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Question Number 8267 by lepan last updated on 04/Oct/16

Show that sinA+sinB=2sin((A+B)/2) cos((A−B)/2).

$${Show}\:{that}\:{sinA}+{sinB}=\mathrm{2}{sin}\frac{{A}+{B}}{\mathrm{2}}\:{cos}\frac{{A}−{B}}{\mathrm{2}}. \\ $$$$ \\ $$

Answered by Yozzias last updated on 04/Oct/16

Let p=(1/2)(A+B) and q= (1/2)(A−B). ∴ 2p=A+B and 2q=A−B. ⇒2p+2q=2A⇒A=p+q anf 2p−2q=2B⇒B=p−q. Using the compound angle formular sin(t±g)=sintcosg±singcost ⇒sinA=sin(p+q)=sinpcosq+cospsinq and sinB=sin(p−q)=sinpcosq−cospsinq. ⇒sinA+sinB=sinpcosq+cospsinq+sinpcosq−cospsinq sinA+sinB=2sinpcosq=2sin((A+B)/2)cos((A−B)/2).

$$\mathrm{Let}\:\mathrm{p}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{A}+\mathrm{B}\right)\:\mathrm{and}\:\mathrm{q}=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{A}−\mathrm{B}\right). \\ $$$$\therefore\:\mathrm{2p}=\mathrm{A}+\mathrm{B}\:\mathrm{and}\:\mathrm{2q}=\mathrm{A}−\mathrm{B}. \\ $$$$\Rightarrow\mathrm{2p}+\mathrm{2q}=\mathrm{2A}\Rightarrow\mathrm{A}=\mathrm{p}+\mathrm{q}\: \\ $$$$\mathrm{anf}\:\mathrm{2p}−\mathrm{2q}=\mathrm{2B}\Rightarrow\mathrm{B}=\mathrm{p}−\mathrm{q}. \\ $$$$\mathrm{Using}\:\mathrm{the}\:\mathrm{compound}\:\mathrm{angle}\:\mathrm{formular} \\ $$$$\mathrm{sin}\left(\mathrm{t}\pm\mathrm{g}\right)=\mathrm{sintcosg}\pm\mathrm{singcost} \\ $$$$\Rightarrow\mathrm{sinA}=\mathrm{sin}\left(\mathrm{p}+\mathrm{q}\right)=\mathrm{sinpcosq}+\mathrm{cospsinq} \\ $$$$\mathrm{and}\:\mathrm{sinB}=\mathrm{sin}\left(\mathrm{p}−\mathrm{q}\right)=\mathrm{sinpcosq}−\mathrm{cospsinq}. \\ $$$$\Rightarrow\mathrm{sinA}+\mathrm{sinB}=\mathrm{sinpcosq}+\mathrm{cospsinq}+\mathrm{sinpcosq}−\mathrm{cospsinq} \\ $$$$\mathrm{sinA}+\mathrm{sinB}=\mathrm{2sinpcosq}=\mathrm{2sin}\frac{\mathrm{A}+\mathrm{B}}{\mathrm{2}}\mathrm{cos}\frac{\mathrm{A}−\mathrm{B}}{\mathrm{2}}. \\ $$

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