Question Number 204293 by depressiveshrek last updated on 11/Feb/24
$$\mathrm{Prove}\:\mathrm{the}\:\mathrm{following}\:\mathrm{trig}\:\mathrm{identity}: \\ $$$$\frac{\mathrm{2sin}\alpha+\mathrm{sin3}\alpha+\mathrm{sin5}\alpha}{\mathrm{cos}\alpha−\mathrm{2cos2}\alpha+\mathrm{cos3}\alpha}=\frac{\mathrm{2cos2}\alpha}{\mathrm{tan}\frac{\alpha}{\mathrm{2}}} \\ $$
Commented by Frix last updated on 11/Feb/24
$$\mathrm{Try}. \\ $$$$\alpha=\frac{\pi}{\mathrm{6}}\:\Rightarrow\:\mathrm{lhs}=−\mathrm{5}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\:\:\:\:\:\mathrm{rhs}=\mathrm{2}+\sqrt{\mathrm{3}} \\ $$$$\alpha=\frac{\pi}{\mathrm{2}}\:\Rightarrow\:\mathrm{lhs}=\mathrm{1}\:\:\:\:\:\mathrm{rhs}=−\mathrm{2} \\ $$$$... \\ $$
Answered by Frix last updated on 11/Feb/24
$$\mathrm{It}'\mathrm{s}\:\mathrm{false}. \\ $$
Commented by depressiveshrek last updated on 11/Feb/24
$$\mathrm{Why}? \\ $$