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anonymous
 one year ago
Can someone help me with Problem Set 1: 1B13
anonymous
 one year ago
Can someone help me with Problem Set 1: 1B13

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The question asks us to solve for cos(Θ1−Θ2), but the answer in the supplemental packet solves for cos(Θ2−Θ1). Is this a mistake?

phi
 one year ago
Best ResponseYou've already chosen the best response.2cos(x) = cos(x) so their switching theta1 and theta2 does not change the answer. (Though it would be nice if they were consistent, so as not to introduce confusion)

phi
 one year ago
Best ResponseYou've already chosen the best response.2Here are more details on their derivation of the cos of a difference of two angles. First, they define a unit length vector with its tail at the origin and its head lying on the unit circle. Based on this definition, we know its i and j components dw:1442327812962:dw we thus have two unit vectors \[ \vec{u_1}= < \cos \theta_1 , \sin \theta_1> \\ \vec{u_2}= < \cos \theta_2 , \sin \theta_2> \] dw:1442328058455:dw

phi
 one year ago
Best ResponseYou've already chosen the best response.2Now use the definition of a dot product \[ \vec{u_1} \cdot \vec{u_2} = u_1 u_2 \cos \phi \] where phi is the angle between the two vectors. we get (note: both vectors have unit length) \[ \cos \theta_1 \cos \theta_2 + \sin \theta_1 \sin \theta_2 = 1 \cdot 1 \cdot \cos \phi \] thus, where \(ϕ=θ_2−θ_1\) we have the result \[ \cos \left(\theta_2  \theta_1 \right)= \cos \theta_1 \cos \theta_2 + \sin \theta_1 \sin \theta_2\]
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