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Trigonometric inequalities

Module by: Sunil Kumar Singh. E-mail the author

Evaluating trigonometric ratios is a direct process in which we make use of known values, trigonometric identities and transformations or even pre-defined trigonometric tables. The evaluation of trigonometric inequalities is somewhat inverse of this process. Consider an inequality :

tan x - 3 tan x - 3

Clearly, we need to know “x” for which this inequality holds. As pointed out earlier, trigonometric functions are “many-one” relation. The value of “x” satisfying given inequality is not an unique interval, but a series of intervals. Incidentally, however, trigonometric function values repeat after certain “period”. So this enables us to define periodic intervals in generic manner for which trigonometric inequality holds.

Figure 1: Intervals satisfying inequality, involving tangent function.
Trigonometric inequality
 Trigonometric inequality  (te1.gif)

We observe that line y = - 3 y = - 3 intersects tangent graph at multiple points. The sections of plots satisfying the inequality are easily identified on the graph and are shown as dark red line.

Solution of trigonometric inequality

Determination of base or fundamental interval is central to solve trigonometric inequality. The function values in this interval is repeated with a periodicity of trigonometric function. The base interval depends on the nature of trigonometric function and inequality in question. The steps to find solution of trigonometric inequality are :

1 : Convert given inequality to trigonometric equation by replacing inequality sign by equality sign.

2 : Solve resulting equation in the interval [0,2π]. There are two solutions. They are the angle values at which trigonometric function has the value which is being compared in the given inequality.

3 : Convert positive angle greater than π to equivalent negative value to account for the fact that basic interval being repeated may lie on negative side of the origin (cosine, secant and tangent function).

4 : Construct base interval between two values, keeping in mind the given inequality. It is always advantageous to draw a rough intersection of graphs of each side of given inequality.

5 : If function asymptotes (tangent, cotangent, secant and cosecant) within the interval constructed, then basic interval is limited by the angle value at which function asymptotes.

6 : Generalize solution by extending base interval with the period of the trigonometric function.

In order to understand the process, let us solve the inequality given by :

tan x - 3 tan x - 3

This example has been selected here as it involves consideration of each step as enumerated above for finding solution of inequality. Corresponding trigonometric equation, in this case, is :

tan x = - 3 tan x = - 3

The acute angle is π/3. Further, tangent function is negative in second and fourth quarter (see sign diagram). Using value diagram in conjunction with sign diagram, solution of given equation in [0, 2π] are :

Figure 2: Tangent function is negative in second and fourth quarter .
Sign and value diagram
 Sign and value diagram  (te2.gif)

x = π - θ = π - π 3 = 2 π 3 x = π - θ = π - π 3 = 2 π 3 x = 2 π - θ = 2 π - π 3 = 5 π 3 x = 2 π - θ = 2 π - π 3 = 5 π 3

Here, second angle is greater than π. Hence, equivalent negative angle is :

y = 5 π 3 - 2 π = - π 3 y = 5 π 3 - 2 π = - π 3

Tangent function, however, is not a continuous function between -π/3 and 2π/3. Tangent values are greater than -√3 for angle greater than -π/3, but value asymptotes to infinity at π/2. This can be verified from the intersection graph.

Figure 3: Intervals satisfying inequality, involving tangent function.
Trigonometric inequality
 Trigonometric inequality  (te1.gif)

Thus, basic interval satisfying inequality is :

- π 3 x < π 2 - π 3 x < π 2

It is also clear that the solution in this interval is repeated with a period of π, which is period of tangent function. Hence, solution of given inequality is :

n π - π 3 x < n π + π 2 ; n Z n π - π 3 x < n π + π 2 ; n Z

Examples

Example 1

Problem : Solve trigonometric inequality given by :

sin x 1 2 sin x 1 2

Solution : The solution of the corresponding equal equation is obtained as :

sin x = 1 2 = sin π 6 sin x = 1 2 = sin π 6

x = π 6 x = π 6

The sine function is positive in first and second quarter. Hence, second angle between “0” and “2π” is :

x = π θ = π π 6 = 5 π 6 x = π θ = π π 6 = 5 π 6

Both angles are less than “π”. Thus, we do not need to convert angle into equivalent negative angle. Further, sine curve is defined for all values of “x”. The base interval, therefore, is :

The valid intervals on sine plot are shown in the figure.

