Identity Crisis

Using a sum-to-product identity,

$\frac { sin\left( \frac { 3\beta }{ 4 } \right) +sin\left( \frac { \beta }{ 4 } \right) }{ cos\left( \frac { 3\beta }{ 4 } \right) +cos\left( \frac { \beta }{ 4 } \right) } \qquad =\qquad \frac { 2sin\left( \frac { \frac { 3\beta }{ 4 } +\frac { \beta }{ 4 } }{ 2 } \right) cos\left( \frac { \frac { 3\beta }{ 4 } -\frac { \beta }{ 4 } }{ 2 } \right) }{ 2cos\left( \frac { \frac { 3\beta }{ 4 } +\frac { \beta }{ 4 } }{ 2 } \right) cos\left( \frac { \frac { 3\beta }{ 4 } -\frac { \beta }{ 4 } }{ 2 } \right) } \\ \qquad \qquad \qquad \qquad =\qquad \frac { sin\left( \frac { \beta }{ 2 } \right) }{ cos\left( \frac { \beta }{ 2 } \right) } \\ \qquad \qquad \qquad \qquad =\qquad tan\left( \frac { \beta }{ 2 } \right) \\ \qquad \qquad \qquad \qquad =\qquad \frac { cos\left( \beta \right) +1 }{ sin\left( \beta \right) }$

Since $y = cos\left( x \right)$ is an even function, $cos\left( -\beta \right) =cos\left( \beta \right)$ . Solving for $sin\left( \beta \right)$ and plugging in the values,

$\qquad \qquad \qquad \qquad =\qquad \frac { \frac { 4 }{ 5 } +1 }{ \frac { 3 }{ 5 } } \\ \qquad \qquad \qquad \qquad =\qquad 3$