Chapter 3.14 - Period of Trigonometric Functions

In the previous section, we completed a discussion on trigonometric identities. We saw some solved example also. In this section, we will see period of trigonometric functions.

Some basics can be written in 9 steps:
1. Fig.3.37 below shows the familiar unit circle. We see that, a ray is making an angle of $\frac{\pi}{6}$ with the positive side of the x axis.

Fig.3.37

• P₁ is the tip of this ray. The coordinates of P₁ are:$\left( \frac{\sqrt 3}{2}, \frac{1}{2}\right)$.
• So we get: $\sin \frac{\pi}{6}=\frac{1}{2}$
2. Another ray is making an angle of $\frac{5\pi}{6}$ with the positive side of the x axis.
• P₂ is the tip of this ray. The coordinates of P₂ are:$\left(- \frac{\sqrt 3}{2}, \frac{1}{2}\right)$.
• So we get: $\sin \frac{5\pi}{6}=\frac{1}{2}$
3. We see that, sine of both $\frac{\pi}{6}$ and $\frac{5\pi}{6}$ is $\frac{1}{2}$.
• Using function notations, we can write:
   ♦ f(x) = sin x is a trigonometric function
   ♦ If the input x is $\frac{\pi}{6}$, we will get an output of $\frac{1}{2}$
   ♦ If the input x is $\frac{5\pi}{6}$, then also we will get the same output of $\frac{1}{2}$
4. Let P₁ be the initial position.
   ♦ The output f(x) here is $\frac{1}{2}$.
• From that initial position, the ray turned through an angle of $\frac{4\pi}{6}$ to reach the second position P₂
   ♦ The output f(x) here is the same $\frac{1}{2}$.
5. From that second position P₂, if the ray turn through $\frac{4\pi}{6}$, will we get the same $\frac{1}{2}$ one more time?
• Let us check. It can be written in 4 steps:
(i) From P₂, if the ray turn through $\frac{4\pi}{6}$, it will reach P₃ shown in the fig.
(ii) At P₃, the total angle x is: $\left(\frac{5\pi}{6}+\frac{4\pi}{6}\right)=\frac{9\pi}{6}=\frac{3\pi}{2}$.
(iii) The coordinates of P₃ are: (0,-1).
• So we get: $\sin \frac{3\pi}{2}= -1$
(iv) This time we do not get the same '$\frac{1}{2}$'. We are getting '-1' instead.
6. Let us write a summary. It can be written in 3 steps:
(i) The initial input x given for f(x) = sin x is $\frac{\pi}{6}$
   ♦ The output is $\frac{1}{2}$
(ii) The next input x given is:
$\left(x+\frac{4\pi}{6}\right)= \left(\frac{\pi}{6}+\frac{4\pi}{6}\right)=\frac{5\pi}{6}$
   ♦ This input x also gives an output of $\frac{1}{2}$
(iii) We get a feeling that, we can repeatedly add $\frac{4\pi}{6}$, all giving the same output $\frac{1}{2}$
• But when $\frac{4\pi}{6}$ is added the second time, the output becomes -1. The output is not $\frac{1}{2}$
• That means, $\frac{4\pi}{6}$ cannot be accepted as the period of the function f(x) = sin x
7. We just saw why $\frac{4\pi}{6}$ has to be discarded. Let us try another initial angle. It can be written in 3 steps:
(i) Let the initial input x given for f(x) = sin x be $\frac{\pi}{3}$
   ♦ The output will be $\frac{\sqrt 3}{2}$
(ii) Let us add $\frac{\pi}{3}$. So the next input x given is:
$\left(x+\frac{\pi}{3}\right)= \left(\frac{\pi}{3}+\frac{\pi}{3}\right)=\frac{2\pi}{3}$
   ♦ This input x also gives an output of $\frac{\sqrt 3}{2}$
(iii) We get a feeling that, we can repeatedly add $\frac{\pi}{3}$, all giving the same output $\frac{\sqrt 3}{2}$
• When $\frac{\pi}{3}$ is added the second time, the input x is:
$\left(\frac{2\pi}{3}+\frac{\pi}{3}\right)=\pi$
• The output f(x) is equal to sin π = 0. The output is not $\frac{\sqrt 3}{2}$
• That means, $\frac{\pi}{3}$ cannot be accepted as the period of the function f(x) = sin x
8. In this way we can try various initial angles. We will find that only 2π is eligible to be accepted as the period of f(x) = sin x
• This is obvious because, 2π indicates one revolution. We have seen that, the sine value repeats after successive revolutions.
• The following table 3.2 is generated using a computer. It clearly shows the effect of the period 2π. (This table can be prepared using standard tables also)

