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!

Trigomometric ratio problem

  1. Oct 9, 2012 #1
    1. The problem statement, all variables and given/known data
    Please see the attached.
    Q1.Why 0° is not a solution of the equation cosθ = 2sinθ - 1?
    What's wrong with my steps?
    Q2:http://www.wolframalpha.com/input/?i=cosx+=+2sinx+-+1
    From the graph,the x-coordinate of the intersection point does NOT equal to the answer
    that I calculated in the attached
    (0° +360°n or 180° + 360°n or 53.1°+360°n or 126.9° + 360°n),
    why?

    2. Relevant equations



    3. The attempt at a solution
    see the attached.
     

    Attached Files:

  2. jcsd
  3. Oct 9, 2012 #2


    Well let's break this up and substitute 0 for θ.

    cos0 = 1
    2sin0 = 0

    So we have 1 = -1, which is clearly not true. I'm really not sure what you were trying to do.
     
  4. Oct 9, 2012 #3
    I mean when we solve cosθ = 2sinθ - 1,we get
    θ = 0° +360°n or 180° + 360°n or 53.1°+360°n or 126.9° + 360°n
    That means when θ equals to the above values,LHS(cosθ) = RHS(2sinθ - 1)
    But why when we substitute 0 into the above equation,LHS =/= RHS?
    And from the graph,the x-coordinate of the intersection point does not equal to the value calculated(0° +360°n or 180° + 360°n or 53.1°+360°n or 126.9° + 360°n),why?
    Thx a lot
     
  5. Oct 10, 2012 #4
    Sorry I have solved Q2 :)
    But I have not solved Q1 yet,hope someone can explain to me why x = 0 degree is not the answer.THX :)
     
  6. Oct 10, 2012 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    θ= 0 is not a solution to cos(θ)= 2sin(θ)- 1 because cos(0)= 1 and 2sin(0)- 1= 0- 1= -1, not 1! You apparently squared both sides- any time you do that, or multiply both sides of an equation, you can introduce new solutions that satisfy the new equation but not the original one. To take an obvious example, if you multiply both sides of the equation x= 1 by x, you get [itex]x^2= x[/itex] which now has both x= 1 and x= 0 as solutions.

    53.1 degrees is (approximately) a solution. 126.9 is NOT because then the left side of the equation is negative while the right side is positive.
     
  7. Oct 25, 2012 #6
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Trigomometric ratio problem
  1. Ratio problem (Replies: 2)

  2. Ratio problem (Replies: 1)

  3. Ratio Problem (Replies: 6)

Loading...