anonymous
  • anonymous
Can someone help me with Problem Set 1: 1B-13
OCW Scholar - Multivariable Calculus
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anonymous
  • anonymous
Can someone help me with Problem Set 1: 1B-13
OCW Scholar - Multivariable Calculus
katieb
  • katieb
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anonymous
  • anonymous
The 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
  • phi
cos(-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
  • phi
Here 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|

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phi
  • phi
Now 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 \[ \cos \left(\theta_2 - \theta_1 \right)= \cos \theta_1 \cos \theta_2 + \sin \theta_1 \sin \theta_2\]
anonymous
  • anonymous
Thank you, Phi!

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