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Tanθ = 5/12 , with θ in Quadrant 1, find sin2θ?

  1. Jul 17, 2011 #1
    1. The problem statement, all variables and given/known data

    Tanθ = 5/12 , with θ in Quadrant 1, find sin2θ?

    2. Relevant equations



    3. The attempt at a solution

    I drew a diagram in quadrant 1, it pretty much looked like a right triangle with a 5 for the length of the sin part and the 12 for the length of the cos part of the tan of 5/12 mentioned in the initial question. Then I noted that sin will be positive in the first quadrant. Now I'm having trouble coming up with the formula to use, I tried Sin2θ=2sinθcosθ which did not work.
     
  2. jcsd
  3. Jul 17, 2011 #2
    What exactly are you using for sine and cosine in the equation? Keep in mind, sine and cosine cannot be less than -1 or greater than 1. 5/12 is the proportion, but that does not mean [itex]\sin(\theta)=5[/itex].

    It might be helpful to consider the definition sine=opposite/hypotenuse. The opposite side of your triangle is 5, as you know. What is the hypotenuse?
     
  4. Jul 17, 2011 #3
    the triangle is 5, 12, 13

    cos(theta) = adj/hyp
    sin(theta) = opp/hyp
    x2= ?
     
  5. Jul 18, 2011 #4

    Mark44

    Staff: Mentor

    So cos(θ) = ?
    And sin(θ) = ?
    You have all the numbers you need to figure these out. Once you get them, you can get sin(2θ) by using the double angle identity.
    It's customary to write two times a variable with the constant first, as in 2x. In any case, your variable is θ, so it would be 2θ. If you write x2 or θ2, people are liable to think you're trying to write x2 or θ2, or that you have a variable named x2.
     
  6. Jul 18, 2011 #5
    okay sorry about that, I meant "times two"
    would that have been the correct next step?
     
  7. Jul 18, 2011 #6
    Can you find [itex]\sin(\theta)[/itex] and [itex]\cos(\theta)[/itex]? If so, you can use the double angle identity for sine to find the answer.
     
  8. Jul 18, 2011 #7

    Mark44

    Staff: Mentor

    No. The steps are laid out in my post and in Strants's post.
     
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