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.

enter image description here

$$\sin \theta = \frac {opp}{hyp} \qquad \cos \theta = \frac {adj}{hyp} \qquad \tan \theta = \frac {opp}{adj}$$

$$\csc \theta = \frac {hyp}{opp} \qquad \sec \theta = \frac {hyp}{adj} \qquad \cot \theta = \frac {adj}{opp}$$

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

enter image description here

$$\sin \theta = \frac {b}{r} \qquad \cos \theta = \frac {a}{r} \qquad \tan \theta = \frac {b}{a}$$

$$\csc \theta = \frac {r}{b} \qquad \sec \theta = \frac {r}{a} \qquad \cot \theta = \frac {a}{b}$$

Now consider another point in $2^{nd}$ quadrant

enter image description here

$$\sin \theta = \frac {b}{r} \qquad\cos \theta = \frac {-a}{r} \qquad \tan \theta = \frac {b}{-a}$$

$$\csc \theta = \frac {r}{b} \qquad \sec \theta = \frac {r}{-a} \qquad \cot \theta = \frac {-a}{b}$$

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 ???

  • $\begingroup$ I think it boils down to the fact that the rays induced by $\phi+2k\pi$ angles are isomorphic and that trigonometric functions deal with ratios of certain segments on these rays (which is well-defined because of the Intersept theorem). Maybe this shortens the argument a fair bit. $\endgroup$
    – chaosflaws
    Sep 21, 2014 at 11:46

2 Answers 2


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.

  • $\begingroup$ Thankyou for your reply @Travis. Is that the only definition we can define trigonometric function or we can define it some other way. $\endgroup$
    – Jasser
    Sep 21, 2014 at 12:15
  • $\begingroup$ There are many ways to define trigonometric functions, though the one you present here is both the most common and (typically) the first encountered. Another very useful definition, encountered in complex analysis, is that $\sin$ and $\cos$ are respectively the real and imaginary parts of the complex exponential map, $z \mapsto e^z$. $\endgroup$ Sep 21, 2014 at 12:19
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    $\begingroup$ (And, you're welcome, I'm glad to help.) $\endgroup$ Sep 21, 2014 at 12:19
  • $\begingroup$ where can i find such generalized (and even different) definitions of different functions normally used in math @Travis. $\endgroup$
    – Jasser
    Sep 21, 2014 at 12:24
  • $\begingroup$ That's a very general question, and there's no single exhaustive answer. Wikipedia is often a productive starting point, but for more details about a particular function you should consult a text that discusses it. $\endgroup$ Sep 21, 2014 at 12:32

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 $f(\theta + 2\pi) = f(\theta)$.


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