A trigonometric inequality: $\cos(\theta) + \sin(\theta) > 0$

Question

How can I find $\theta$ such that $$\cos(\theta)+\sin(\theta)>0$$

Answer 1

$\sin \theta + \cos \theta = \sqrt 2 \sin (\theta + \frac{\pi}{4})$

Sine is positive in the first and second quadrants.

So the problem reduces to finding $\theta$ such that $0 < (\theta + \frac{\pi}{4}) < \pi$, giving $-\frac{\pi}{4} <\theta < \frac{3\pi}{4}$

Answer 2

In $1^{st}$ quadrant both $\sin \theta$ and $\cos \theta$ are positive.

In $2^{nd}$ quadrant from $90^{\circ}$ to $135^{\circ}$ degrees $\sin \theta > \cos \theta$ and $\sin\theta$ is positive while $\cos\theta$ is negative.

In $3^{rd}$ quadrant both are negative.

In $4^{th}$ quadrant from $315^{\circ}$ to $360^{\circ}$ $\cos \theta > \sin \theta$ and $\cos \theta$ is positive while $\sin \theta$ is negative

So your answer is $(0^{\circ}, 135^{\circ}) \cup (315^{\circ}, 360 ^{\circ})$

Answer 3

One calculus answer would go something like this:

$a)$ Find all the zeroes of the function

$b)$ Find the critical values

$c)$ If your function has zeroes at $x = a$ and $x = b$ and is continuous on $[a, b]$, if you have a relative minimum on $[a, b]$ then your function is $<0$ on that interval, and if it has a relative maximum then it is $>0$ on that interval.

So lets start with the first step:

$a)$ Let $$f(\theta) = \sin \theta + \cos \theta$$

$$0 = \sin \theta + \cos \theta$$

$$-\sin \theta = \cos \theta$$

$$\tan \theta = -1$$

From what you know about the unit circle, this happens at $\theta = \dfrac{3\pi}{4}+n\pi$. Now, for step $b$:

$b)$ $$f'(\theta) = \cos \theta - \sin \theta$$

Can you take it from here?

Answer 4

You can transform the inequation in

$$\sin(\theta)>-\sin(\frac\pi2-\theta)=\sin(\theta-\frac\pi2).$$

By continuity, the truth value of the inequality can only change at the roots*.

From the fact that angles of the same sine are equal or supplementary, with a $2k\pi$ indeterminacy, you obtain

$$\theta_k=\theta_k-\frac\pi2+2k\pi\lor \theta_k=\pi-\theta_k+\frac\pi2+2k\pi,$$

where the second equality gives

$$\theta_k=\left(\frac34+k\right)\pi.$$

Now, given that $\theta_{-1}<0<\theta_0$ and $\cos(0)+\sin(0)>0$, the inequation is verified between roots of odd and even $k$.

$$\left(\frac34+2n-1\right)\pi<\theta<\left(\frac34+2n\right)\pi.$$

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*For complete rigor, we must check that the roots do not have an even multiplicity, which would mean no change of sign. This is indeed the case as the derivatives do not coincide at the roots $$\cos(\theta_k)\ne\cos(\theta_k-\frac\pi2).$$