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