1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Inverse trigonometry question

  1. May 22, 2016 #1
    1. The problem statement, all variables and given/known data
    If 1/2<=x<=1
    , then cos-1[ x/2 + (√(3-3x2))/2] + cos-1 x is equal to?



    2. Relevant equations


    3. The attempt at a solution
    Substituting x=cosθ
    = cos-1(cosθ) + cos-1 [ cos(θ)/2 + (√(3(1-cos2(θ)))/2]
    = θ + cos-1 [cos (π/3) cos θ + sin (π/3) sinθ]
    = θ + cos-1 [cos (θ -(π/3)]
    = 2θ -(π/3)
    = 2 cos-1x -(π/3)
    But the correct answer according to the book is π/3. Please help me understand what I've done wrong.
     
    Last edited: May 22, 2016
  2. jcsd
  3. May 22, 2016 #2

    Mark44

    Staff: Mentor

    Then use parentheses to indicate this, as in √(3-3x2)/2. From what you wrote I can't tell if you mean this:
    $$\frac{\sqrt{3 - 3x^2}} 2$$

    or this:$$\sqrt{\frac{3 - 3x^2}{2}}$$

    Also, what is the question? The problem statement should include what it is you need to do.
     
  4. May 22, 2016 #3
    It is the first one. Square root for the numerator.
    We are asked to find what that expression is equal to. I've reworded it now.
     
  5. May 22, 2016 #4

    Mark44

    Staff: Mentor

    How did you get the term on the right, above? I don't see where cos(π/3) and sin(π/3) come in.

    I haven't worked the problem yet, but I suspect that your substitution x = cos(θ) isn't the best one to use.
     
  6. May 22, 2016 #5
    1/2= cosπ/3
    √3/2 =sinπ/3
    So I thought i'll plug those in and use the cos(x+y)=cosxcosy -sinxsiny.
     
  7. May 22, 2016 #6

    Mark44

    Staff: Mentor

    My question is this: How does cosθ/2 + √(3(1-cos2θ))/2 become cos π/3 cos θ + sinπ/3 sinθ?

    Also, you are continuing to write things ambiguously. When you write cosθ/2 do you mean ##\cos(\frac{\theta}{2})## or ##\frac{\cos(\theta)}{2}##? And the same question for the sine term.
    Use parentheses to make things clear!
     
  8. May 22, 2016 #7
    it's ##\frac{\cos(\theta)}{2}## .
    I'm sorry, I've made more corrections to the question now. Is it clear?
     
  9. May 22, 2016 #8

    Mark44

    Staff: Mentor

    It's clearer now, but I still don't understand something.
    How does this -- cos(θ)/2 + (√(3(1-cos2(θ)))/2 -- turn into this -- cos π/3 cos θ + sinπ/3 sinθ?
     
  10. May 22, 2016 #9
    cos(θ) * 1/2 + sin(θ) *√3/2 ====> cos (θ) cos (π/3) + sin (θ) sin(π/3).
    as √(1-cos2θ) =sin(θ)
    Few more corrections added to the question.
     
    Last edited: May 22, 2016
  11. May 22, 2016 #10

    Mark44

    Staff: Mentor

    OK, now I see. One thing you didn't take into account when you went from ##\sqrt{1 - \cos^2(\theta)}\frac{\sqrt{3}} 2## to ##\sin(\theta) \frac{\sqrt{3}} 2## is that ##\sqrt{1 - \cos^2(\theta)}## isn't equal to ##\sin(\theta)## -- it's equal to ##|\sin(\theta)|##.
    If ##0 \le \theta \le \frac{\pi} 3## then ##\sin(\theta) \ge 0## but if ##-\frac{\pi} 3 \le \theta \le 0##, then ##\sin(\theta) \le 0##.

    I'm not sure if that's where the problem lies -- I still haven't worked through this problem. I've just been trying to follow what you're doing.
     
  12. May 22, 2016 #11
    Well according tot he question 1/2<=x<=1
    So, (π/3) ≤(cos-1x or θ) ≤ 0 ? That doesn't seem right. The inequality signs seem to be reversed. Am I missing something?
     
