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Inverse Trig Functions Help

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By McGraw-Hill Professional
Updated on Oct 4, 2011

Introduction to Inverse Trig Functions

Only one-to-one functions can have inverses, and the trigonometric functions are certainly not one to one. But we can limit their domains and force them to be one to one. Limiting the sine function to the interval from x = − π /2 to x = π /2 makes f ( x ) = sin x one to one. The graph in Figure 13.44 passes the Horizontal Line Test.

Fig. 13.44 .

The domain of this function is [− π /2, π /2], and the range is [−1, 1]. If we limit the cosine function to the interval from x = 0 to x = π , we have another one-to-one function. Its graph is shown in Figure 13.45. The domain of this function is [0, π ] and the range is [−1, 1].

Fig. 13.45 .

By limiting the tangent function from x = − π /2 to x = π /2, f ( x ) = tan x is one to one. Its domain is (− π /2, π /2), and its range is all real numbers.

Fig. 13.46 .

Inverse Trigonometric Functions - Two Notations

There are two notations for inverse trigonometric functions. One uses “−1,” and the other uses the letters arc . For example, the inverse sine function is noted as sin −1 or arcsin. Remember that for any function f ( x ) and its inverse f −1 ( x ), f ( f −1 ( x )) = x and f −1 ( f ( x )) = x . In other words, a function evaluated at its inverse “cancels” itself.

The x and y values are reversed for inverse functions. For example, if (4, 9) is a point on the graph of f ( x ), then (9, 4) is a point on the graph of f −1 ( x ). This means that the y-values for the inverse trigonometric functions are angles. Though we will need to use a calculator to evaluate most of these functions, we can find a few of them without a calculator. For , ask yourself what angle (between 0 and π ) has a cosine of 1/2? Because cos π /3 = 1/2, . When evaluating inverse trigonometric functions, we need to keep in mind what their range is. The domain of f ( x ) = sin( x ) is [− π /2, π /2] (Quadrants I and IV), so the range of y = sin −1 x is [− π /2, π /2]. The domain of f( x ) = cos x is [0, π ], so the range of y = cos −1 x is [0, π ] (Quadrants I and II). And the domain of f ( x ) = tan x is (− π /2, π , 2), so the range of y = tan −1 x is (− π /2, π /2) (Quadrants I and IV).

EXAMPLES

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• cos −1 (−1)

cos −1 (−1) = π because cos π = −1.

• tan −1 (1/3)

None of the important angles between − π /2 and π /2 has a tangent of 1/3, so we need to use a calculator to get an approximation: tan −1 (1/3) ≈ 0.32175.

• sin −1 (cos π /6)

, so we need to replace cos π /6 with . This gives us . Because , .

• cos(tan −1 (−1))

What angle in the interval (− π /2, π /2) has a tangent of −1? That would be − π /4, so tan −1 (−1) = − π /4.

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