Question Number 10825 by Saham last updated on 26/Feb/17
$$\mathrm{Given}\:\mathrm{that}\: \\ $$$$\mathrm{sin}\left(\mathrm{x}\right)\:−\:\mathrm{sin}\left(\mathrm{y}\right)\:=\:\mathrm{sin}\left(\theta\right) \\ $$$$\mathrm{cos}\left(\mathrm{x}\right)\:+\:\mathrm{cos}\left(\mathrm{y}\right)\:=\:\mathrm{cos}\left(\theta\right) \\ $$$$\mathrm{Show}\:\mathrm{that}\: \\ $$$$\mathrm{cos}\left(\mathrm{x}\:+\:\mathrm{y}\right)\:=\:−\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Answered by mrW1 last updated on 26/Feb/17
![(sin(x) − sin(y))^2 = sin^2 (θ) sin^2 (x)−2sin (x)sin (y)+sin^2 (y)=sin^2 (θ) ...(i) (cos(x) + cos(y))^2 = cos^2 (θ) cos^2 (x)+2cos (x)cos (y)+cos^2 (y)=cos^2 (θ) ...(ii) (i)+(ii): sin^2 (x)+cos^2 (x)+2cos (x)cos (y)−2sin (x)sin (y)+sin^2 (y)+cos^2 (y)=sin^2 (θ)+cos^2 (θ) 1+2[cos (x)cos (y)−sin (x)sin (y)]+1=1 1+2cos (x+y)+1=1 ⇒cos (x+y)=−(1/2)](https://www.tinkutara.com/question/Q10831.png)
$$\left(\mathrm{sin}\left(\mathrm{x}\right)\:−\:\mathrm{sin}\left(\mathrm{y}\right)\right)^{\mathrm{2}} \:=\:\mathrm{sin}^{\mathrm{2}} \left(\theta\right) \\ $$$$\mathrm{sin}^{\mathrm{2}} \:\left({x}\right)−\mathrm{2sin}\:\left({x}\right)\mathrm{sin}\:\left({y}\right)+\mathrm{sin}^{\mathrm{2}} \:\left({y}\right)=\mathrm{sin}^{\mathrm{2}} \:\left(\theta\right)\:\:\:…\left({i}\right) \\ $$$$ \\ $$$$\left(\mathrm{cos}\left(\mathrm{x}\right)\:+\:\mathrm{cos}\left(\mathrm{y}\right)\right)^{\mathrm{2}} \:=\:\mathrm{cos}^{\mathrm{2}} \left(\theta\right) \\ $$$$\mathrm{cos}^{\mathrm{2}} \:\left({x}\right)+\mathrm{2cos}\:\left({x}\right)\mathrm{cos}\:\left({y}\right)+\mathrm{cos}^{\mathrm{2}} \:\left({y}\right)=\mathrm{cos}^{\mathrm{2}} \:\left(\theta\right)\:\:\:…\left({ii}\right) \\ $$$$ \\ $$$$\left({i}\right)+\left({ii}\right): \\ $$$$\mathrm{sin}^{\mathrm{2}} \:\left({x}\right)+\mathrm{cos}^{\mathrm{2}} \:\left({x}\right)+\mathrm{2cos}\:\left({x}\right)\mathrm{cos}\:\left({y}\right)−\mathrm{2sin}\:\left({x}\right)\mathrm{sin}\:\left({y}\right)+\mathrm{sin}^{\mathrm{2}} \:\left({y}\right)+\mathrm{cos}^{\mathrm{2}} \:\left({y}\right)=\mathrm{sin}^{\mathrm{2}} \:\left(\theta\right)+\mathrm{cos}^{\mathrm{2}} \left(\theta\right) \\ $$$$\mathrm{1}+\mathrm{2}\left[\mathrm{cos}\:\left({x}\right)\mathrm{cos}\:\left({y}\right)−\mathrm{sin}\:\left({x}\right)\mathrm{sin}\:\left({y}\right)\right]+\mathrm{1}=\mathrm{1} \\ $$$$\mathrm{1}+\mathrm{2cos}\:\left({x}+{y}\right)+\mathrm{1}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{cos}\:\left({x}+{y}\right)=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by Saham last updated on 27/Feb/17
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$