Question

If sin (α + β) = 1 and sin (α − β) \[= \frac{1}{2}\], where 0 ≤ α, \[\beta \leq \frac{\pi}{2}\], then find the values of tan (α + 2β) and tan (2α + β).

Answer 1

Given:
\[\sin (\alpha + \beta) = 1\text{ and }\sin (\alpha - \beta) = \frac{1}{2}\]
\[ \Rightarrow \alpha + \beta = 90^\circ . . . (1) \]
\[and \alpha - \beta = 30^\circ . . . (2) \]
By adding eq (1) and eq (2) we get:
\[ 2\alpha = 120^\circ\]
\[ \Rightarrow \alpha = 60^\circ\]
By subtracting eq (2) from eq (1), we get:
\[ 2\beta = 60^\circ\]
\[ \Rightarrow \beta = 30^\circ\]
Therefore, 
\[\tan(\alpha + 2\beta) = \tan \left( 60^\circ + 2 \times 30^\circ \right) = \tan 120^\circ = - \sqrt{3}\]
\[\tan(2\alpha + \beta) = \tan \left( 2 \times 60^\circ + 30^\circ \right) = \tan 150^\circ = - \frac{1}{\sqrt{3}}\]