Why is that for any trigonometric function $f, f(2\pi + \theta )=f(\theta )$ for any value of $\theta$ [proof reading]

Question

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 ???

Answer 1

Yes, this is a common way to define trigonometric functions, and if you take this to be your definition your proof is essentially correct. Strictly speaking one probably ought to define the basic trigonometric functions for a particular radius, as in principle one must check that the ratios are independent of the radius $r$ of the circle.

It's worth noting that if one takes a circle of radius $1$, then $\sin \theta$ and $\cos \theta$ can be regarded as the (signed) lengths of the horizontal and vertical legs of the triangle you construct, rather than as merely ratios. In fact, with a little more work, and can also realize the other four standard trigonometric functions as lengths on a diagram defined by a unit circle and a given angle. Constructing these lengths is an enjoyable exercise.

Answer 2

The key point of your argument is that a $2\pi$ rotation yields a point coterminal with the original point.

If an angle $\theta$ is in standard position, the vertex is at the origin, the initial side of the angle lies on the positive x-axis, and the terminal side of the angle intersects the unit circle at the point $(\cos\theta, \sin\theta)$. Since coterminal angles intersect the unit circle at the same point, the values of $\cos\theta$ and $\sin\theta$ are the same for coterminal angles. Since the measures of coterminal angles differ by integer multiples of $2\pi$,

\begin{align*} \sin(\theta + 2\pi) & = \sin\theta\\ \cos(\theta + 2\pi) & = \cos\theta \end{align*}

Since the other trigonometric functions are defined in terms of sine and cosine, if $f$ is a trigonometric function, then $f(\theta + 2\pi) = f(\theta)$.