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Question Number 1808 by alib last updated on 04/Oct/15

Prove sin (a)+sin (a+((14π)/3))+sin (a−((8π)/3))=0/

$$\boldsymbol{\mathcal{P}}{rove}\:\:\mathrm{sin}\:\left({a}\right)+\mathrm{sin}\:\left({a}+\frac{\mathrm{14}\pi}{\mathrm{3}}\right)+\mathrm{sin}\:\left({a}−\frac{\mathrm{8}\pi}{\mathrm{3}}\right)=\mathrm{0}/ \\ $$$$ \\ $$

Answered by 112358 last updated on 05/Oct/15

Using the compound angle formula sin(x±y)=sinxcosy±cosxsiny we get sin(a+((14π)/3))=sin(a)cos(((14π)/3))+cos(a)sin(((14π)/3)) ((14π)/3)=4π+((2π)/3) so that sin(((14π)/3))=sin4πcos((2π)/3)+sin((2π)/3)cos4π =0+((√3)/2) sin(14π/3)=(√3)/2 Using cos(x±y)=cosxcosy∓sinxsiny we obtain cos((14π)/3)=cos4πcos((2π)/3)−sin4πsin((2π)/3) =−(1/2)−0 cos(14π/3)=1/2 ∴ sin(a+((14π)/3))=(1/2)(−sina+(√3)cosa) Next we get sin(a−((8π)/3))=sin(a)cos((8π)/3)−sin((8π)/3)cos(a) =sin(a)cos(2π+((2π)/3))−sin(2π+((2π)/3))cos(a) =−(1/2)sina−((√3)/2)cosa =−(1/2)(sina+(√3)cosa) The left−hand expression becomes sin(a)+(1/2)(−sina+(√3)cosa)−(1/2)(sina+(√3)cosa)=sina−sina=0

$${Using}\:{the}\:{compound}\:{angle}\:{formula} \\ $$$${sin}\left({x}\pm{y}\right)={sinxcosy}\pm{cosxsiny}\:{we}\:{get} \\ $$$${sin}\left({a}+\frac{\mathrm{14}\pi}{\mathrm{3}}\right)={sin}\left({a}\right){cos}\left(\frac{\mathrm{14}\pi}{\mathrm{3}}\right)+{cos}\left({a}\right){sin}\left(\frac{\mathrm{14}\pi}{\mathrm{3}}\right) \\ $$$$\:\frac{\mathrm{14}\pi}{\mathrm{3}}=\mathrm{4}\pi+\frac{\mathrm{2}\pi}{\mathrm{3}}\:{so}\:{that}\: \\ $$$${sin}\left(\frac{\mathrm{14}\pi}{\mathrm{3}}\right)={sin}\mathrm{4}\pi{cos}\frac{\mathrm{2}\pi}{\mathrm{3}}+{sin}\frac{\mathrm{2}\pi}{\mathrm{3}}{cos}\mathrm{4}\pi \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{0}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${sin}\left(\mathrm{14}\pi/\mathrm{3}\right)=\sqrt{\mathrm{3}}/\mathrm{2} \\ $$$${Using}\:{cos}\left({x}\pm{y}\right)={cosxcosy}\mp{sinxsiny} \\ $$$${we}\:{obtain}\: \\ $$$${cos}\frac{\mathrm{14}\pi}{\mathrm{3}}={cos}\mathrm{4}\pi{cos}\frac{\mathrm{2}\pi}{\mathrm{3}}−{sin}\mathrm{4}\pi{sin}\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{0} \\ $$$${cos}\left(\mathrm{14}\pi/\mathrm{3}\right)=\mathrm{1}/\mathrm{2} \\ $$$$\therefore\:{sin}\left({a}+\frac{\mathrm{14}\pi}{\mathrm{3}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(−{sina}+\sqrt{\mathrm{3}}{cosa}\right) \\ $$$${Next}\:{we}\:{get} \\ $$$${sin}\left({a}−\frac{\mathrm{8}\pi}{\mathrm{3}}\right)={sin}\left({a}\right){cos}\frac{\mathrm{8}\pi}{\mathrm{3}}−{sin}\frac{\mathrm{8}\pi}{\mathrm{3}}{cos}\left({a}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={sin}\left({a}\right){cos}\left(\mathrm{2}\pi+\frac{\mathrm{2}\pi}{\mathrm{3}}\right)−{sin}\left(\mathrm{2}\pi+\frac{\mathrm{2}\pi}{\mathrm{3}}\right){cos}\left({a}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−\frac{\mathrm{1}}{\mathrm{2}}{sina}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{cosa} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−\frac{\mathrm{1}}{\mathrm{2}}\left({sina}+\sqrt{\mathrm{3}}{cosa}\right) \\ $$$${The}\:{left}−{hand}\:{expression}\:{becomes} \\ $$$${sin}\left({a}\right)+\frac{\mathrm{1}}{\mathrm{2}}\left(−{sina}+\sqrt{\mathrm{3}}{cosa}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left({sina}+\sqrt{\mathrm{3}}{cosa}\right)={sina}−{sina}=\mathrm{0} \\ $$

Commented by alib last updated on 05/Oct/15

so the equation is not equal to zero?

$${so}\:{the}\:{equation}\:{is}\:{not}\:{equal}\:{to}\:{zero}? \\ $$

Commented by 112358 last updated on 05/Oct/15

Mistake corrected.

$${Mistake}\:{corrected}. \\ $$

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