# Periodicity of Trigonometric Functions

## Solved Problems

### 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
$\cos \theta = \cos \left( {\theta + {{360}^\circ} n} \right),\;\text{where } n \in \mathbb{Z}.$
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
$\tan \theta = \tan \left( {\theta + {{180}^\circ} n} \right),\;\text{where } n \in \mathbb{Z},$
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,
$\cot \theta = \cot \left( {\theta + {{180}^\circ} n} \right),\;\text{where } n \in \mathbb{Z}.$
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. $$\sin \frac{{17\pi }}{3}$$
2. $$\cos \left({- \frac{{38\pi }}{3}}\right)$$
3. $$\sec {\frac{{43\pi }}{6}}$$
4. $$\csc \left({- \frac{{27\pi }}{4}}\right)$$

Solution.

1. We express the angle $$\frac{{17\pi }}{3}$$ 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 $$\frac{{5\pi }}{3}$$ lies in the $$4\text{th}$$ quadrant and has a reference angle of $$\frac{{\pi }}{3}.$$ 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 $${- \frac{{38\pi }}{3}}$$ 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 $$\frac{{4\pi }}{3}$$ is in the $$3\text{rd}$$ quadrant, in which cosine has negative values. The reference angle for $$\frac{{4\pi }}{3}$$ is $$\frac{{\pi }}{3}.$$ 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 $${\frac{{7\pi }}{6}}$$ lies in the $$3\text{rd}$$ quadrant where secant is negative. Its reference angle is equal to $${\frac{{\pi }}{6}},$$ so we have
$\sec \frac{{43\pi }}{6} = \sec \frac{{7\pi }}{6} = - \sec \frac{\pi }{6} = - \frac{2}{{\sqrt 3 }}.$
4. The angle $$\left({- \frac{{27\pi }}{4}}\right)$$ 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 $${\frac{{5\pi }}{4}}$$ is in the $$3\text{rd}$$ quadrant, in which cosecant is negative. The reference angle of $${\frac{{5\pi }}{4}}$$ is $${\frac{{\pi }}{4}}.$$ 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 $$\frac{{3\pi }}{4}$$ lies in the $$2\text{nd}$$ quadrant, in which cotangent is negative. The reference angle of $$\frac{{3\pi }}{4}$$ is equal to $$\frac{{\pi }}{4},$$ 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 $$\frac{{2\pi }}{3}$$ is $$\frac{{\pi }}{3}.$$ 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:

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