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State the domain and range of the sin, cos, tan, functions. State the domain
and range of the inverse sin, cos, and tangent functions.

Sagot :

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;

  • All real numbers.

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

  • The range is [-π/2, π/2]

arccos(x)

Similarly;

  • The domain of arccos (x) is (-1, 1)

  • The range is [0, π]

arctan(x)

The input of arctan(x) is tan(x)

tan(x) = sin(x)/cos(x)

The domain of arctan(x) is therefore;

  • {-∞, ∞}

  • The range is [-π/2, π/2]

Learn more about the domain and range of functions here:

https://brainly.com/question/1942755

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