If sin(α+β)=1 and sin(α−β)=1/2, Find tan(α+2β) and tan(2α+β)
Question
If
\[ \sin(\alpha+\beta)=1 \]
and
\[ \sin(\alpha-\beta)=\frac12 \]
where
\[ 0\le \alpha,\beta \le \frac{\pi}{2} \]
find:
\[ \tan(\alpha+2\beta) \quad \text{and} \quad \tan(2\alpha+\beta) \]
Solution
Since
\[ \sin(\alpha+\beta)=1 \]
\[ \alpha+\beta=\frac{\pi}{2} \]
Also,
\[ \sin(\alpha-\beta)=\frac12 \]
\[ \alpha-\beta=\frac{\pi}{6} \]
Adding,
\[ 2\alpha=\frac{\pi}{2}+\frac{\pi}{6} =\frac{2\pi}{3} \]
\[ \alpha=\frac{\pi}{3} \]
Subtracting,
\[ 2\beta=\frac{\pi}{2}-\frac{\pi}{6} =\frac{\pi}{3} \]
\[ \beta=\frac{\pi}{6} \]
Now,
\[ \alpha+2\beta = \frac{\pi}{3}+\frac{\pi}{3} = \frac{2\pi}{3} \]
\[ \tan(\alpha+2\beta) = \tan\frac{2\pi}{3} = -\sqrt3 \]
Also,
\[ 2\alpha+\beta = \frac{2\pi}{3}+\frac{\pi}{6} = \frac{5\pi}{6} \]
\[ \tan(2\alpha+\beta) = \tan\frac{5\pi}{6} = -\frac{1}{\sqrt3} \]
Therefore,
\[ \boxed{\tan(\alpha+2\beta)=-\sqrt3} \]
and
\[ \boxed{\tan(2\alpha+\beta)=-\frac{1}{\sqrt3}} \]