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Inverse trigonometric functions review

Review your knowledge of the inverse trigonometric functions, arcsin(x), arccos(x), & arctan(x).

What are the inverse trigonometric functions?

arcsin(x), or sin1(x), is the inverse of sin(x).
arccos(x), or cos1(x), is the inverse of cos(x).
arctan(x), or tan1(x), is the inverse of tan(x).

Range of the inverse trig functions

RadiansDegrees
π2arcsin(x)π290arcsin(x)90
0arccos(x)π0arccos(x)180
π2<arctan(x)<π290<arctan(x)<90
The trigonometric functions aren't really invertible, because they have multiple inputs that have the same output. For example, sin(0)=sin(π)=0. So what should be sin1(0)?
In order to define the inverse functions, we have to restrict the domain of the original functions to an interval where they are invertible. These domains determine the range of the inverse functions.
The value from the appropriate range that an inverse function returns is called the principal value of the function.
Want to learn more about arcsin(x)? Check out this video.
Want to learn more about arccos(x)? Check out this video.
Want to learn more about arctan(x)? Check out this video.

Check your understanding

Problem 1
The sine value of all options is 0.98. Which is the principal value of arcsin(0.98)?
All measures are in radians.
Choose 1 answer:

Want to try more problems like this? Check out this exercise.

Want to join the conversation?

  • duskpin ultimate style avatar for user Monika M
    Can the range of arctan also be from 0 to 180? If we divide the circle into quadrants where 0-90 is 1st, 90-180 2nd, 180-270 3rd and 270-360 4th quadrant than the 1st and 3rd have the same tan and the 2nd and 4th have the same tan. As I understand it range is determined so there is only one value for each tan and 1st and 2nd quadrant don't give the same tan.
    (31 votes)
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    • mr pink red style avatar for user andrewp18
      Right, but if we define the range as [0, π], then we would actually have to exclude π/2 because tan π/2 is undefined. Also we would exclude either 0 or π because both of those angles have the same tangent. Therefore, we would have to write: (0, π/2) ∪ (π/2, π] or [0, π/2) ∪ (π/2, π). Since it is simpler to write (-π/2, π/2), this is what we use for the range of arctan.
      (90 votes)
  • male robot hal style avatar for user Nate Hong
    I don't understand this particular unit
    (49 votes)
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  • leaf green style avatar for user Jeremiah Barro
    I've gone through Algebra I & II and half of Geometry and all the Trig up until here, and I've understood everything, but this makes zero sense to me. I get the domains of each function, but how should we know how many radians cos−1(0.32)? Or how can we know what tan−1(−21) is and whether it's inside the domain of pi/2 and -pi/2? And why are the radians not in terms of pi? What's 1.62 radians? Are we supposed to assume there are roughly 6.28 radians in a circle and estimate whether each of the answers would be inside of the domain of the function?
    (17 votes)
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    • male robot donald style avatar for user Venkata
      A lot of questions here! Let me address them one at a time.

      "how should we know how many radians cos−1(0.32)"

      We can't, at least not without a calculator. You can approximate the value using something known as a Taylor Polynomial, but at this stage, punch it into a calculator.

      " Or how can we know what tan−1(−21) is and whether it's inside the domain of pi/2 and -pi/2"

      Pretty much the same thing. Use a calculator. As for the "range" (not domain. The domain of arctan(x) is all reals), find the answer first, which comes to -1.5232. Now, the range of arctan(x) is (-1.57,1.57). Clearly, our answer falls in this interval. That's how you check for it.

      "And why are the radians not in terms of pi? What's 1.62 radians?"

      Thing is, they are in terms of pi. For example, if my question was "what is the value of arccos(0)?", usually you'd say pi/2 radians. But, one can also say 1.57 radians, and they'd be right. You'll mostly deal with radians having pi in them (because it helps in scaling a big fraction to a manageable one) but if you see a decimal, you'd probably want to use a calculator.

      "Are we supposed to assume there are roughly 6.28 radians in a circle and estimate whether each of the answers would be inside of the domain of the function?"

