# 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.

• The inequality in the title differs from the one in the body. You may want to fix that. Jul 8, 2016 at 11:15

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 $y = 2\pi k + \frac{3\pi}{4}$.

$$|\sin{x} + \cos{x} |> 1$$ 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 $0<x <\pi/2$

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 $1 + \sin x (1-\sin x)> 1$.

• How would this approach work if $\sin x\cdot\cos x<0$? Jul 8, 2016 at 11:04
• Haven't figured that out yet! Jul 8, 2016 at 11:19
• @TZakrevskiy: Omar Nagib has answered your doubt. The claim is false if $x>pi/2$. I checked with calclulator (radians) $\sin(2) + \cos (2)\approx 0.49315$ Jul 8, 2016 at 11:30

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?