The domain and range of trigonometric functions are as follows;
Function Domain Range
sin(x). All real numbers [-1, 1]
cos(x). All real numbers [-1, 1]
tan(x). {x| x ≠ -3•π/2, -π/2, π/2, } R
The domain and range of the inverse sine, cosine, and tangent are;
Function Domain Range
arcsin(x). (-1, 1) [-π/2, π/2]
arccos(x). (-1, 1) [0, π]
arctan(x) (-∞, ∞). [-π/2, π/2]
What are the domains and ranges of the trigonometric functions?
sin(x)
The domain of sin(x) is the set of possible values for x, for which sin(x) is defined, which is;
The range of sin(x) is the set of possible values output values of sin(x) is, which is; -1 ≤ sin(x) ≤ 1
- The range is therefore; [-1, 1]
cos(x)
Similarly;
- The domain of cos(x) is the set of all real numbers
- The range of cos(x) is; -1 ≤ cos(x) ≤ 1
tan(x)
tan(x) = sin(x)/cos(x), however, cos(x) = 0, when x = -3•π/2, -π/2, π/2, 3•π/2...
The domain of tan(x) is therefore;
- {x| x ≠ -3•π/2, -π/2, π/2, 3•π/2...}
sin(x) increases as cos(x) decreases, therefore;
The range of tan(x) = sin(x)/cos(x) is therefore;
- -∞ < tan(x) < ∞ which is {-∞, ∞}
arcsin(x)
The input of arcsin(x) is sin(x)
sin(x) = opposite/hypotenuse
Given that the magnitude of hypotenuse > opposite, the ratio, opposite/hypotenuse is < 1 in magnitude
The domain of arcsin(x) is therefore;
- -1 < sin(x) < 1, which is (-1, 1)
The range is the set of possible outputs, which is [-π/2, π/2], given that sin(x) is positive and negative from -π/2, π/2
arccos(x)
Similarly;
- The domain of arccos (x) is (-1, 1)
arctan(x)
The input of arctan(x) is tan(x)
tan(x) = sin(x)/cos(x)
The domain of arctan(x) is therefore;
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