Precalculus

Trigonometry

Trigonometry Logo

Periodicity of Trigonometric Functions

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. \(\sin \frac{{17\pi }}{3}\)
  2. \(\cos \left({- \frac{{38\pi }}{3}}\right)\)
  3. \(\sec {\frac{{43\pi }}{6}}\)
  4. \(\csc \left({- \frac{{27\pi }}{4}}\right)\)

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
    \[\cos \theta = \cos \left( {\theta + {{360}^\circ} n} \right),\;\text{where } n \in \mathbb{Z}.\]
    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:

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