Question Number 101382 by student work last updated on 02/Jul/20
Answered by Rio Michael last updated on 02/Jul/20
$$\mathrm{we}\:\mathrm{know}\:\mathrm{or}\:\mathrm{we}\:\mathrm{are}\:\mathrm{to}\:\mathrm{prove}\:\mathrm{that}\:\:\mathrm{sin}\:{A}\:+\:\mathrm{sin}{B}\:+\:\mathrm{sin}\:{C}\:=\:\mathrm{4}\:\mathrm{cos}\frac{{A}}{\mathrm{2}}\:\mathrm{cos}\frac{{B}}{\mathrm{2}}\:\mathrm{cos}\frac{{C}}{\mathrm{2}}\:\left(\left(\mathrm{1}\right)\right. \\ $$$$\mathrm{but}\:{A}\:+\:{B}\:+\:{C}\:=\:\pi\left(\mathrm{180}°\right) \\ $$$$\Rightarrow\:{A}\:+\:{B}\:=\:\pi−{C}\:\mathrm{and}\:\:\frac{{A}\:+{B}}{\mathrm{2}}\:=\:\frac{\pi\:−\:{C}}{\mathrm{2}} \\ $$$$ \\ $$$$\Rightarrow\:\:\frac{{A}\:+\:{B}}{\mathrm{2}}\:=\:\frac{\pi}{\mathrm{2}}−\frac{{C}}{\mathrm{2}} \\ $$$$\mathrm{so}\:\mathrm{sin}\:\left(\frac{{A}\:+\:{B}}{\mathrm{2}}\right)\:=\:\mathrm{cos}\left(\frac{\pi}{\mathrm{2}}\:−\frac{{C}}{\mathrm{2}}\right)\:\:=\:\mathrm{cos}\:\left(\frac{{C}}{\mathrm{2}}\right)\:\left(\mathrm{2}\right) \\ $$$$\mathrm{also}\:\mathrm{sin}\:{C}\:=\:\mathrm{2}\:\mathrm{sin}\:\left(\frac{{C}}{\mathrm{2}}\right)\:\mathrm{cos}\:\left(\frac{{C}}{\mathrm{2}}\right)……\left(\mathrm{3}\right) \\ $$$$\mathrm{but}\:\mathrm{sin}\:{A}+\:\mathrm{sin}\:{B}\:+\:\mathrm{sin}{C}\:\:=\:\mathrm{2}\:\mathrm{cos}\left(\frac{{C}}{\mathrm{2}}\right)\mathrm{cos}\:\left(\frac{{A}\:−\:{B}}{\mathrm{2}}\right)\:+\:\mathrm{2}\:\mathrm{sin}\:\left(\frac{{C}}{\mathrm{2}}\right)\:\mathrm{cos}\left(\frac{{C}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:=\:\mathrm{2}\:\mathrm{cos}\:\left(\frac{{C}}{\mathrm{2}}\right)\left[\mathrm{cos}\:\left(\frac{{A}\:−\:{B}}{\mathrm{2}}\right)\:+\:\mathrm{sin}\:\left(\frac{{C}}{\mathrm{2}}\right)\right] \\ $$$$\:\:\:\:\:=\:\mathrm{2}\:\mathrm{cos}\:\left(\frac{{C}}{\mathrm{2}}\right)\left[\mathrm{cos}\:\left(\frac{{A}−{B}}{\mathrm{2}}\right)\:+\:\mathrm{sin}\:\left(\frac{\pi−\left({A}\:+\:{B}\right)}{\mathrm{2}}\right)\right]\:\:\mathrm{since}\:{A}\:+\:{B}\:+\:{C}\:=\:\pi \\ $$$$\:\:\:\:\:=\:\mathrm{2}\:\mathrm{cos}\left(\frac{{C}}{\mathrm{2}}\right)\left[\mathrm{cos}\:\left(\frac{{A}−{B}}{\mathrm{2}}\right)\:+\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\frac{{A}\:+\:{B}}{\mathrm{2}}\right)\right]\: \\ $$$$\:\:\:\:\:=\:\mathrm{2}\:\mathrm{cos}\:\left(\frac{{C}}{\mathrm{2}}\right)\left[\mathrm{cos}\:\left(\frac{{A}−{B}}{\mathrm{2}}\right)\:+\:\mathrm{cos}\:\left(\frac{{A}\:+\:{B}}{\mathrm{2}}\right)\right] \\ $$$$\:\:\:\:\:=\:\mathrm{2}\:\mathrm{cos}\:\left(\frac{{C}}{\mathrm{2}}\right)\left[\mathrm{2cos}\:\left(\frac{\frac{{A}\:−{B}}{\mathrm{2}}\:+\:\frac{{A}\:+{B}}{\mathrm{2}}}{\mathrm{2}}\right)\:\mathrm{cos}\left(\frac{\frac{{A}−{B}}{\mathrm{2}}−\frac{{A}\:+{B}}{\mathrm{2}}}{\mathrm{2}}\right)\right] \\ $$$$\:\:\:\:\:=\:\mathrm{2}\:\mathrm{cos}\:\left(\frac{{C}}{\mathrm{2}}\right)\left[\mathrm{2cos}\left(\frac{{A}}{\mathrm{2}}\right)\mathrm{cos}\:\left(−\frac{{B}}{\mathrm{2}}\right)\right] \\ $$$$\:\:\:\:\:=\:\mathrm{4}\:\mathrm{cos}\:\left(\frac{{A}}{\mathrm{2}}\right)\mathrm{cos}\:\left(\frac{{B}}{\mathrm{2}}\right)\:\mathrm{cos}\:\left(\frac{{C}}{\mathrm{2}}\right)\:\mathrm{since}\:\mathrm{cos}\:{x}\:\mathrm{is}\:\mathrm{an}\:\mathrm{even}\:\mathrm{function}.\:{QED} \\ $$
Answered by 1549442205 last updated on 02/Jul/20
$$\boldsymbol{\mathrm{sinA}}+\boldsymbol{\mathrm{sinB}}=\mathrm{2}\boldsymbol{\mathrm{sin}}\frac{\boldsymbol{\mathrm{A}}+\boldsymbol{\mathrm{B}}}{\mathrm{2}}\boldsymbol{\mathrm{cos}}\frac{\boldsymbol{\mathrm{A}}−\boldsymbol{\mathrm{B}}}{\mathrm{2}}=\mathrm{2}\boldsymbol{\mathrm{cos}}\frac{\boldsymbol{\mathrm{C}}}{\mathrm{2}}\boldsymbol{\mathrm{cos}}\frac{\boldsymbol{\mathrm{A}}−\boldsymbol{\mathrm{B}}}{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{sinC}}=\mathrm{2}\boldsymbol{\mathrm{sin}}\frac{\boldsymbol{\mathrm{C}}}{\mathrm{2}}\boldsymbol{\mathrm{cos}}\frac{\boldsymbol{\mathrm{C}}}{\mathrm{2}}=\mathrm{2}\boldsymbol{\mathrm{cos}}\frac{\boldsymbol{\mathrm{C}}}{\mathrm{2}}\boldsymbol{\mathrm{cos}}\frac{\boldsymbol{\mathrm{A}}+\boldsymbol{\mathrm{B}}}{\mathrm{2}}.\boldsymbol{\mathrm{Hence}}, \\ $$$$\mathrm{s}\boldsymbol{\mathrm{inA}}+\boldsymbol{\mathrm{sinB}}+\boldsymbol{\mathrm{si}}\mathrm{n}\boldsymbol{\mathrm{C}}=\mathrm{2}\boldsymbol{\mathrm{cos}}\frac{\boldsymbol{\mathrm{C}}}{\mathrm{2}}\left(\boldsymbol{\mathrm{cos}}\frac{\boldsymbol{\mathrm{A}}−\boldsymbol{\mathrm{B}}}{\mathrm{2}}+\boldsymbol{\mathrm{cos}}\frac{\boldsymbol{\mathrm{A}}+\boldsymbol{\mathrm{B}}}{\mathrm{2}}\right) \\ $$$$=\mathrm{2}\boldsymbol{\mathrm{cos}}\frac{\boldsymbol{\mathrm{C}}}{\mathrm{2}}.\mathrm{2}\boldsymbol{\mathrm{cos}}\frac{\frac{\boldsymbol{\mathrm{A}}−\boldsymbol{\mathrm{B}}}{\mathrm{2}}+\frac{\boldsymbol{\mathrm{A}}+\boldsymbol{\mathrm{B}}}{\mathrm{2}}}{\mathrm{2}}\boldsymbol{\mathrm{cos}}\frac{\frac{\boldsymbol{\mathrm{A}}−\boldsymbol{\mathrm{B}}}{\mathrm{2}}−\frac{\boldsymbol{\mathrm{A}}+\boldsymbol{\mathrm{B}}}{\mathrm{2}}}{\mathrm{2}} \\ $$$$=\mathrm{4}\boldsymbol{\mathrm{cos}}\frac{\boldsymbol{\mathrm{C}}}{\mathrm{2}}\boldsymbol{\mathrm{cos}}\frac{\boldsymbol{\mathrm{A}}}{\mathrm{2}}\boldsymbol{\mathrm{cos}}\frac{−\boldsymbol{\mathrm{B}}}{\mathrm{2}}=\mathrm{4}\boldsymbol{\mathrm{cos}}\frac{\boldsymbol{\mathrm{A}}}{\mathrm{2}}\boldsymbol{\mathrm{cos}}\frac{\boldsymbol{\mathrm{B}}}{\mathrm{2}}\boldsymbol{\mathrm{cos}}\frac{\boldsymbol{\mathrm{C}}}{\mathrm{2}} \\ $$$$\left(\boldsymbol{\mathrm{q}}.\boldsymbol{\mathrm{e}}.\boldsymbol{\mathrm{d}}\right) \\ $$