# Periodicity of Trigonometric Functions

### Periodic Processes and Functions

We often encounter periodic phenomena in the nature, technology, and human society. Recall the $$24\text{-hour}$$ day-night cycle, or tidal cycles caused by the Moon revolving around the Earth.

Another example is a pendulum. When a pendulum makes one complete swing over and back in $$T$$ seconds, the deflection of the pendulum from its equilibrium position will be the same at times $$t,$$ $$t + T,$$ $$t + 2T,$$ etc.

Periodic processes are described using periodic functions.

A positive real number $$T$$ is called the period of a function $$f$$ if

for all values of $$t$$ from the domain of $$f.$$

If $$T$$ is a period of a function $$f,$$ then the product $$nT,$$ where $$n \in \mathbb{Z},$$ is also a period of the function $$f:$$

$f\left( t \right) = f\left( {t + nT} \right).$

In particular for $$n = -1,$$ we have

$f\left( {t – T} \right) = f\left( t \right).$

The least positive period of a function is called the fundamental period or simply the period of the function.

### Period of Sine and Cosine Functions

The sine and cosine functions are periodic, with period $$2\pi.$$

Indeed, consider two points $$M\left( \theta \right)$$ and $$N\left( {\theta + 2\pi } \right)$$ lying on the unit circle.

These points coincide and have the same coordinates. Since the point $$M\left( \theta \right)$$ has coordinates $$\cos\theta$$ and $$\sin\theta,$$ we can write

These relationships show that $$2\pi$$ is one of the periods of sine and cosine.

Prove that $$2\pi$$ is the least period for these functions. By contradiction, suppose there is a period $$T$$ less than $$2\pi$$ for the cosine function. Then we have

$\cos \left( {\theta + T} \right) = \cos \theta .$

This identity is valid for any $$\theta,$$ so let $$\theta = 0:$$

$\cos T = \cos 0 = 1.$

The equation $$\cos T = 1$$ has the following solutions: $$T = 0, 2\pi, 4\pi, 6\pi, \ldots$$ However by our assumption, $$0 \lt T \lt 2\pi.$$ We have here a contradiction. Hence, the equation $$\cos T = 1$$ is false, and the cosine function does not have positive periods less than $$2\pi.$$

The proof for the sine function is carried out in the same way.

### Period of Other Trigonometric Functions

The tangent function has a period of $$\pi:$$

The tangent function is defined for any angles $$\theta$$ except the values where $$\cos \theta = 0,$$ that is, the values $$\large{\frac{\pi }{2}}\normalsize + \pi n,$$ $$n \in \mathbb{Z}.$$

Similarly, the period of the cotangent function is also $$\pi:$$

The cotangent function is the quotient of cosine and sine. Its domain contains all angles $$\theta$$ except the points $$\pi n, n \in \mathbb{Z},$$ where the sine function is equal to zero.

The secant and cosecant are the reciprocal functions of cosine and sine, respectively. Therefore, the secant function is periodic, with period $$2\pi:$$

It is defined for all real numbers $$\theta$$ except the points $$\large{\frac{\pi }{2}}\normalsize + \pi n,$$ $$n \in \mathbb{Z}.$$

Cosecant also has a period of $$2\pi:$$

Cosecant is defined for all values of $$\theta$$ except $$\pi n, n \in \mathbb{Z}.$$

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Calculate exact values of trigonometric functions:
1. $$\sin 750^\circ$$
2. $$\cos \left({-765^\circ}\right)$$
3. $$\tan 2760^\circ$$
4. $$\cot \left({-225^\circ}\right)$$

### Example 2

Calculate exact values of trigonometric functions:
1. $$\sin \large{\frac{{17\pi }}{3}}\normalsize$$
2. $$\cos \left({- \large{\frac{{38\pi }}{3}}\normalsize}\right)$$
3. $$\sec {\large{\frac{{43\pi }}{6}}\normalsize}$$
4. $$\csc \left({- \large{\frac{{27\pi }}{4}}\normalsize}\right)$$

### Example 3

Find the value of the expression $\cos {630^\circ} – \sin {1470^\circ} – \cot {1125^\circ}.$

### Example 4

Find the value of the expression $\tan {1800^\circ} – \sin {495^\circ} + \cos {945^\circ}.$

