$\sin2x>0$$ Now $\sin2x$ is just a sine function but which is $\pi$ periodic, then this function is positive for $$x+ \pi k $$ where $ 0<x <\pi/2$

A: I am restricting to those $x$ when $\sin x$ and $\cos x$ are both positive.
Remember for a positive number less than $1$ its square root is bigger than itself.
Now $$\sin x + \cos x = \sin x +\sqrt{1-\sin ^2 x} > \sin x + (1 -\sin ^2 x ) $$
Note that the last term is $1 + \sin x (1-\sin x)> 1$.
A: There is an inequality that is very useful in this case: if $a,b \in \mathbb{R}$ then $$|a\sin(x)+b\cos(x)|\leq \sqrt{a^2+b^2}.$$
To prove this one can find use the Cauchy-Schwarz inequality : $$|a\sin x+b \cos x| = |(a,b)\cdot (\sin x,\cos x)|\leq \sqrt{a^2+b^2}\sqrt{\sin^2 x+\cos^2 x}=\sqrt{a^2+b^2}$$
You are being asked to solve $|\sin(x)-\cos(x)|<1$. What happens, for example, if you take $a=\frac{1}{\sqrt{2}}$ and $b=-\frac{1}{\sqrt{2}}$ in the inequality proven above?