Figure 4: Intervals satisfying inequality
Trigonometric inequality
 Trigonometric inequality  (te4.gif)

π 6 x 5 π 6 π 6 x 5 π 6

The periodicity of sine function is “2π”. Hence, we add “2nπ” on either side of the base interval :

2 n π + π 6 x 2 n π + 5 π 6 , n Z 2 n π + π 6 x 2 n π + 5 π 6 , n Z

Example 2

Problem : Solve trigonometric inequality given by :

sin x > cos x sin x > cos x

Solution : In order to solve this inequality, it is required to convert it in terms of inequality of a single trigonometric function.

sin x > cos x sin x > cos x

sin x cos x > 0 sin x cos x > 0

sin x cos π 4 cos x sin π 4 > 0 sin x cos π 4 cos x sin π 4 > 0

sin x π 4 > 0 sin x π 4 > 0

Let y = x π / 4 y = x π / 4 . Then,

sin y > 0 sin y > 0

Thus, we see that problem finally reduces to solving trigonometric sine inequality. The solution of the corresponding equality is obtained as :

sin y = 0 = sin 0 sin y = 0 = sin 0

y = 0 y = 0

The second angle between “0” and “2π” is “π”. The base interval, therefore, is :

0 < y < π 0 < y < π

The periodicity of sine function is “2π”. Hence, we add “2nπ” on either side of the base interval :

2 n π < y < 2 n π + π , n Z 2 n π < y < 2 n π + π , n Z

Now substituting for y = x π / 4 y = x π / 4 , we have :

2 n π < x π 4 < 2 n π + π , n Z 2 n π < x π 4 < 2 n π + π , n Z

2 n π + π 4 < x < 2 n π + 5 π 4 , n Z 2 n π + π 4 < x < 2 n π + 5 π 4 , n Z

Example 3

Problem : If domain of a function, “f(x)”, is [0,1], then find the domain of the function given by :

f 2 sin x 1 f 2 sin x 1

Solution : The domain of the function is given here. We need to find the domain when argument (input) to the function is a trigonometric expression. The given domain is :

0 x 1 0 x 1

Changing argument of the function, the domain becomes :

0 2 sin x 1 1 1 2 sin x 2 1 / 2 sin x 1 0 2 sin x 1 1 1 2 sin x 2 1 / 2 sin x 1

However, the range of sinx is [-1,1]. It means that the above interval is equivalent to a trigonometric inequality given by :

sin x 1 2 sin x 1 2

The sine function is positive in first and second quadrant. Two values of “x” between “0” and “2π” are :

π 6 , π π 6 π 6 , π π 6

π 6 , 5 π 6 π 6 , 5 π 6

The value of “x” satisfying above equation :

2 n π + π / 6 < = x < = 2 n π + 5 π / 6, n Z 2 n π + π / 6 < = x < = 2 n π + 5 π / 6, n Z

Hence, required domain is :

[ 2 n π + π 6 , 2 n π + 5 π 6 ] , n Z [ 2 n π + π 6 , 2 n π + 5 π 6 ] , n Z

Example 4

Problem : Find the domain of the function given by :

f x = log e 1 [ cos x ] [ sin x ] f x = log e 1 [ cos x ] [ sin x ]

Solution : The function is a logarithmic function, which is valid for all positive values of its argument. Also, the argument of logarithmic function is in rational form, having denominator as a square root. We have to find values of “x” for which the expression within the square root is a positive number. It means that :

[ cos x ] [ sin x ] > 0 [ cos x ] [ sin x ] > 0

[ cos x ] > [ sin x ] [ cos x ] > [ sin x ]

In order to evaluate this inequality, we determine modulus of two trigonometric functions in four quadrants at all bounding values of angle “x” and also at intermediate angles. The values are shown in the figure :

Figure 5: The values of modulus of sine and cosine functions in four quadrants are shown.
Modulus of trigonometric functions
 Modulus of trigonometric functions   (t3a.gif)

It is clear that [ cos x ] > [ sin x ] [ cos x ] > [ sin x ] is true in the fourth quadrant. Hence, domain of the function is :

Domain = π 2 x 0 = [ - π 2, 0 ] Domain = π 2 x 0 = [ - π 2, 0 ]

Exercise

Exercise 1

Solve the inequality :

tan x < 1 tan x < 1

Solution

Hints : Here, corresponding trigonometric equation is :

tan x = 1 = tan π 4 tan x = 1 = tan π 4

The other solution in [0,2π] is :

x = π + π 4 = 5 π 4 x = π + π 4 = 5 π 4

Corresponding negative angle :

y = 5 π 4 2 π = - 3 π 4 y = 5 π 4 2 π = - 3 π 4

However, tangent function asymptotes at - π/2. Hence, basic interval is :

