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Finding angle inverse cosine

  1. Oct 1, 2015 #1
    • Member warned about posting without the HW template
    Not sure if I'm doing this right. I have an angle theta to find but the cosine has been squared. I brought over inverse cosine to multiply to leave theta on its own. I was told the answer should be closer to 37 degrees? Am I doing something wrong here?

    IMAG0042.jpg
     
  2. jcsd
  3. Oct 1, 2015 #2

    jedishrfu

    Staff: Mentor

    Did you calculate it via

    arccos(sqrt(50/79.57) ?
     
  4. Oct 1, 2015 #3

    Mark44

    Staff: Mentor

    On the 4th line from the bottom you have ##\frac{\sigma_n}{\sigma_x} = \cos^2(\theta)##
    This is equivalent to ##\cos(\theta) = \pm \sqrt{\frac{\sigma_n}{\sigma_x}}##
    If you take the inverse cosine of both sides, you can isolate ##\theta##. You are NOT multiplying by inverse cosine to get this.

    There are many values of ##\theta## that satisfy the last equation above. One that I get is around 37.6°. Please show us the calculation you did.
     
  5. Oct 1, 2015 #4

    jedishrfu

    Staff: Mentor

    Remember the notation ##cos^{-1}(\theta)## is not the same as ##1/cos(\theta)##.
     
  6. Oct 1, 2015 #5
    Thanks for helping I worked it out as the same 37.56 degrees, my maths is a little rusty at the moment!
     
  7. Oct 1, 2015 #6

    Mark44

    Staff: Mentor

    Keep in mind that +/-. There's another value around 142°.
     
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