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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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Questions & Answers

find the domain and range of f(x)= 4x-7/x²-6x+8
Nick Reply
find the range of f(x)=(x+1)(x+4)
Jane Reply
-1, -4
Marcia
That's domain. The range is [-9/4,+infinity)
Jacob
If you're using calculus to find the range, you have to find the extrema through the first derivative test and then substitute the x-value for the extrema back into the original equation.
Jacob
Good morning,,, how are you
Harrieta Reply
d/dx{1/y - lny + X^3.Y^5}
mogomotsi Reply
How to identify domain and range
Umar Reply
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Akpevwe
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Harrieta
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Dr
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velocity
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Dr
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Dr
Smarter
Adri
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Dr
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Dr
:(
Shun
was up
Dr
hello
Adarsh
is it chatting app?.. I do not see any calculus here. lol
Adarsh
Find the arc length of the graph of f(x) = In (sinx) on the interval [Π/4, Π/2].
mukul Reply
Sand falling freely from a lorry form a conical shape whose height is always equal to one-third the radius of the base. a. How fast is the volume increasing when the radius of the base is (1m) and increasing at the rate of 1/4cm/sec Pls help me solve
ade
show that lim f(x) + lim g(x)=m+l
BARNABAS Reply
list the basic elementary differentials
Chio Reply
Differentiation and integration
Okikiola Reply
yes
Damien
proper definition of derivative
Syed Reply
the maximum rate of change of one variable with respect to another variable
Amdad
terms of an AP is 1/v and the vth term is 1/u show that the sum of uv terms is 1/2(uv+1)
Inembo Reply
what is calculus?
BISWAJIT Reply
calculus is math that studies the change in math, such as the rate and distance,
Tamarcus
what are the topics in calculus
Augustine
what is limit of a function?
Geoffrey Reply
what is x and how x=9.1 take?
Pravin Reply
what is f(x)
Inembo Reply
the function at x
Marc
also known as the y value so I could say y=2x or f(x)= 2x same thing just using functional notation your next question is what is dependent and independent variables. I am Dyslexic but know math and which is which confuses me. but one can vary the x value while y depends on which x you use. also
Marc
up domain and range
Marc
enjoy your work and good luck
Marc
I actually wanted to ask another questions on sets if u dont mind please?
Inembo
I have so many questions on set and I really love dis app I never believed u would reply
Inembo
Hmm go ahead and ask you got me curious too much conversation here
Adri
am sorry for disturbing I really want to know math that's why *I want to know the meaning of those symbols in sets* e.g n,U,A', etc pls I want to know it and how to solve its problems
Inembo
and how can i solve a question like dis *in a group of 40 students, 32 offer maths and 24 offer physics and 4 offer neither maths nor physics , how many offer both maths and physics*
Inembo
next questions what do dy mean by (A' n B^c)^c'
Inembo
The sets help you to define the function. The function is like a magic box where you put inside stuff(numbers or sets) and you get out the stuff but in different shapes (forms).
Adri
I dont understand what you wanna say by (A' n B^c)^c'
Adri
(A' n B (rise to the power of c)) all rise to the power of c
Inembo
Aaaahh
Adri
Ok so the set is formed by vectors and not numbers
Adri
A vector of length n
Adri
But you can make a set out of matrixes as well
Adri
I I don't even understand sets I wat to know d meaning of all d symbolsnon sets
Inembo
Wait what's your math level?
Adri
High-school?
Adri
yes
Inembo
am having big problem understanding sets more than other math topics
Inembo
So f:R->R means that the function takes real numbers and provides real numer. For ex. If f(x) =2x this means if you give to your function a real number like 2,it gives you also a real number 2times2=4
Adri
pls answer this question *in a group of 40 students, 32 offer maths and 24 offer physics and 4 offer neither maths nor physics , how many offer both maths and physics*
Inembo
If you have f:R^n->R^n you give to your function a vector of length n like (a1,a2,...an) where all a1,.. an are reals and gives you also a vector of length n... I don't know if i answering your question. Otherwise on YouTube you havr many videos where they explain it in a simple way
Adri
I would say 24
Adri
Offer both
Adri
Sorry 20
Adri
Actually you have 40 - 4 =36 who offer maths or physics or both.
Adri
I know its 20 but how to prove it
Inembo
You have 32+24=56who offer courses
Adri
56-36=20 who give both courses... I would say that
Adri
solution: In a question involving sets and Venn diagram, the sum of the members of set A + set B - the joint members of both set A and B + the members that are not in sets A or B = the total members of the set. In symbolic form n(A U B) = n(A) + n (B) - n (A and B) + n (A U B)'.
Mckenzie
In the case of sets A and B use the letters m and p to represent the sets and we have: n (M U P) = 40; n (M) = 24; n (P) = 32; n (M and P) = unknown; n (M U P)' = 4
Mckenzie
Now substitute the numerical values for the symbolic representation 40 = 24 + 32 - n(M and P) + 4 Now solve for the unknown using algebra: 40 = 24 + 32+ 4 - n(M and P) 40 = 60 - n(M and P) Add n(M and P), as well, subtract 40 from both sides of the equation to find the answer.
Mckenzie
40 - 40 + n(M and P) = 60 - 40 - n(M and P) + n(M and P) Solution: n(M and P) = 20
Mckenzie
thanks
Inembo
Simpler form: Add the sums of set M, set P and the complement of the union of sets M and P then subtract the number of students from the total.
Mckenzie
n(M and P) = (32 + 24 + 4) - 40 = 60 - 40 = 20
Mckenzie
Practice Key Terms 5

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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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