      Pretty much, actually. Knowing the decimal values of the multiples of pi isn't bad. So, if you know that the range of a function is (-1.57,1.57) and your answer comes to 1.5, you can be sure that you're right. But from my experience, in higher classes, you'll rarely see values of radians in decimals, as fractions of pi are much easier to deal with.
      (41 votes)
  • piceratops seedling style avatar for user Shreya V
    How do you find the sin(arctan(2)) without using a calculator? 2 is not a value on the unit circle, so I'm not sure how to do this problem by hand. Another problem like this would be cos(arcsin(1/4)). Is there a way to do these types of problems using right triangles? If so, how?
    Thank you!
    (8 votes)
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    • mr pink red style avatar for user andrewp18
      sin(arctan 2)
      Let's take a look at what this question is really asking for. First, we are taking the arctangent of 2. That is, we are finding some angle 𝜃 such that tan 𝜃 = 2. Then, after finding that angle, we are taking the sine of that angle! In other words, if we have an angle 𝜃 such that tan 𝜃 = 2, we must find sin 𝜃.

      This is possible to do with a right triangle. If 𝜃 is an angle in a right triangle such that the opposite side is 2 and the adjacent side is 1, then tan 𝜃 = 2. Then, the sine of this angle is 2/ℎ such that ℎ is the hypotenuse of the triangle. By the Pythagorean theorem, the hypotenuse is √5. Thus, sin 𝜃 = 2/√5 and we have:
      sin(arctan 2) = 2/√5
      You can use a similar argument for the second question. You could have also solved this using the Pythagorean trig identities (which are basically condensed versions of the process we used above).
      (17 votes)
  • primosaur ultimate style avatar for user Matthew
    Are there inverse trigonometric functions for the reciprocal trigonometric functions (secant,cosecant, and cotangent)?
    (5 votes)
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  • hopper cool style avatar for user ethan.fys07
    How did you guys figure out the trigonometric functions?
    By using calculaters?
    (2 votes)
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  • blobby green style avatar for user Chris Easterday
    All I can figure out is what quadrant arcsin (0.98) is in. I don't understand where the 1.37 comes from.
    (3 votes)
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    • duskpin ultimate style avatar for user Sílvio Filho
      1.37 is the angle in radians (in degrees it is approximately 78.52º) in which its sine is equal to 0.98 and, in fact, it is also the first angle that has sine of 0.98 if you follow the trigonometric circle counterclockwise from 0 radians (0º) to 2pi radians (360º). Therefore, it is the principal value of arcsin(0.98), which can also be written as sin^-1(0.98).
      (10 votes)
  • purple pi teal style avatar for user Marvin Pangilinan
    why are they call arc? I looked it up on wiki and cant find more other source :(
    (1 vote)
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  • leafers seed style avatar for user Alex Cain
    Hello.

    For example, I know how to solve for arcsine(45 degrees), arcsin(3pi/4), and arcsin(square root of 3 / 2). But when it comes to solving for arcsin(0.98), I have no idea how to deal with the number 0.98 and I'm totally confused about how the process works. When I read the explanation down below the exercise 1, it didn't tell me how to get the range of [-pi/2, pi/2]. How would you know the range when you are dealing with the number 0.98? And can somebody explain how the process works to get the answer?

    Best regards,
    Alex
    (3 votes)
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    • leaf green style avatar for user kubleeka
      I think you are confusing sine and arcsine. The domain of arcsin(x) is [-1, 1], and neither 45º nor 3π/4 is in that set. However, sin(45º) and sin(3π/4) are both √2/2.

      At this level, the only way to compute arcsin(0.98) is with a calculator. In calculus, you'll develop techniques to compute it by hand, but that's beyond the scope of precalculus.

      Remember what happens to a functions graph when inverted: it's reflected across the line y=x. If we do this to sin(x), we get a wave going up the y-axis. But this isn't a function. So instead of reflecting the entirety of sin(x), we just reflect the part in [-π/2, π/2]. This smaller piece passes the vertical line test, and it's the largest possible piece that does so.

      So because the domain of sin(x) is [-π/2, π/2], this is also the range of arcsin(x).
      (2 votes)
  • male robot johnny style avatar for user Caleb
    how I can recognize where domain and range is, also I don't get the "principal" value in the question.
    (2 votes)
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    • duskpin ultimate style avatar for user AHsciencegirl
      The domains for arcsine and arctangent are both -pi/2 > x < pi/2. The domain for arccosine is 0 > x < pi. (Sal explains how those domains are derived in the videos "intro to arcsine/arctangent/arccosine" in this playlist)

      The term "principal" in this context roughly means 'possible.' That is, which option is possible given the correct domains. Technically any of the options would give the same output but they must first be reduced to the 'principal' answer that is within the domain.
      (3 votes)