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Yoloswag097
Solve 1- sin 2 theta/ 1-cos theta= -cos theta
1-(sin 2 theta) / (1-cos theta) = -cos theta
1+cosx=sin2x/(1-cosx) 1-(cosx)^2=2sinxcosx (sinx)^2=2sinxcosx sinx=2cosx tanx=2 x=arctan2 x=63.43 degree or 1.107 radians
\[1 - \frac{sin(2\theta)}{1-cos(\theta)} = -cos(\theta)\] \[1 +cos(\theta) = \frac{sin(2\theta)}{1-cos(\theta)}\] \[(1 +cos(\theta))(1-cos(\theta)) = \frac{sin(2\theta)}{1-cos(\theta)}*(1-cos(\theta))\] \[1 -cos^2(\theta) = sin(2\theta)\] note that \[1 = sin^2(\theta) + cos^2(\theta) \] therefore \[1 -cos^2(\theta) = sin^2(\theta)\] so plug that back into the orginal equation \[sin^2(\theta) = sin(2\theta)\] also note that \[2cos(\theta)sin(\theta) = sin(2\theta)\] therefore \[sin^2(\theta) = 2cos(\theta)sin(\theta)\] divide by \[sin(\theta)\] and you get \[sin(\theta)=2cos(\theta)\] divide by \[cos(\theta)\]and you get \[tan(\theta)=2\] do a little \[tan^{-1}(\theta)=tan^{-1}(2)\] and you've solved for theta