Precalculus

Trigonometry

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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.

The Moon revolving around the Earth
Figure 1.

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

Definition of a periodic function

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.

Proof of sine and cosine periodicity
Figure 2.

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

Periodicity of cosine
Periodicity of sine

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 \(\pi:\)

Periodicity of tangent

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 \(\pi:\)

Periodicity of cotangent

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 \(2\pi:\)

Periodicity of secant

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 \(2\pi:\)

Periodicity of cosecant

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:

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