+26400 In the unit circle we find by using the periphery angle:
\(\tan{( \frac{\theta}{2} ) } = \frac{ \sin{( \theta ) }}{ 1+\cos{(\theta ) }} \)
49.
\(\begin{array}{rcll} \tan{( \frac{\theta}{2} ) } &=& \frac{ \sin{( \theta ) }}{ 1+\cos{(\theta ) }} \qquad \sin{( \theta ) } = \frac35 \qquad \cos{( \theta ) } = \frac45 \\ \tan{( \frac{\theta}{2} ) } &=& \frac{ \frac35 }{ 1+\frac45 } \\ \tan{( \frac{\theta}{2} ) } &=& \frac35 \cdot \frac{ 1 }{ 1+\frac45 } \\ \tan{( \frac{\theta}{2} ) } &=& \frac35 \cdot \frac{ 1 }{ \frac{5+4}{5} } \\ \tan{( \frac{\theta}{2} ) } &=& \frac35 \cdot \frac{ 5 }{ 9 } \\ \tan{( \frac{\theta}{2} ) } &=& \frac39 \\ \tan{( \frac{\theta}{2} ) } &=& \frac13 \\ \end{array}\)
+26400 In the unit circle we find by using the periphery angle:
\(\tan{( \frac{\theta}{2} ) } = \frac{ \sin{( \theta ) }}{ 1+\cos{(\theta ) }} \)
49.
\(\begin{array}{rcll} \tan{( \frac{\theta}{2} ) } &=& \frac{ \sin{( \theta ) }}{ 1+\cos{(\theta ) }} \qquad \sin{( \theta ) } = \frac35 \qquad \cos{( \theta ) } = \frac45 \\ \tan{( \frac{\theta}{2} ) } &=& \frac{ \frac35 }{ 1+\frac45 } \\ \tan{( \frac{\theta}{2} ) } &=& \frac35 \cdot \frac{ 1 }{ 1+\frac45 } \\ \tan{( \frac{\theta}{2} ) } &=& \frac35 \cdot \frac{ 1 }{ \frac{5+4}{5} } \\ \tan{( \frac{\theta}{2} ) } &=& \frac35 \cdot \frac{ 5 }{ 9 } \\ \tan{( \frac{\theta}{2} ) } &=& \frac39 \\ \tan{( \frac{\theta}{2} ) } &=& \frac13 \\ \end{array}\)
