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anonymous
 4 years ago
Solve 1 sin 2 theta/ 1cos theta= cos theta
anonymous
 4 years ago
Solve 1 sin 2 theta/ 1cos theta= cos theta

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0use parenthesis please

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.01(sin 2 theta) / (1cos theta) = cos theta

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.01+cosx=sin2x/(1cosx) 1(cosx)^2=2sinxcosx (sinx)^2=2sinxcosx sinx=2cosx tanx=2 x=arctan2 x=63.43 degree or 1.107 radians

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[1  \frac{sin(2\theta)}{1cos(\theta)} = cos(\theta)\] \[1 +cos(\theta) = \frac{sin(2\theta)}{1cos(\theta)}\] \[(1 +cos(\theta))(1cos(\theta)) = \frac{sin(2\theta)}{1cos(\theta)}*(1cos(\theta))\] \[1 cos^2(\theta) = sin(2\theta)\] 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
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