"If $|\sin x + \cos x |=|\sin x|+|\cos x| (\sin x, \cos x \neq 0)$, in which quadrant does $x$ lie?"

Question

"If $|\sin x + \cos x |=|\sin x|+|\cos x| (\sin x, \cos x \neq 0)$, in which quadrant does $x$ lie?"

What I did: For this equality, signs of $\sin x$ and $\cos x$ must be same. Therefore $\sin x\cos x>0$ Multiplying and dividing by $2$, $\sin 2x>0$ $$0<2x<\pi$$ $$0<x<\frac{π}{2}$$

That gives the value of $x$ in the first quadrant.

But the sign of $\sin$ and $\cos$ functions are also the same in the third quadrant, so where did I go wrong in my solution?

Thanks!

Answer 1

Let $\theta = 2x$.

If $0 < x < 2\pi$, then $0 < 2x < 4\pi$, so you need to check for angles between $0$ and $4\pi$ in which $\sin\theta > 0$.

$$\sin\theta > 0 \implies 0 < \theta < \pi~\vee~2\pi < \theta < 3\pi$$

Replace $\theta$ by $2x$. The inequality

$$0 < \theta < \pi \implies 0 < 2x < \pi \implies 0 < x < \frac{\pi}{2}$$

while the inequality

$$2\pi < \theta < 3\pi \implies 2\pi < 2x < 3\pi \implies \pi < x < \frac{3\pi}{2}$$

so we obtain angles in the first and third quadrants as solutions.

Answer 2

$$\sin2x>0\iff \color{green}{2k\pi}<2x<\pi\color{green}{+2k\pi}\iff \color{green}{k\pi}<x<\frac\pi2\color{green}{+k\pi}.$$

Answer 3

We know that $|a+b|=|a|+|b|$ iff $a$ and $b$ share the same sign. Now if $(a,b)$ is a point in the coordinate system satisfying that property is must be in quadrant $1$ or $3$.