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 one year 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[1cos(θ)])^2 + (1/2sin(θ))^2
3. Square root of: (sin^2(θ/2))^2 + (1/2sin(θ))^2
I'm stuck here.
 one year 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[1cos(θ)])^2 + (1/2sin(θ))^2 3. Square root of: (sin^2(θ/2))^2 + (1/2sin(θ))^2 I'm stuck here.

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JohnM
 one year ago
Best ResponseYou've already chosen the best response.0Note: 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.

larrymud
 one year ago
Best ResponseYou've already chosen the best response.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

larrymud
 one year ago
Best ResponseYou've already chosen the best response.1The second line in the above answer should have sin^2 * cos^2, not '+', sorry.
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