FIND DOMAIN AND RANGE OF TRIGONOMETRIC FUNCTIONS

Example 1 :

Find the domain and range of the following functions

f(x)  =  1/(2 - sin 3x)

Solution :

In order to find the domain, let us equate the denominator to 0.

2 - sin 3x  =  0

sin 3x  =  2

This is impossible, because the minimum value of sin is -1 and maximum value is 1. So we will not the above situation at any more.

So, the domain is all real values.

Range for sin function is between -1 and 1.

-1 ≤  sin 3x  ≤  1

Multiply by negative sign, through out

1 ≤  -sin 3x  ≤  -1

 -1  ≤  -sin 3x  ≤  1

Adding 2 through out, we get

 -1 + 2  ≤  -sin 3x + 2  ≤  1 + 2

1  ≤ 2 - sin 3x  ≤  3

3  ≤ (2 - sin 3x)  ≤  1

Taking reciprocals, we get

(1/3) ≤ 1/(2 - sin 3x)  ≤  (1/1)

(1/3) ≤ 1/(2 - sin 3x)  ≤  1

So, the range is [1/3, 1]

Example 2 :

Find the range of the following function.

f(x)  =  1/(1 - 2 cos x) 

Solution :

Generally range for the cos function lies between -1 and 1

-1 ≤  cos x  ≤  1

Multiply by -2 throughout, we get

2  ≤  -2 cos x  ≤  -2

-2  ≤  -2 cos x  ≤  2

Add 1 though out the line

(-2 + 1)  ≤  (-2 cos x + 1)  ≤  (2 + 1)

-1 ≤  1 - 2 cos x  ≤  3

Taking reciprocal, we get

-1/1 ≤ 1/(1 - 2 cos x)  ≤  (1/3)

-1 ≤ 1/(1 - 2 cos x)  ≤  (1/3)

So, the range is [-1, 1/3]

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