Question Number 110254 by ZiYangLee last updated on 28/Aug/20
$$\mathrm{Given}\:\mathrm{tan}\:\alpha\:\mathrm{and}\:\mathrm{tan}\:\beta\:\mathrm{are}\:\mathrm{the}\:\mathrm{two}\:\mathrm{roots}\: \\ $$$$\mathrm{of}\:\mathrm{2}{x}^{\mathrm{2}} −{x}−\mathrm{2}=\mathrm{0},\:\mathrm{then} \\ $$$$\mathrm{sin}\left(\mathrm{2}\alpha+\mathrm{2}\beta\right)+\mathrm{cos}\left(\mathrm{2}\alpha+\mathrm{2}\beta\right)+\mathrm{tan}\left(\mathrm{2}\alpha+\mathrm{2}\beta\right)=? \\ $$
Answered by som(math1967) last updated on 28/Aug/20
$$\mathrm{tan}\alpha+\mathrm{tan}\beta=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{tan}\alpha\mathrm{tan}\beta=\frac{−\mathrm{2}}{\mathrm{2}}=−\mathrm{1} \\ $$$$\mathrm{tan}\left(\alpha+\beta\right)=\frac{\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{1}−\left(−\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{sin2}\left(\alpha+\beta\right)+\mathrm{cos2}\left(\alpha+\beta\right)+\mathrm{tan}\:\mathrm{2}\left(\alpha+\beta\right) \\ $$$$\frac{\mathrm{2tan}\left(\alpha+\beta\right)}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \left(\alpha+\beta\right)}+\frac{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \left(\alpha+\beta\right)}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \left(\alpha+\beta\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:+\frac{\mathrm{2tan}\left(\alpha+\beta\right)}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \left(\alpha+\beta\right)} \\ $$$$\mathrm{now}\:\mathrm{put}\:\mathrm{tan}\left(\alpha+\beta\right)=\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Commented by ZiYangLee last updated on 28/Aug/20
$$\mathrm{Thanks}! \\ $$