### Example 5

Simplify the expression $\frac{{\sin \left( { – \frac{{13\pi }}{2}} \right) + \tan \left( { – 7\pi } \right)}}{{\cos \left( { – 7\pi } \right) + \cot \left( { – \frac{{65\pi }}{4}} \right)}}.$

### Example 6

Simplify the expression $\frac{{\cos \left( { – 3\pi } \right) + \sin {\frac{{8\pi }}{3}} }}{{\tan {\frac{{9\pi }}{4}} + \cot {\frac{{13\pi }}{6}} }}.$

### Example 1.

Calculate exact values of trigonometric functions:
1. $$\sin 750^\circ$$
2. $$\cos \left({-765^\circ}\right)$$
3. $$\tan 2760^\circ$$
4. $$\cot \left({-225^\circ}\right)$$

Solution.

1. We represent the angle $$750^\circ$$ in the form ${{750^\circ} = {30^\circ} + {720^\circ} }={ {30^\circ} + {360^\circ} \times 2.}$ The sine function has a period of $$2\pi$$ or $$360^\circ.$$ Therefore, ${\sin {750^\circ} }={ \sin \left( {{{30}^\circ} + {{360}^\circ} \times 2} \right) }={ \sin {30^\circ} }={ \frac{{1}}{2}.}$
2. The angle $${-765^\circ}$$ can be written as follows: ${- {765^\circ} = {315^\circ} – {1080^\circ} }={ {315^\circ} – {360^\circ} \times 3.}$ The period of cosine is $$360^\circ.$$ Therefore, we use the identity ${\cos \theta = \cos \left( {\theta + {{360}^\circ} n} \right),\;}\kern0pt{\text{where } n \in \mathbb{Z}.}$ This yields: ${\cos \left( { – {{765}^\circ}} \right) }={ \cos \left( {{{315}^\circ} – {{360}^\circ} \times 3} \right) }={ \cos {315^\circ}.}$ The angle $$315^\circ$$ is in the $$4\text{th}$$ quadrant where cosine is positive. The reference angle of $$315^\circ$$ is equal to ${360^\circ} – {315^\circ} = {45^\circ}.$ Hence, ${\cos \left( { – {{765}^\circ}} \right) = \cos {315^\circ} }={ \cos {45^\circ} }={ \frac{{\sqrt 2 }}{2}.}$
3. We can represent the angle $$2760^\circ$$ as ${{2760^\circ} = {60^\circ} + {2700^\circ} }={ {60^\circ} + {180^\circ} \times 15.}$ The period of tangent is $$\pi$$ or $$180^\circ.$$ Using the identity ${\tan \theta = \tan \left( {\theta + {{180}^\circ} n} \right),\;}\kern0pt{\text{where } n \in \mathbb{Z},}$ we get ${\tan {2760^\circ} }={ \tan \left( {{{60}^\circ} + {{180}^\circ} \times 15} \right) }={ \tan {60^\circ} }={ {\sqrt 3}.}$
4. The angle $${-225^\circ}$$ is written as ${ – {225^\circ} = {135^\circ} – {360^\circ} }={ {135^\circ} – {180^\circ} \times 2.}$ Recall that cotangent is a periodic function with a period of $$\pi = 180^\circ,$$ that is, ${\cot \theta = \cot \left( {\theta + {{180}^\circ} n} \right),\;}\kern0pt{\text{where } n \in \mathbb{Z}.}$ Substituting the values above, we have ${\cot \left( { – {{225}^\circ}} \right) }={ \cot \left( {{{135}^\circ} – {{180}^\circ} \times 2} \right) }={ \cot {135^\circ}.}$ The angle $$135^\circ$$ lies in quadrant $$II,$$ in which cotangent is negative. The reference angle for $$135^\circ$$ is equal to $$45^\circ.$$ Then ${\cot \left( { – {{225}^\circ}} \right) }={ \cot {135^\circ} }={ – \cot {45^\circ} }={ – 1.}$

### Example 2.

Calculate exact values of trigonometric functions:
1. $$\sin \large{\frac{{17\pi }}{3}}\normalsize$$
2. $$\cos \left({- \large{\frac{{38\pi }}{3}}\normalsize}\right)$$
3. $$\sec {\large{\frac{{43\pi }}{6}}\normalsize}$$
4. $$\csc \left({- \large{\frac{{27\pi }}{4}}\normalsize}\right)$$

Solution.

