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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.

Trigonometric inequality

Intervals satisfying inequality, involving tangent function.

Thus, basic interval satisfying inequality is :

- 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

Examples

Problem : Solve trigonometric inequality given by :

sin x 1 2

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

sin x = 1 2 = sin 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

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.

Trigonometric inequality

Intervals satisfying inequality

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

Problem : Solve trigonometric inequality given by :

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 > 0

sin x cos π 4 cos x sin π 4 > 0

sin x π 4 > 0

Let y = x π / 4 . Then,

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

y = 0

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

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

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

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

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

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

Changing argument of the function, the domain becomes :

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

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 , 5 π 6

The value of “x” satisfying above equation :

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

Problem : Find the domain of the function given by :

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 ]

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 :

Modulus of trigonometric functions

The values of modulus of sine and cosine functions in four quadrants are shown.

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

Domain = π 2 x 0 = [ - π 2, 0 ]

Exercise

Solve the inequality :

tan x < 1

Hints : Here, corresponding trigonometric equation is :

tan x = 1 = tan π 4

The other solution in [0,2π] is :

x = π + π 4 = 5 π 4

Corresponding negative angle :

y = 5 π 4 2 π = - 3 π 4

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

- 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

Find domain of function :

y = log e cos x

Argument of logarithmic function is positive. Hence,

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

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 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

Find domain of function :

f(x) = sin x - 1

Expression under square root is non-negative. Hence,

sin x - 1 0 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

The domain of given function, therefore, is :

x = { x : n π + - 1 n π / 2 ; n Z }

Find domain of function :

f x = 1 sin x + 3 sin x

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

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

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

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

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

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 sin 0 sin x sin 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

The second inequality is :

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

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

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 tan 0 tan x tan π 4 0 x 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

The domain of given function, therefore, is :

n π x n π + π 4 n Z

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Source:  OpenStax, Functions. OpenStax CNX. Sep 23, 2008 Download for free at http://cnx.org/content/col10464/1.64
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