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Evaluating an Inverse Trigonometric Function

  1. Oct 14, 2013 #1
    1. The problem statement, all variables and given/known data
    Evaluate sin^-1(cos70°)


    2. Relevant equations



    3. The attempt at a solution

    sin^-1(cos70°)=θ
    sinθ=cos70°
    sinθ=1/2
    sinθ=∏/3
     
  2. jcsd
  3. Oct 14, 2013 #2
    First use cos(90° - θ ) = sinθ and then sin-1(sinθ) = θ
     
  4. Oct 15, 2013 #3

    Mark44

    Staff: Mentor

    Also, cos(70°) ≠ 1/2 and there is no real number θ for which sin(θ) = ##\pi/3## > 1.
     
  5. Oct 15, 2013 #4
    Could you possibly explain the cos(90°-θ)=sinθ?
     
  6. Oct 15, 2013 #5
    Ah a trig identity I see.
    So cos(90°-70°)=sinθ
    cos(20°)=sinθ
    0.94=sinθ

    sin^-1(sinθ)=θ
    sin^-1(0.94)=θ
    0.94=θ
     
    Last edited: Oct 15, 2013
  7. Oct 15, 2013 #6

    Mark44

    Staff: Mentor

    No, cos(90°-70°) = cos(20°) = sin(70°).
    In general, cos(90° - θ) = sin(θ).
    Most of the above makes no sense, with each line having little or no relation to the preceding line. Your value for θ is incorrect, and you could easily check that it is incorrect by using a calculator.
     
  8. Oct 15, 2013 #7
    Alright. So if cos(90°-x)=sinx, which in my case is 70°, I still do not understand why Tanya then encourages me to use sin^-1(sinθ)=θ. How is cos70° to be replaced by sinθ? By the trig identities, sin(90°-x)=cosx, it is sin20° which is equivalent to cos70°. Would my equation then be accurately represented by sin^-1(sin20°)=θ and then simply by using cancellation identities reduced to 20°=θ?
     
  9. Oct 15, 2013 #8

    Mark44

    Staff: Mentor

    Because f-1(f(x)) = x, as long as x is in the domain of f. That's a basic property of a function and its inverse. It's also true that f(f-1(y)) = y, when y is in the domain of f-1. In short, the composition of a function and its inverse "cancels."
    There's really no need for you to use θ, here, and I think that it is confusing you.
    Not equivalent to - equal to.
    cos(70°) = cos(90° - 20°) = sin(20°)
    Make it simpler by getting rid of θ.

    sin-1(sin(20°)) = ?
     
  10. Oct 15, 2013 #9
    sin^-1(sin(20°))=20°=∏/9
     
  11. Oct 15, 2013 #10

    Mark44

    Staff: Mentor

    Yes.
     
  12. Oct 15, 2013 #11
    I apologize for being so confused, independent study is not being kind to me. Thank you very much, your help is immensely appreciated.
     
  13. Oct 15, 2013 #12

    Mark44

    Staff: Mentor

    Sure, you're welcome!
     
  14. Oct 16, 2013 #13
    Thanks Mark for nicely guiding the OP.
     
  15. Oct 16, 2013 #14

    Mark44

    Staff: Mentor

    That's what they pay me for!

    Oh, wait - I'm an unpaid volunteer.
     
  16. Oct 17, 2013 #15
    and an excellent mentor :smile:
     
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