1. We express the angle $$\large{\frac{{17\pi }}{3}}\normalsize$$ in the form ${\frac{{17\pi }}{3} = \frac{{5\pi }}{3} + \frac{{12\pi }}{3} }={ \frac{{5\pi }}{3} + 4\pi }={ \frac{{5\pi }}{3} + 2\pi \times 2.}$ Given that the sine is a periodic function, with period $$2\pi,$$ we obtain ${\sin \frac{{17\pi }}{3} = \sin \left( {\frac{{5\pi }}{3} + 2\pi \times 2} \right) }={ \sin \frac{{5\pi }}{3}.}$ The angle $$\large{\frac{{5\pi }}{3}}\normalsize$$ lies in the $$4\text{th}$$ quadrant and has a reference angle of $$\large{\frac{{\pi }}{3}}\normalsize.$$ The sine function is negative in this quadrant. Then ${\sin \frac{{17\pi }}{3} = \sin \frac{{5\pi }}{3} }={ – \sin \frac{\pi }{3} }={ – \frac{{\sqrt 3 }}{2}.}$
2. We represent the negative angle $${- \large{\frac{{38\pi }}{3}}\normalsize}$$ in the following way: ${- \frac{{38\pi }}{3} }={ \frac{{4\pi }}{3} – \frac{{42\pi }}{3} }={ \frac{{4\pi }}{3} – 14\pi }={ \frac{{4\pi }}{3} – 2\pi \times 7.}$ The period of cosine is $$2\pi.$$ Then ${\cos \left( { – \frac{{38\pi }}{3}} \right) }={ \cos \left( {\frac{{4\pi }}{3} – 2\pi \times 7} \right) }={ \cos \frac{{4\pi }}{3}.}$ The angle $$\large{\frac{{4\pi }}{3}}\normalsize$$ is in the $$3\text{rd}$$ quadrant, in which cosine has negative values. The reference angle for $$\large{\frac{{4\pi }}{3}}\normalsize$$ is $$\large{\frac{{\pi }}{3}}\normalsize.$$ Thus, ${\cos \left( { – \frac{{38\pi }}{3}} \right) }={ \cos \frac{{4\pi }}{3} }={ – \cos \frac{\pi }{3} }={ – \frac{1}{2}.}$
3. Here we have: ${\frac{{43\pi }}{6} = \frac{{7\pi }}{6} + \frac{{36\pi }}{6} }={ \frac{{7\pi }}{6} + 6\pi }={ \frac{{7\pi }}{6} + 2\pi \times 3.}$ The period of secant is $$2\pi.$$ Therefore, we reduce the angle value: ${\sec \frac{{43\pi }}{6} = \sec \left( {\frac{{7\pi }}{6} + 2\pi \times 3} \right) }={ \sec \frac{{7\pi }}{6}.}$ The angle $${\large{\frac{{7\pi }}{6}}\normalsize}$$ lies in the $$3\text{rd}$$ quadrant where secant is negative. Its reference angle is equal to $${\large{\frac{{\pi }}{6}}\normalsize},$$ so we have ${\sec \frac{{43\pi }}{6} = \sec \frac{{7\pi }}{6} }={ – \sec \frac{\pi }{6} }={ – \frac{2}{{\sqrt 3 }}.}$
4. The angle $$\left({- \large{\frac{{27\pi }}{4}}\normalsize}\right)$$ can be written as ${- \frac{{27\pi }}{4} = \frac{{5\pi }}{4} – \frac{{32\pi }}{4} }={ \frac{{5\pi }}{4} – 8\pi }={ \frac{{5\pi }}{4} – 2\pi \times 4.}$ Given that the period of cosecant is $$2\pi,$$ we get ${\csc \left( { – \frac{{27\pi }}{4}} \right) }={ \csc \left( {\frac{{5\pi }}{4} – 2\pi \times 4} \right) }={ \csc \frac{{5\pi }}{4}.}$ The angle $${\large{\frac{{5\pi }}{4}}\normalsize}$$ is in the $$3\text{rd}$$ quadrant, in which cosecant is negative. The reference angle of $${\large{\frac{{5\pi }}{4}}\normalsize}$$ is $${\large{\frac{{\pi }}{4}}\normalsize}.$$ Then we have ${\csc \left( { – \frac{{27\pi }}{4}} \right) }={ \csc \frac{{5\pi }}{4} }={ – \csc \frac{\pi }{4} }={ – \sqrt 2 .}$

### Example 3.

Find the value of the expression $\cos {630^\circ} – \sin {1470^\circ} – \cot {1125^\circ}.$

Solution.

