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Definition

The inverse sine function, denoted sin −1 or arcsin, and the inverse cosine function, denoted cos −1 or arccos, are defined on the domain D = { x | 1 x 1 } as follows:

sin −1 ( x ) = y if and only if sin ( y ) = x and π 2 y π 2 ; cos −1 ( x ) = y if and only if cos ( y ) = x and 0 y π .

The inverse tangent function, denoted tan −1 or arctan, and inverse cotangent function, denoted cot −1 or arccot, are defined on the domain D = { x | < x < } as follows:

tan −1 ( x ) = y if and only if tan ( y ) = x and π 2 < y < π 2 ; cot −1 ( x ) = y if and only if cot ( y ) = x and 0 < y < π .

The inverse cosecant function, denoted csc −1 or arccsc, and inverse secant function, denoted sec −1 or arcsec, are defined on the domain D = { x | | x | 1 } as follows:

csc −1 ( x ) = y if and only if csc ( y ) = x and π 2 y π 2 , y 0 ; sec −1 ( x ) = y if and only if sec ( y ) = x and 0 y π , y π / 2 .

To graph the inverse trigonometric functions, we use the graphs of the trigonometric functions restricted to the domains defined earlier and reflect the graphs about the line y = x ( [link] ).

An image of six graphs. The first graph is of the function “f(x) = sin inverse(x)”, which is an increasing curve function. The function starts at the point (-1, -(pi/2)) and increases until it ends at the point (1, (pi/2)). The x intercept and y intercept are at the origin. The second graph is of the function “f(x) = cos inverse (x)”, which is a decreasing curved function. The function starts at the point (-1, pi) and decreases until it ends at the point (1, 0). The x intercept is at the point (1, 0). The y intercept is at the point (0, (pi/2)). The third graph is of the function f(x) = tan inverse (x)”, which is an increasing curve function. The function starts close to the horizontal line “y = -(pi/2)” and increases until it comes close the “y = (pi/2)”. The function never intersects either of these lines, it always stays between them - they are horizontal asymptotes. The x intercept and y intercept are both at the origin. The fourth graph is of the function “f(x) = cot inverse (x)”, which is a decreasing curved function. The function starts slightly below the horizontal line “y = pi” and decreases until it gets close the x axis. The function never intersects either of these lines, it always stays between them - they are horizontal asymptotes. The fifth graph is of the function “f(x) = csc inverse (x)”, a decreasing curved function. The function starts slightly below the x axis, then decreases until it hits a closed circle point at (-1, -(pi/2)). The function then picks up again at the point (1, (pi/2)), where is begins to decrease and approach the x axis, without ever touching the x axis. There is a horizontal asymptote at the x axis. The sixth graph is of the function “f(x) = sec inverse (x)”, an increasing curved function. The function starts slightly above the horizontal line “y = (pi/2)”, then increases until it hits a closed circle point at (-1, pi). The function then picks up again at the point (1, 0), where is begins to increase and approach the horizontal line “y = (pi/2)”, without ever touching the line. There is a horizontal asymptote at the “y = (pi/2)”.
The graph of each of the inverse trigonometric functions is a reflection about the line y = x of the corresponding restricted trigonometric function.

Go to the following site for more comparisons of functions and their inverses.

When evaluating an inverse trigonometric function, the output is an angle. For example, to evaluate cos −1 ( 1 2 ) , we need to find an angle θ such that cos θ = 1 2 . Clearly, many angles have this property. However, given the definition of cos −1 , we need the angle θ that not only solves this equation, but also lies in the interval [ 0 , π ] . We conclude that cos −1 ( 1 2 ) = π 3 .

We now consider a composition of a trigonometric function and its inverse. For example, consider the two expressions sin ( sin −1 ( 2 2 ) ) and sin −1 ( sin ( π ) ) . For the first one, we simplify as follows:

sin ( sin −1 ( 2 2 ) ) = sin ( π 4 ) = 2 2 .

For the second one, we have

sin −1 ( sin ( π ) ) = sin −1 ( 0 ) = 0 .

The inverse function is supposed to “undo” the original function, so why isn’t sin −1 ( sin ( π ) ) = π ? Recalling our definition of inverse functions, a function f and its inverse f −1 satisfy the conditions f ( f −1 ( y ) ) = y for all y in the domain of f −1 and f −1 ( f ( x ) ) = x for all x in the domain of f , so what happened here? The issue is that the inverse sine function, sin −1 , is the inverse of the restricted sine function defined on the domain [ π 2 , π 2 ] . Therefore, for x in the interval [ π 2 , π 2 ] , it is true that sin −1 ( sin x ) = x . However, for values of x outside this interval, the equation does not hold, even though sin −1 ( sin x ) is defined for all real numbers x .

What about sin ( sin −1 y ) ? Does that have a similar issue? The answer is no . Since the domain of sin −1 is the interval [ −1 , 1 ] , we conclude that sin ( sin −1 y ) = y if −1 y 1 and the expression is not defined for other values of y . To summarize,

sin ( sin −1 y ) = y if −1 y 1

and

sin −1 ( sin x ) = x if π 2 x π 2 .

Similarly, for the cosine function,

cos ( cos −1 y ) = y if −1 y 1

and

cos −1 ( cos x ) = x if 0 x π .

Similar properties hold for the other trigonometric functions and their inverses.

Evaluating expressions involving inverse trigonometric functions

Evaluate each of the following expressions.

  1. sin −1 ( 3 2 )
  2. tan ( tan −1 ( 1 3 ) )
  3. cos −1 ( cos ( 5 π 4 ) )
  4. sin −1 ( cos ( 2 π 3 ) )
  1. Evaluating sin −1 ( 3 / 2 ) is equivalent to finding the angle θ such that sin θ = 3 / 2 and π / 2 θ π / 2 . The angle θ = π / 3 satisfies these two conditions. Therefore, sin −1 ( 3 / 2 ) = π / 3 .
  2. First we use the fact that tan −1 ( −1 / 3 ) = π / 6 . Then tan ( π / 6 ) = −1 / 3 . Therefore, tan ( tan −1 ( −1 / 3 ) ) = −1 / 3 .
  3. To evaluate cos −1 ( cos ( 5 π / 4 ) ) , first use the fact that cos ( 5 π / 4 ) = 2 / 2 . Then we need to find the angle θ such that cos ( θ ) = 2 / 2 and 0 θ π . Since 3 π / 4 satisfies both these conditions, we have cos ( cos −1 ( 5 π / 4 ) ) = cos ( cos −1 ( 2 / 2 ) ) = 3 π / 4 .
  4. Since cos ( 2 π / 3 ) = −1 / 2 , we need to evaluate sin −1 ( −1 / 2 ) . That is, we need to find the angle θ such that sin ( θ ) = −1 / 2 and π / 2 θ π / 2 . Since π / 6 satisfies both these conditions, we can conclude that sin −1 ( cos ( 2 π / 3 ) ) = sin −1 ( −1 / 2 ) = π / 6 .
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Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
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