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Could you explain finding continuity of the function on this example:

$$\ f(x) = \begin{cases} \sin(x+a), & x \in \left(n \pi, (n+\frac{1}{2})\pi\right] \\ \cos(x+b), & x \in \left((n+\frac{1}{2})\pi, (n+1)\pi\right] \end{cases} ?$$

I got confused so I would appreciate your help. I got messed at trigonometrric function.

Brian61354270
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    Welcome to MSE. Please include where you got confused, so that people can work with you accordingly. (Please add those in the body of the question instead of commenting.) – Lee David Chung Lin Feb 16 '20 at 09:20

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In order to assure of continuity, we have to obtain:

$$\ \sin(x+a) = \cos(x+b) $$ From formulas of the sum of the angles in sin and cos, we have $$\ \sin(x+a) = \cos(x+b) $$ $$\ \sin(n\pi+a) = \cos(n\pi+b) $$ fpr x = $n\pi$ $$\ \sin(n\pi+a) = \sin(n\pi)\cos(a)+\sin(a)\cos(n\pi) = (-1)^n\sin(a) $$ $$\ \cos(n\pi+b) = \cos(n\pi)\cos(a)-\sin(a)\cos(n\pi) = (-1)^n\cos(b) $$ $$\ sina=cosb $$ Similarly we act for $\ x=n\pi+\frac{\pi}{2}$ and get $\cos(a)=-\sin(b)$

$$\ \sin(x+a) = \cos(x+b) $$ is satisfied when $\ a = \frac{\pi}{2} + b + 2k\pi $ or $\ a = \frac{\pi}{2} - b + 2k\pi $

$\cos(a)=-\sin(b)$ is satisfied when $\ a = \frac{\pi}{2} + b + 2k\pi $ or $\ a = \frac{3\pi}{2} - b + 2k\pi $

Therefore the function is continuous everywhere if $\ a = \frac{\pi}{2} + b + 2k\pi $,

continuous in x = $\ n\pi $ and not in $\ x=n\pi+\frac {\pi}{2}$, if $\ a = \frac{\pi}{2} - b + 2k\pi $

continuous in $\ x=n\pi+\frac{\pi}{2}$ and not in x = $\ n\pi $, if $\ a = \frac{3\pi}{2} - b + 2k\pi = a = -\frac{\pi}{2} - b + 2k\pi $

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