Cosine and sine are periodic functions, with period $$360^\circ.$$ Therefore

${\cos {630^\circ} }={ \cos \left( {{{270}^\circ} + {{360}^\circ}} \right) }={ \cos {270^\circ} }={ 0,}$

${\sin {1470^\circ} }={ \sin \left( {{{30}^\circ} + {{1440}^\circ}} \right) }={ \sin \left( {{{30}^\circ} + {{360}^\circ} \times 4} \right) }={ \sin {30^\circ} }={ \frac{1}{2}.}$

The period of cotangent is $$180^\circ.$$ Hence,

${\cot {1125^\circ} }={ \cot \left( {{{45}^\circ} + {{1080}^\circ}} \right) }={ \cot \left( {{{45}^\circ} + {{180}^\circ} \times 6} \right) }={ \cot {45^\circ} }={ 1.}$

Substitute these values into the initial expression:

${\cos {630^\circ} – \sin {1470^\circ} – \cot {1125^\circ} }={ 0 – \frac{1}{2} – 1 }={ – \frac{3}{2}.}$

### Example 4.

Find the value of the expression $\tan {1800^\circ} – \sin {495^\circ} + \cos {945^\circ}.$

Solution.

Calculate the tangent:

${\tan {1800^\circ} }={ \tan \left( {{0^\circ} + {{180}^\circ} \times 10} \right) }={ \tan {0^\circ} }={ 0.}$

The sine function is given by

${\sin {495^\circ} }={ \sin \left( {{{135}^\circ} + {{360}^\circ}} \right) }={ \sin {135^\circ}.}$

The angle $$135^\circ$$ lies in the $$2\text{nd}$$ quadrant where sine is positive, and its reference angle is equal to $$45^\circ.$$ Then

${\sin {495^\circ} = \sin {135^\circ} }={ \sin {45^\circ} }={ \frac{{\sqrt 2 }}{2}.}$

Determine the value of cosine:

${\cos {945^\circ} }={ \cos \left( {{{45}^\circ} + {{900}^\circ}} \right) }={ \cos \left( {{{225}^\circ} + {{360}^\circ} \times 2} \right) }={ \cos {225^\circ}.}$

The angle $$225^\circ$$ is in the $$3\text{rd}$$ quadrant, in which cosine is negative. Therefore,

${\cos {945^\circ} = \cos {225^\circ} }={ – \cos {45^\circ} }={ – \frac{{\sqrt 2 }}{2}.}$

Thus,

${\tan {1800^\circ} – \sin {495^\circ} + \cos {945^\circ} }={ 0 – \frac{{\sqrt 2 }}{2} – \frac{{\sqrt 2 }}{2} }={ – \sqrt 2 .}$

### Example 5.

Simplify the expression $\frac{{\sin \left( { – \frac{{13\pi }}{2}} \right) + \tan \left( { – 7\pi } \right)}}{{\cos \left( { – 7\pi } \right) + \cot \left( { – \frac{{65\pi }}{4}} \right)}}.$

Solution.

We calculate each term separately:

${\sin \left( { – \frac{{13\pi }}{2}} \right) }={ \sin \left( {\frac{{3\pi }}{2} – \frac{{16\pi }}{2}} \right) }={ \sin \left( {\frac{{3\pi }}{2} – 8\pi } \right) }={ \sin \left( {\frac{{3\pi }}{2} – 2\pi \times 4} \right) }={ \sin \frac{{3\pi }}{2} }={ – 1,}$

${\tan \left( { – 7\pi } \right) }={ \tan \left( {0 – \pi \times 7 } \right) }={ \tan 0 }={ 0,}$

${\cos \left( { – 7\pi } \right) = \cos \left( {\pi – 8\pi } \right) }={ \cos \left( {\pi – 2\pi \times 4} \right) }={ \cos \pi }={ – 1,}$

