Periodicity of Trigonometric Functions

Solved Problems

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
   \[\cot \left( { - {{225}^\circ}} \right) = \cot {135^\circ} = - \cot {45^\circ} = - 1.\]

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
   \[\csc \left( { - \frac{{27\pi }}{4}} \right) = \csc \frac{{5\pi }}{4} = - \csc \frac{\pi }{4} = - \sqrt 2 .\]

Solution.

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

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

Substitute these values into the initial expression:

Solution.

Calculate the tangent:

The sine function is given by

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

Determine the value of cosine:

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

Thus,

Solution.

We calculate each term separately:

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

Hence,

Solution.

Find the value of each term:

The reference angle for \(\frac{{2\pi }}{3}\) is \(\frac{{\pi }}{3}.\) Then

The other terms are given by

Substitute the found values: