Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Inverse trignometric functions

  1. May 24, 2005 #1
    Solve equation for Theta. Round to three significant digits.

    tan theta = -2.7; 90<theta<180

    I did the following:

    Theta = (Tan^-1)(-2.7)
    Theta = -70 degrees

    Now what reference angle can I use to get an angle lower than 180 and higher than 90?? I really dont understand how am I supposed to find reference angles... I added 180 and 70=250 but that didnt make any sense :p
    I subtracted 180-70= 110 degrees.. 110 could be the answer .. If so, can anyone explain why is it the answer ?

    is this correct?

    Thanks much
    Last edited: May 24, 2005
  2. jcsd
  3. May 24, 2005 #2
    Your angle is -70 degrees, which is the same as 290 degrees. Try working at it from here. You are on the right train of thought, just made a mistake.
  4. May 24, 2005 #3
    is it 70 degrees since 360-290???

    Damn.. I dont know the actual AIM of this is.. what am I exactly supposed to find??? I already know that angle ( -70degres) so whats the point of this question?

    Last edited: May 24, 2005
  5. May 24, 2005 #4
    [tex] tan^{-1} (-2.7) = -70 \ deg [/tex]

    -70 + 360 = 290, same angle.
    The reference angle is the angle it makes with the X axis between 0 and 90 degrees, in this case its 70 degrees. To find theta in the specificed range (90 to 180 deg), you add multiples of 180 until you reach the range you desire.
  6. May 25, 2005 #5


    User Avatar
    Science Advisor
    Homework Helper

    110 degrees is correct because the problem established the solution interval from 90 to 180 degrees, and because the tangent function is periodic with a period of 180 degrees. You need to understand that the inverse trig functions can only return one value, even though there are many angles that have the same trig value. There is a technical reason having to do with the definition of a function. A function must always have one and only one output value for every input value.

    Technically, the direct trig functions (sin, cos, tan, and their reciprocals csc, sec, cot) do not have inverses. To have an inverse, a function must have a one-to-one correspondence between input and output values. No two input values can have the same output value. There are no periodic functions that satisfy this condition. However, there are related functions that can be defined that do satisfy the condition, and those functions serve as the basis for defining what are called the inverse trig functions.

    For sine and tangent, the related functions are the functions that are equal to the sine and tangent functions in the interval from -90 to +90 degrees inclusive and exist nowhere else. For the cosine the related function is the function that equals the cosine function in the interval from 0 to 180 degrees inclusive and exists nowhere else. These related functions satisfy the one-to-one condition, and they have inverses that we call the inverse trig functions. When you use the calculator to find an inverse trig function value, it must return a value between -90 and +90 for the inverse sine or tangent, and between 0 and 180 for the inverse cosine.

    Technically, the solution to your problem is not

    [tex] \theta = tan^{-1}(-2.7)[/tex]

    Rather, the solution is

    [tex] \theta = tan^{-1}(-2.7) +n \cdot 180^o[/tex]

    where n is an integer that must satisfy the stated interval condition, which in this problem means n = 1

    For similar problems involving angles for which the sine or cosine are given as x, such that [tex] -1 \le x \le +1[/tex] , the solutions should be written as

    [tex] \theta = \pm cos^{-1}(x) +n \cdot 360^o[/tex]


    [tex] \theta = sin^{-1}(x) +n \cdot 360^o, \ \ \ \ \ \ 180^o - sin^{-1}(x) +n \cdot 360^o[/tex]

    with n chosen to satisfy the interval requirements for the solution. If the solution interval is only 90 degrees, usually only one of the two forms of each solution is valid. For larger solution intervals both forms are needed because the sine and cosine functions have two input values with the same output value in every 360 degree period.
    Last edited: May 25, 2005
  7. May 25, 2005 #6
    You really like to be thorough don't you. :)
  8. May 25, 2005 #7
    Ok , once you have known that it should be -70 degress and because you have to find out an angle between 90 and 180 , remember the following rule:

    In Ist quadrant: sin and cos are positive
    IInd quadrant: sin is positive ,cos is -ve
    III rd quadrant: Both are negative
    IV quadrant: sin is negative and cos is positive

    Now because in your question tan is negative , you muct look II and IV quadrants , cuz thats where tan is negative.
    In IInd quadrant : 180-Q
    In IV quadrant: 360-Q
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook