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FromTheMoon [43]
3 years ago
8

How do you prove cosx = 1-tan^2(x/2)/1+tan^2(x/2)?

Mathematics
1 answer:
nikdorinn [45]3 years ago
8 0
\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)}}

now, let's plug that in the right-hand-side expression,

\bf cos(x)=\cfrac{1-tan^2\left( \frac{x}{2} \right)}{1+tan^2\left( \frac{x}{2} \right)}\\\\
-------------------------------\\\\
\cfrac{1-tan^2\left( \frac{x}{2} \right)}{1+tan^2\left( \frac{x}{2} \right)}\implies \cfrac{1-\frac{sin^2(x)}{1+2cos(x)+cos^2(x)}}{1+\frac{sin^2(x)}{1+2cos(x)+cos^2(x)}}
\\\\\\
\cfrac{\frac{1+2cos(x)+cos^2(x)~-~sin^2(x)}{1+2cos(x)+cos^2(x)}}{\frac{1+2cos(x)+cos^2(x)~+~sin^2(x)}{1+2cos(x)+cos^2(x)}}

\bf \cfrac{1+2cos(x)+cos^2(x)~-~sin^2(x)}{\underline{1+2cos(x)+cos^2(x)}}\cdot \cfrac{\underline{1+2cos(x)+cos^2(x)}}{1+2cos(x)+cos^2(x)~+~sin^2(x)}
\\\\\\
\cfrac{1+2cos(x)+cos^2(x)~-~sin^2(x)}{1+2cos(x)+cos^2(x)~+~sin^2(x)}

\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)
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