Trigonometric Equations and Inequalities
5.0 Period of trigonometric function
5.0 Period of trigonometric function
The period of any trigonometric function $f(x)$ is defined as the distance in $x$ it takes for the pattern to repeat itself or in other words we can say the length of one complete cycle of a trigonometric function is the period.
Let us recall three basic trigonometric functions and plot them one by one on graph to find the distance in $X$-axis after which it repeat its pattern.
1. $y=sinx$: As it is clear from the graph that the similar pattern is repeated after every $2\pi $ distance on $X$-axis (both positive and negative direction), we can say that the period of the function $y=sinx$ is $2\pi $ radian or ${360^ \circ }$.
2. $y=cosx$: As it is clear from the graph that the similar pattern is repeated after every $2\pi $ distance on $X$-axis (both positive and negative direction), we can say that the period of the function $y=cosx$ is $2\pi $ radian or ${360^ \circ }$.
3. $y=tanx$: As it is clear from the graph that the similar pattern is repeated after every $\pi $ distance on $X$-axis (both positive and negative direction), we can say that the period of the function $y=tanx$ is $\pi $ radian or ${180^ \circ }$.
Note:
1. If the trigonometric functions are not in their standard form $sinx, cosx, tanx$. Let us assume the given function is $y=sin(kx)$ where $k>0$ then the graph makes $k$ complete cycles between $0$ and $2\pi $ and the period can be find outby dividing the period of the function in the standard form by $k$ i.e., Period of the function $y=sin(kx)$ is $$\frac{{2\pi }}{k}$$ Similarly, the period of function $y=cos(kx)$ is $$\frac{{2\pi }}{k}$$ and for the function $y=tan(kx)$ is $$\frac{{\pi }}{k}$$
2. If the trigonometric function consists of more than one function i.e., sum or difference of two or more trigonometric functions then find the period of each trigonometric function and then find the L.C.M. of all the individual periods to find the period of the combined trigonometric function.
For example: Period of the function $\sin x + \sin 2x + \sin \frac{x}{2} < 1$ is calculated first by calculating the period of individual functions i.e., period of $sinx$ is $2\pi$, period of $sin2x$ is $\frac{{2\pi }}{2} = \pi $ and period of $\sin \frac{x}{2}$ is $\frac{{2\pi }}{{\frac{1}{2}}} = 4\pi $. Therefore, the combined period is calculated by taking the L.C.M. of all the three periods which gives us $4\pi$.
Example 13. Solve the inequality $\sin x > 0$.
Solution:
Step 1: Transform the equation: The given equation contains only one trigonometric function and is already in standard form.
Step 2: Find common period: Period of $sinx$ is $2\pi$.
Step 3: Solve the equation: Assume $sinx=0$, with in one revolution of the unit radius, we get $x=0$ and $x=\pi$. ( As there is only one function we can skip the third step in this case)
Step 4: Solve the inequality: As there is no interval given, let us assume with in one revolution of the unit radius, we get $$0<x<\pi$$ Now, we add the sine period calculated in step 2 to the results to remove our assumption of one revolution of unit radius, adding $2n\pi$ both sides, we get $$0+2n\pi<x<\pi+2n\pi$$
Example 14. Solve the inequality $$\cos \left( {2x + \frac{\pi }{4}} \right) < \frac{1}{2}$$ with in period of $2\pi$.
Solution:
Step 1: Transform the equation: The given equation contains only one trigonometric function and is already in standard form.
Step 2: Find common period: Period of $\cos \left( {2x + \frac{\pi }{4}} \right)$ is ${\frac{{2\pi }}{2} = \pi }$.
Step 3: Solve the equation: As there is only one function we can skip the third step in this case.
Step 4: Solve the inequality: As there is interval given i.e., $2\pi$, let us assume with in one revolution of the unit radius, we get $$\begin{equation} \begin{aligned} \frac{\pi }{3} < 2x + \frac{\pi }{4} < \frac{{5\pi }}{3} \\ \frac{\pi }{3} - \frac{\pi }{4} < 2x < \frac{{5\pi }}{3} - \frac{\pi }{4} \\ \frac{\pi }{{12}} < 2x < \frac{{17\pi }}{{12}} \\ \frac{\pi }{{24}} < x < \frac{{17\pi }}{{24}} \\\end{aligned} \end{equation} $$
Example 15. Solve the inequality $$\sin x + \sin 3x < - \sin 2x$$ for $(0 < x < 2\pi )$.
Solution: Let us proceed step by step-
Step 1: Transform the equation: $\sin x + \sin 2x + \sin 3x < 0$
Step 2: Find common period: $$\frac{{2\pi }}{1}\;;\;\frac{{2\pi }}{2}\;;\;\frac{{2\pi }}{3} \Rightarrow \frac{{L.C.M}}{{H.C.F}} = \frac{{2\pi }}{1} = 2\pi $$
Step 3: Solve the equation: $$\begin{equation} \begin{aligned} \sin x + \sin 2x + \sin 3x = 0 \\ \sin 2x(2\cos x + 1) = 0 \\ f(x).g(x) = 0 \\\end{aligned} \end{equation} $$ Solve for $x$ by putting $f(x)$ and $g(x)$ equals to $0$, we get $$\begin{equation} \begin{aligned} \sin 2x = 0 \Rightarrow x = 0,\pi ,2\pi ,... \\ 2\cos x + 1 = 0 \Rightarrow \cos x = \frac{{ - 1}}{2} \Rightarrow x = \frac{{2\pi }}{3},\frac{{4\pi }}{3},... \\\end{aligned} \end{equation} $$ As it is given in the question that $x$ varies from $0$ to $2\pi$, we take only those values lies with in the given interval.
Step 4: Solve the inequality: In the inequality $$\begin{equation} \begin{aligned} \sin 2x(2\cos x + 1) < 0 \\ f(x).g(x) < 0 \\\end{aligned} \end{equation} $$ we need those values of $x$ in which the multiplication of two functions $f(x)$ and $g(x)$ becomes negative. Let us take small intervals on the basis of values found out in step 3.
| Interval | $f(x)$ | $g(x)$ | $f(x).g(x)$ |
| $0 < x < \frac{\pi }{2}$ | $+$ | $+$ | $+$ |
| $\frac{\pi }{2} < x < \frac{{2\pi }}{3}$ | $-$ | $+$ | $-$ |
| $\frac{{2\pi }}{3} < x < \pi $ | $-$ | $-$ | $+$ |
| $\pi < x < \frac{{4\pi }}{3}$ | $+$ | $-$ | $-$ |
| $\frac{{4\pi }}{3} < x < \frac{{3\pi }}{2}$ | $+$ | $+$ | $+$ |
| $\frac{{3\pi }}{2} < x < 2\pi $ | $-$ | $+$ | $-$ |
From the table, we can say that the solution set of given trigonometric inequality is $$\left( {\frac{\pi }{2},\frac{{2\pi }}{3}} \right) \cup \left( {\pi ,\frac{{4\pi }}{3}} \right) \cup \left( {\frac{{3\pi }}{2},2\pi } \right)$$
