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JohnM
Group Title
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
 one year ago
JohnM Group Title
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
 one year ago

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JohnM Group TitleBest ResponseYou've already chosen the best response.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.
 one year ago

larrymud Group TitleBest 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
 one year ago

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

JohnM Group TitleBest ResponseYou've already chosen the best response.0
Of course! Thanks.
 one year ago
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