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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

I need help under implicit differentiation
Uchenna Reply
What is derivative of antilog x dx ?
Tanmay Reply
what's the meaning of removable discontinuity
Brian Reply
what's continuous
Brian
an area under a curve is continuous because you are looking at an area that covers a range of numbers, it is over an interval, such as 0 to 4
Lauren
using product rule x^3,x^5
Kabiru
please help me to calculus
World Reply
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Sunny
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Scott
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Scott
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Bilal
help the integral of x^2/lnxdx
Levis
also find the value of "X" from the equation that follow (x-1/x)^4 +4(x^2-1/x^2) -6=0 please guy help
Levis
Use integration by parts. Let u=lnx and dv=x2dx Then du=1xdx and v=13x3. ∫x2lnxdx=13x3lnx−∫(13x3⋅1x)dx ∫x2lnxdx=13x3lnx−∫13x2dx ∫x2lnxdx=13x3lnx−19x3+C
Bilal
itz 1/3 and 1/9
Bilal
now you can find the value of X from the above equation easily
Bilal
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usman
usman from where do you need help?
Levis
thanks Bilal
Levis
integrate e^cosx
Uchenna
-sinx e^x
Leo
Do we ask only math question? or ANY of the question?
Levis Reply
yh
Gbesemete
How do i differentiate between substitution method, partial fraction and algebraic function in integration?
usman
you just have to recognize the problem. there can be multiple ways to solve 1 problem. that's the hardest part about integration
Lauren
test
MOHAMMAD
we asking the question cause only the question will tell us the right answer
Sunny
find integral of sin8xcos12xdx
Levis Reply
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Bilal
well find the integral of x^x
Levis
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Levis
Levis I am sorry
Bilal
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Levis
x^x ... no anti derivative for this function... but we can find definte integral numerically.
Bilal
thank you Bilal Kumhar then how we may find definite integral let say x^x,3,5?
Levis
evaluate 5-×square divided by x+2 find x as limit approaches infinity
Michagaye Reply
i have not understood
Leo
The answer is 0
Michael
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Sunny
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Michagaye
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Sunny
The denominator is the aggressive one
Sunny
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Harold
or should I say any prime number greater then 11 ?
Harold
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Harold
I think as limit Approach infinity then X=0
Levis
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No Reply
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Leonito
Fond the value of the six trigonometric function of an angle theta, which terminal side passes through the points(2x½-y)²,4
albert Reply
What's f(x) ^x^x
Emeka Reply
What's F(x) =x^x^x
Emeka
are you asking for the derivative
Leo
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rd
if your asking for derivative dy/dz=x^2/2(lnx-1/2)
Levis
iam sorry f(x)=x^x it means the output(range ) depends to input(domain) value of x by the power of x that is to say if x=2 then x^x would be 2^2=4 f(x) is the product of X to the power of X its derivatives is found by using product rule y=x^x introduce ln each side we have lny=lnx^x =lny=xlnx
Levis
the derivatives of f(x)=x^x IS (1+lnx)*x^x
Levis
what is a maximax
Chinye Reply
A maxima in a curve refers to the maximum point said curve. The maxima is a point where the gradient of the curve is equal to 0 (dy/dx = 0) and its second derivative value is a negative (d²y/dx² = -ve).
Viewer
what is the limit of x^2+x when x approaches 0
Dike Reply
it is 0 because 0 squared Is 0
Leo
0+0=0
Leo
simply put the value of 0 in places of x.....
Tonu
the limit is 2x + 1
Nicholas
the limit is 0
Muzamil
limit s x
Bilal
The limit is 3
Levis
Leo we don't just do like that buddy!!! use first principle y+∆y=x+∆x ∆y=x+∆x-y ∆y=(x+∆x)^2+(x+∆x)-x^2+x on solving it become ∆y=3∆x+∆x^2 as ∆x_>0 limit=3 if you do by calculator say plugging any value of x=0.000005 which approach 0 you get 3
Levis
find derivatives 3√x²+√3x²
Care Reply
3 + 3=6
mujahid
How to do basic integrals
dondi Reply
the formula is simple x^n+1/n+1 where n IS NOT EQUAL TO 1 And n stands for power eg integral of x^2 x^2+1/2+1 =X^3/3
Levis
write something lmit
ram Reply
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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