Mathematics, 29.07.2019 19:40 gbprulesmile
How do you prove cosx = 1-tan^2(x/2)/1+tan^2(x/2)?
Answers: 1
![\bf tan\left(\cfrac{{{ \theta}}}{2}\right)= \begin{cases} \pm \sqrt{\cfrac{1-cos({{ \theta}})}{1+cos({{ \theta}})}} \\ \quad \\ \boxed{\cfrac{sin({{ \theta}})}{1+cos({{ \theta}})}} \\ \quad \\ \cfrac{1-cos({{ \theta}})}{sin({{ \theta}})} \end{cases}\\\\ -------------------------------\\\\ tan^2\left( \frac{x}{2} \right)\implies \left[ \cfrac{sin(x)}{1+cos(x)} \right]^2\implies \cfrac{sin^2(x)}{[1+cos(x)]^2} \\\\\\ \boxed{\cfrac{sin^2(x)}{1+2cos(x)+cos^2(x)}}](/tpl/images/0147/8307/5134e.png)
![\bf -------------------------------\\\\ recall\qquad sin^2(\theta)+cos^2(\theta)=1\\\\ -------------------------------\\\\ \cfrac{\boxed{sin^2(x)+cos^2(x)}+2cos(x)+cos^2(x)~-~sin^2(x)}{1+2cos(x)+\boxed{1}} \\\\\\ \cfrac{cos^2(x)+2cos(x)+cos^2(x)}{2+2cos(x)}\implies \cfrac{2cos(x)+2cos^2(x)}{2+2cos(x)} \\\\\\ \cfrac{\underline{2} cos(x)~\underline{[1+cos(x)]}}{\underline{2}~\underline{[1+cos(x)]}}\implies cos(x)](/tpl/images/0147/8307/b0904.png)
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