Question

Let cos(α+β)=4/5 then let sin(α- β)=5/13, where 0≤ α, β ≤ π/4. Then prove that tan 2α =56/33

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \cos \left(\alpha +\beta \right)=\frac{4}{5} \\ \Rightarrow \sin \left(\alpha +\beta \right)=\sqrt{1-{\left(\frac{4}{5}\right)}^2}=\frac{3}{5} \\ \mathrm{And} \\ \sin \left(\alpha -\beta \right)=\frac{5}{13} \\ \Rightarrow \cos \left(\alpha -\beta \right)=\sqrt{1-{\left(\frac{5}{13}\right)}^2}=\frac{12}{13} \\ \mathrm{Now}, \\ \sin 2\alpha =\sin \left[\left(\alpha +\beta \right)+\left(\alpha -\beta \right)\right] \\ =\sin \left(\alpha +\beta \right)\cos \left(\alpha -\beta \right)+\sin \left(\alpha -\beta \right)\cos \left(\alpha +\beta \right) \\ =\frac{3}{5}\times \frac{12}{13}+\frac{5}{13}\times \frac{4}{5} \\ =\frac{36}{65}+\frac{20}{65}=\frac{56}{65} \\ \therefore \cos 2\alpha =\sqrt{1-{\sin }^{2}2\alpha } \\ =\sqrt{1-{\left(\frac{56}{65}\right)}^2}=\sqrt{1-\frac{3136}{4225}}=\sqrt{\frac{1089}{4225}}=\frac{33}{65} \\ \mathrm{Hence},\tan 2\alpha =\frac{\sin 2\alpha }{\cos 2\alpha }=\frac{{\displaystyle \frac{56}{65}}}{{\displaystyle \frac{33}{65}}}=\frac{56}{33} \end{gathered}$$