Table 3.2

◼ In the table, we see that:
• sine of $\frac{\pi}{6}$ is 0.5
• When 2π is added, x becomes $\frac{13\pi}{6}$. But the sine remains at 0.5
• When another 2π is added, x becomes $\frac{25\pi}{6}$. But the sine remains at 0.5
• When another 2π is added, x becomes $\frac{37\pi}{6}$. But the sine remains at 0.5
9. We can write: f(x) = sin x = sin (2nπ+x) where n is any integer.
• The smallest possible positive value of 2nπ is obtained when n = +1
• When n = +1, we get: 2nπ = 2pi
• This smallest possible positive value is the period.

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• In the same way, the function f(x) = cos x can be analyzed.
• We will find that for the cosine function also, the period is 2π. The last row of the above table 3.2 demonstrates this result. The reader may write all the steps for the cosine function in his/her own note books.
Hint:
   ♦ Let initial value of x be $\frac{\pi}{3}$. Then $\cos \frac{\pi}{3}=\frac{1}{2}$
   ♦ Add $\frac{\pi}{3}$ to x.
         ✰ Then $\cos \left(\frac{\pi}{3}+\frac{\pi}{3} \right)=\cos \frac{2\pi}{3}=\frac{1}{2}$
   ♦ It appears that, the period of the cosine function is $\frac{\pi}{3}$.
   ♦ But $\cos \left(\frac{2\pi}{3}+\frac{\pi}{3} \right)=\cos \frac{3\pi}{3}=\cos \pi = -1$
   ♦ It is not equal to $\frac{1}{2}$

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Next we will see the period of tangent function. It can be written in 9 steps:
1. Fig.3.38 below shows the familiar unit circle. We see that, a ray is making an angle of $\frac{\pi}{6}$ with the positive side of the x axis.