${\cot \left( { – \frac{{65\pi }}{4}} \right) }={ \cot \left( {\frac{{3\pi }}{4} – \frac{{68\pi }}{4}} \right) }={ \cot \left( {\frac{{3\pi }}{4} – 17\pi } \right) }={ \cot \frac{{3\pi }}{4}.}$

The angle $$\large{\frac{{3\pi }}{4}}\normalsize$$ lies in the $$2\text{nd}$$ quadrant, in which cotangent is negative. The reference angle of $$\large{\frac{{3\pi }}{4}}\normalsize$$ is equal to $$\large{\frac{{\pi }}{4}}\normalsize,$$ so

${\cot \left( { – \frac{{65\pi }}{4}} \right) }={ \cot \frac{{3\pi }}{4} }={ – \cot \frac{\pi }{4} }={ – 1.}$

Hence,

${\frac{{\sin \left( { – \frac{{13\pi }}{2}} \right) + \tan \left( { – 7\pi } \right)}}{{\cos \left( { – 7\pi } \right) + \cot \left( { – \frac{{65\pi }}{4}} \right)}} }={ \frac{{ – 1 + 0}}{{ – 1 – 1}} }={ \frac{1}{2}.}$

### Example 6.

Simplify the expression $\frac{{\cos \left( { – 3\pi } \right) + \sin {\frac{{8\pi }}{3}} }}{{\tan {\frac{{9\pi }}{4}} + \cot {\frac{{13\pi }}{6}} }}.$

Solution.

Find the value of each term:

${\cos \left( { – 3\pi } \right) = \cos \left( {\pi – 4\pi } \right) }={ \cos \left( {\pi – 2\pi \times 2} \right) }={ \cos \pi }={ – 1,}$

${\sin \frac{{8\pi }}{3} }={ \sin \left( {\frac{{2\pi }}{3} + \frac{{6\pi }}{3}} \right) }={ \sin \left( {\frac{{2\pi }}{3} + 2\pi } \right) }={ \sin \frac{{2\pi }}{3}.}$

The reference angle for $$\large{\frac{{2\pi }}{3}}\normalsize$$ is $$\large{\frac{{\pi }}{3}}\normalsize.$$ Then

${\sin \frac{{8\pi }}{3} = \sin \frac{{2\pi }}{3} }={ \sin \frac{\pi }{3} }={ \frac{{\sqrt 3 }}{2}.}$

The other terms are given by

${\tan \frac{{9\pi }}{4} }={ \tan \left( {\frac{\pi }{4} + \frac{{8\pi }}{4}} \right) }={ \tan \left( {\frac{\pi }{4} + 2\pi } \right) }={ \tan \frac{\pi }{4} }={ 1,}$

${\cot \frac{{13\pi }}{6} }={ \cot \left( {\frac{\pi }{6} + \frac{{12\pi }}{6}} \right) }={ \cot \left( {\frac{\pi }{6} + 2\pi } \right) }={ \cot \frac{\pi }{6} }={ \sqrt 3 .}$

Substitute the found values:

${\frac{{\cos \left( { – 3\pi } \right) + \sin \left( {\frac{{8\pi }}{3}} \right)}}{{\tan \left( {\frac{{9\pi }}{4}} \right) + \cot \left( {\frac{{13\pi }}{6}} \right)}} }={ \frac{{ – 1 + \frac{{\sqrt 3 }}{2}}}{{1 + \sqrt 3 }} }={ \frac{{ – 2 + \sqrt 3 }}{{2\left( {1 + \sqrt 3 } \right)}} }={ \frac{{\left( { – 2 + \sqrt 3 } \right)\left( {1 – \sqrt 3 } \right)}}{{2\left( {1 + \sqrt 3 } \right)\left( {1 – \sqrt 3 } \right)}} }={ \frac{{ – 2 + \sqrt 3 + 2\sqrt 3 – 3}}{{2\left( {{1^2} – {{\left( {\sqrt 3 } \right)}^2}} \right)}} }={ \frac{{3\sqrt 3 – 5}}{{2\left( {1 – 3} \right)}} }={ \frac{{5 – 3\sqrt 3 }}{4}.}$