Here was the question asked to me :: Why is that for any trigonometric function $f, f(2\pi + \theta )=f(\theta )$ for any value of $\theta$
I spontaneously said that it was because of their very definition.
Here is the complete description of answer i provided.
Normally we define trigonometric functions for acute angles using right angled triangle as follows.
$$\sin \theta = \frac {opp}{hyp} \qquad \cos \theta = \frac {adj}{hyp} \qquad \tan \theta = \frac {opp}{adj}$$
$$\csc \theta = \frac {hyp}{opp} \qquad \sec \theta = \frac {hyp}{adj} \qquad \cot \theta = \frac {adj}{opp}$$
But we can extend our definitions for even other angles using point $P(a,b)$ on cartesian plane as follows
Definition of trigonometric functions as i remember
Consider a point in cartesian plane in $1^{st}$ quadrant as shown
$$\sin \theta = \frac {b}{r} \qquad \cos \theta = \frac {a}{r} \qquad \tan \theta = \frac {b}{a}$$
$$\csc \theta = \frac {r}{b} \qquad \sec \theta = \frac {r}{a} \qquad \cot \theta = \frac {a}{b}$$
Now consider another point in $2^{nd}$ quadrant
$$\sin \theta = \frac {b}{r} \qquad\cos \theta = \frac {-a}{r} \qquad \tan \theta = \frac {b}{-a}$$
$$\csc \theta = \frac {r}{b} \qquad \sec \theta = \frac {r}{-a} \qquad \cot \theta = \frac {-a}{b}$$
As you can see we consider the signed values of x and y in the definition keeping in mind that r is always greater than zero.
similarly we can extend this definition (in other quadrants) for other angles.
so if we rotate the line passing through origin and $P(a,b)$ through an angle of $360^o$ we again reach the same point and hence $f(2\pi + \theta )=f(\theta )$ where f is any trigonometric function.
Is my arguement for the question given to me correct ???