- π 2 , π 4 - π 2 , π 4

Further, tangent function has period of π. The general solution is :

n π - π 2 < x < n π + π 4 ; n Z n π - π 2 < x < n π + π 4 ; n Z

Exercise 2

Find domain of function :

y = log e cos x y = log e cos x

Solution

Argument of logarithmic function is positive. Hence,

cos x > 0 cos x > 0

The angles of corresponding trigonometric equation sinx = 0 in the interval [0,2π] are 3π/2 and π/2. In terms of negative angles are -π/2 and π/2. Cosine function is positive in first and fourth quadrant,

- π 2 < x < π 2 - π 2 < x < π 2

Since cosine values are repeated after “2π”, we can write general inequality as :

2 n π π 2 < x < 2 n π + π 2 ; n Z 2 n π π 2 < x < 2 n π + π 2 ; n Z 4 n - 1 π 2 < x < 4 n + 1 π 2 ; n Z 4 n - 1 π 2 < x < 4 n + 1 π 2 ; n Z

The domain of given function, therefore, is :

4 n - 1 π 2 < x < 4 n + 1 π 2 ; n Z 4 n - 1 π 2 < x < 4 n + 1 π 2 ; n Z

Exercise 3

Find domain of function :

f(x) = sin x - 1 f(x) = sin x - 1

Solution

Expression under square root is non-negative. Hence,

sin x - 1 0 sin x - 1 0 sin x 1 sin x 1

But value of sine function can not be greater than 1. Thus, we need to only evaluate equation sinx=1 to find the domain. Here,

sin x = 1 = sin π 2 sin x = 1 = sin π 2

The domain of given function, therefore, is :

x = { x : n π + - 1 n π / 2 ; n Z } x = { x : n π + - 1 n π / 2 ; n Z }

Exercise 4

Find domain of function :

f x = 1 sin x + 3 sin x f x = 1 sin x + 3 sin x

Solution

We treat this function as addition of two individual functions. Expression under square root in the first function is non-negative. However, radical appears in denominator. As such,

sin x > 0 sin x > 0

The angles of corresponding trigonometric equation sinx = 0 in the interval [0,2π] are 0 and π. Since sine function is positive in first and second quadrant,

0 < x < π 0 < x < π

Since sine values are repeated after “2π”, we can write general inequality as :

0 + 2 n π < x < 2 n π + π ; n Z 0 + 2 n π < x < 2 n π + π ; n Z

On the other hand, domain of second function is real number set R. Note that it is not even radical. Now, domain of addition of two functions is intersection of individual domains. Hence, domain of given function is :

2 n π < x < 2 n + 1 π ; n Z 2 n π < x < 2 n + 1 π ; n Z

Exercise 5

A function f(x) is defined in [0,1]. Determine range of function definition f(sinx).

Solution

The domain of f(x) is [0,1]. Here, argument of function changes from “x” to “sinx”. It changes the input value to function for x, but values in themselves lie within the domain interval of the function. It means that :

0 sin x 1 0 sin x 1 sin 0 sin x sin π 2 sin 0 sin x sin π 2 0 x π / 2 0 x π / 2

This is now a problem of solving two inequalities. The first inequality is : sinx≥0

As solved earlier, the solution is :

2 n π < x < 2 n + 1 π ; n Z 2 n π < x < 2 n + 1 π ; n Z

The second inequality is :

sin x 1 sin x 1

Since value of sinx can not exceed value of 1, we conclude that values of x which satisfies inequality sinx≥0 also satisfies the inequality sinx ≤1. Hence, domain of given function is :

2 n π < x < 2 n + 1 π ; n Z 2 n π < x < 2 n + 1 π ; n Z

Exercise 6

A function f(x) is defined in [0,1]. Determine range of function definition f(tanx).

Solution

The domain of f(x) is [0,1]. Here, argument of function changes from “x” to “tanx”. It changes the input value to function for x, but values in themselves lie within the domain interval of the function. It means that :

0 tan x 1 0 tan x 1 tan 0 tan x tan π 4 tan 0 tan x tan π 4 0 x π 4 0 x π 4 0 + n π x n π + π 4 0 + n π x n π + π 4

Since tangent values are repeated after “π”, we can write general inequality as :

n π x n π + π 4 n Z n π x n π + π 4 n Z

The domain of given function, therefore, is :

n π x n π + π 4 n Z n π x n π + π 4 n Z

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