Question Number 17828 by Abbas-Nahi last updated on 11/Jul/17
Answered by Tinkutara last updated on 11/Jul/17
$$\mathrm{sin}\:\left(\alpha\:+\:\beta\right)\:=\:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\beta\:+\:\mathrm{cos}\:\alpha\:\mathrm{sin}\:\beta \\ $$$$\mathrm{sin}\:\left(\alpha\:−\:\beta\right)\:=\:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\beta\:−\:\mathrm{cos}\:\alpha\:\mathrm{sin}\:\beta \\ $$$$\mathrm{sin}\:\left(\alpha\:+\:\beta\right)\:+\:\mathrm{sin}\:\left(\alpha\:−\:\beta\right)\:=\:\mathrm{2}\:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\beta \\ $$$$\mathrm{Let}\:\alpha\:+\:\beta\:=\:{x}\:\mathrm{and}\:\alpha\:−\:\beta\:=\:{y} \\ $$$$\mathrm{so}\:\mathrm{that}\:\alpha\:=\:\frac{{x}\:+\:{y}}{\mathrm{2}}\:\mathrm{and}\:\beta\:=\:\frac{{x}\:−\:{y}}{\mathrm{2}}. \\ $$$$\boldsymbol{\mathrm{sin}}\:\boldsymbol{{x}}\:+\:\boldsymbol{\mathrm{sin}}\:\boldsymbol{{y}}\:=\:\mathrm{2}\:\boldsymbol{\mathrm{sin}}\:\frac{\boldsymbol{{x}}\:+\:\boldsymbol{{y}}}{\mathrm{2}}\:\boldsymbol{\mathrm{cos}}\:\frac{\boldsymbol{{x}}\:−\:\boldsymbol{{y}}}{\mathrm{2}} \\ $$
Commented by Abbas-Nahi last updated on 11/Jul/17
$$\boldsymbol{{thank}}\:\boldsymbol{{you}}\:\boldsymbol{{M}}{r}\:{Tinkutara} \\ $$