Fig.3.38

• P₁ is the tip of this ray. The coordinates of P₁ are:$\left( \frac{\sqrt 3}{2}, \frac{1}{2}\right)$.
• So we get: $\tan \frac{\pi}{6}=\frac{\frac{1}{2}}{\frac{\sqrt 3}{2}}=\frac{1}{\sqrt 3}$
2. Another ray is making an angle of $\frac{7\pi}{6}$ with the positive side of the x axis.
• P₂ is the tip of this ray. The coordinates of P₂ are:$\left(- \frac{\sqrt 3}{2}, -\frac{1}{2}\right)$.
• So we get: $\tan \frac{7\pi}{6}=\frac{\frac{1}{2}}{\frac{\sqrt 3}{2}}=\frac{1}{\sqrt 3}$.
3. We see that, tangent of both $\frac{\pi}{6}$ and $\frac{7\pi}{6}$ is $\frac{1}{\sqrt 3}$.
• Using function notations, we can write:
   ♦ f(x) = tan x is a trigonometric function
   ♦ If the input x is $\frac{\pi}{6}$, we will get an output of $\frac{1}{\sqrt 3}$
   ♦ If the input x is $\frac{7\pi}{6}$, then also we will get the same output of $\frac{1}{\sqrt 3}$
4. Let P₁ be the initial position.
   ♦ The output f(x) here is $\frac{1}{\sqrt 3}$.
• From that initial position, the ray turned through an angle of $\frac{7\pi}{6}-\frac{\pi}{6}=\pi$ to reach the second position P₂
   ♦ The output f(x) here is the same $\frac{1}{\sqrt 3}$.
5.From that second position P₂, if the ray turn through π, will we get the same $\frac{1}{\sqrt 3}$ one more time?
• Let us check. It can be written in 5 steps:
(i) From P₂, if the ray turn through π, it will reach back at the initial position P₁.
(ii) When this point is reached, the total angle is:
$\left(\frac{7\pi}{6}+\pi\right)=\frac{13\pi}{6}$
(iii) The coordinates here are already known.
• So we get: $\tan \frac{13\pi}{6}=\frac{1}{\sqrt 3}$
(iv) This time also, we get the same '$\frac{1}{\sqrt 3}$'.
(v) If from this third point, we make another turn of π, we will reach the fourth point, which will be same as the second point. There also, we will get:
$\tan \frac{19\pi}{6}=\frac{1}{\sqrt 3}$
6. Let us write a summary. It can be written in 3 steps:
(i) The initial input x given for f(x) = sin x is $\frac{\pi}{6}$
   ♦ The output is $\frac{1}{\sqrt 3}$
(ii) The next input x given is:
$\left(x+\pi\right)= \left(\frac{\pi}{6}+\pi \right)=\frac{7\pi}{6}$
   ♦ This input x also gives an output of $\frac{1}{\sqrt 3}$
(iii) We get a feeling that, we can repeatedly add π, all giving the same output $\frac{1}{\sqrt 3}$
• When π is added the second time, then also output becomes $\frac{1}{\sqrt 3}$.
• That means, π is a possible candidate, who may get accepted as the period of the function f(x) = tan x
7. We just saw that π has the possibility to become the period. Let us try another initial angle. It can be written in 3 steps:
(i) Let the initial input x given for f(x) = tan x be $\frac{\pi}{3}$
   ♦ The output will be √3
(ii) Let us add π. So the next input x given is:
$\left(x+\pi\right)= \left(\frac{\pi}{3}+\pi \right)=\frac{4\pi}{3}$
• The output f(x) is equal to $\tan \frac{4\pi}{3}$ = √3. The output does not change.
(iii) We get a feeling that, we can repeatedly add π, all giving the same output √3
• When π is added the second time, the input x is:
$\left(\frac{4\pi}{3}+\pi \right)=\frac{7\pi}{3}$
• The output f(x) is equal to $\tan \frac{7\pi}{3}$ = √3. The output does not change.
(iv) When π is added the third time, the input x is:
$\left(\frac{7\pi}{3}+\pi \right)=\frac{10\pi}{3}$
• The output f(x) is equal to $\tan \frac{10\pi}{3}$ = √3. The output does not change.
8. In this way we can try various initial angles. We will find that π is eligible to be accepted as the period of f(x) = tan x
• Note that, 2π is also eligible. But the smallest one is accepted as the period.
• The following table 3.3 is generated using a computer. It clearly shows the effect of the period π. $\frac{\pi}{5}$ is taken as the initial angle. (This table can be prepared using standard tables also)

Table 3.3

◼ In the table, we see that:
• tangent of $\frac{\pi}{5}$ is 0.7265
• When π is added, x becomes $\frac{6\pi}{5}$. But the tangent remains at 0.7265
• When another π is added, x becomes $\frac{11\pi}{5}$. But the tangent remains at 0.7265
• When another π is added, x becomes $\frac{16\pi}{5}$. But the tangent remains at 0.7265
9. We can write: f(x) = tan x = tan (nπ+x) where n is any integer.
• The smallest possible positive value of nπ is obtained when n = +1
• When n = +1, we get: nπ = π
• This smallest possible positive value is the period.

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So we have seen the periods of sine, cosine and tangent functions. In the next section, we will see trigonometric equations.

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