Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

JohnM

  • 3 years ago

Session 20 - Examples: Velocity, Speed and Arc Length In #2 in this PDF how does 2a<sin^2(θ/2), sin(θ/2)cos(θ/2)> get reduced to 2a (sin^2(θ/2))^1/2? I assume we start by squaring sin^2(θ/2) and sin(θ/2)cos(θ/2) and then taking the square root of the sums, yes? If yes I can reduce sin^2(θ/2) down to itself but I cannot seem to get rid of sin(θ/2)cos(θ/2)? My steps are: 1. Take the square root of the sum of (sin^2(θ/2))^2 + ((sin(θ/2)cos(θ/2))^2 2. Square root of: (1/2[1-cos(θ)])^2 + (1/2sin(θ))^2 3. Square root of: (sin^2(θ/2))^2 + (1/2sin(θ))^2 I'm stuck here.

  • This Question is Closed
  1. JohnM
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Note: I'm leaving out 2a until the end, after I find a ways to solve the problem of taking the square root of the sum of the squares. In other words, I'm OK with the 2a.

  2. larrymud
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    [sin(θ/2)]^2 +[sin(θ/2)cos(θ/2)]^2 =sin(θ/2)^4 +sin(θ/2)^2+cos(θ/2)^2 =sin(θ/2)^2 * [sin(θ/2)^2 + cos(θ/2)^2] =sin(θ/2)^ * 1

  3. larrymud
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    The second line in the above answer should have sin^2 * cos^2, not '+', sorry.

  4. JohnM
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Of course! Thanks.

  5. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy