deletions | additions       diff --git a/section_Analysis_1_Direct_Fit__.tex b/section_Analysis_1_Direct_Fit__.tex  index cf6b009..ce499d2 100644  --- a/section_Analysis_1_Direct_Fit__.tex  +++ b/section_Analysis_1_Direct_Fit__.tex 
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$$V_{c}=\frac{1}{L}\times \frac{\Delta\theta}{\Delta B}=\frac{1}{0.1m}\times 2.07\frac{radians}{T}=20.7\frac{radians}{T*m}$$  2. Slope Fit\newline  Initially, %Initially,  we fit our data to the form $V=V_{0}(sin(\theta_{1}+\theta_{2}))^2$, which using trigonometry, is $V=V_{0}\frac{(1-cos(\theta_{1}+\theta_{2})^2)}{2}$ (the actual equation that we fit to). However, we should have been fitting to the form $V=V_{0}cos(\theta_{1}-\theta_{2})^2$, which is $V=V_{0}\frac{(1+cos(\theta_{1}-\theta_{2})^2)}{2}$. Now, $$sin(\theta_{1}+\theta_{2})^2=sin2(θ1-(-θ2))$$  2π-θ2=θ2’ %$$sin(\theta_{1}+\theta_{2})^2=sin2(θ1-(-θ2))$$  %2π-θ2=θ2’  to account for the incorrect sign in our original equation. θ2’-π=θ0 %θ2’-π=θ0  to change from sine to cosine, so we do end up with the form V=V0(½)(1+cos(2(θ1-θ0)))  which %V=V0(½)(1+cos(2(θ1-θ0)))  %which  is V=V0cos2(θ1-θ2), giving us the form we should have initially fit to. As shown in Figure 2:  $$V=0.000443B+0.127$$