  13. May 22, 2016 #12

    Mark44

    Staff: Mentor

    I wrote ##-\frac{\pi} 3 \le \theta \le 0##

    In your substitution, ##x = \cos(\theta)##, and you're given that 1/2 <= x <= 1. If you graph ##x = \cos(\theta)##, the usual thing to do would be to have x on the vertical axis, and ##\theta## the horizontal axis. If ##1/2 \le x \le 1##, then one arch of the cosine graph gives the interval for ##\theta## that I wrote.
     
  14. May 22, 2016 #13
    Looking at the cos-1 graph, the corresponding interval for 1/2≤x≤1 is 0≤θ≤π/3. How is it −π3≤θ≤0 ?The cos-1 graph's range doesn't even cover those values.
     
  15. May 23, 2016 #14

    Mark44

    Staff: Mentor

    The graph of x = cos(θ), with θ on the horizontal axis and x on the vertical axis, isn't one-to-one, so it doesn't have an inverse that is itself a function. To make x = cos(θ) one-to-one, the domain must be restricted. The usual convention is to restrict the domain to [0, π], but that interval is somewhat arbitrary.

    But your substitution was x = cos(θ), with the condition that 1/2 ≤ x ≤ 1. If you look at the graph of x = cos(θ) (which looks exactly like the graph of y = cos(x) ), when 1/2 ≤ x ≤ 1, the x values under the arch closest to the origin lie between -π/3 and +π/3. There are many other concave down arches for whiich 1/2 ≤ x ≤ 1, but I was talking about the one closest to the origin.
     
  16. May 23, 2016 #15
    I am sorry, but I just can't get it. "x = cos(θ) (which looks exactly like the graph of y = cos(x)" , yes but with the axis reversed. So in the cos-1, graph, are you asking me to pick another section where the function is invertible? ie [0, -π]? ie. other half of the concave section where -π/3≤x≤0 also lies?
    How does all this finally tie up with the question.?
     
    Last edited: May 23, 2016
  17. May 23, 2016 #16

    Mark44

    Staff: Mentor

    This side discussion is about my response in post #10.
    ##\sqrt{1 - \cos^2(\theta)} = \sqrt{\sin^2(\theta)}##
    The last expression above isn't equal to ##\sin(\theta)##; it's equal to ##|\sin(\theta)|##. You have to take into account the possibility that ##\sin(\theta)## could be negative.
     
  18. May 23, 2016 #17
    Yes, I get that, but even if sin θ is negative , it results in the final answer to be 2θ + (π/3). Which is 2cos-1(x) + (π/3).
    So considering both cases of sin(θ) being positive or negative doesn't give me the final answer as π/3.
    It seems as though the only way is to write this bit as,
    cos ( (π/3) - θ).
    Coming to think of it, how do we decide whether to choose cos( (π/3) -θ) or cos ( (π/3) +θ) from that previous expression?
     
  19. May 23, 2016 #18

    Mark44

    Staff: Mentor

    cos(A)cos(B) + sin(A)sin(B) = cos(A - B)
     
  20. May 23, 2016 #19

    ehild

    User Avatar
    Homework Helper
    Gold Member

    which is the same as cos(B-A). The range of the arccos function is [0, pi], so you have to choose the non-negative one from A-B and B-A.
    As the range of the arccos function is [0, pi], 0.5≤x≤1 means 0.5≤cos(θ)≤1, that is 0≤θ≤π/3. In that range of θ, sin(θ)≥0, so
    ##\sqrt{1-\cos^2(\theta)}=\sin(\theta)##. There is only one possible value for
    ## \cos^{-1} (\frac{x}{2}+\frac{\sqrt{3}}{2}\sqrt{1-\cos^2(x)})##.
     
  21. May 24, 2016 #20
    Finally......
    I shall summarize it again, please check if it's right.
    √(1-cos2θ) = +sin(θ) , as (π/3)≤θ≤0.
    This part is wrong, as the range of arccos is [0,π] and since (π/3)≤θ≤0, θ-(π/3) is -ve and I can write only the +ve (π/3)-θ.
    Just one more question.
    Why are the inequality signs being reversed?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Inverse trigonometry question
Loading...