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

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1. anonymous

2. anonymous

1-(sin 2 theta) / (1-cos theta) = -cos theta

3. anonymous

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

4. anonymous

$1 - \frac{sin(2\theta)}{1-cos(\theta)} = -cos(\theta)$ $1 +cos(\theta) = \frac{sin(2\theta)}{1-cos(\theta)}$ $(1 +cos(\theta))(1-cos(\theta)) = \frac{sin(2\theta)}{1-cos(\theta)}*(1-cos(\theta))$ $1 -cos^2(\theta) = sin(2\theta)$ 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

5. anonymous

Thank you(:

6. Swag

good job