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Question Number 161698 by cortano last updated on 21/Dec/21
sin (x+y)=sin x+sin y
$$\:\:\mathrm{sin}\:\left({x}+{y}\right)=\mathrm{sin}\:{x}+\mathrm{sin}\:{y} \\ $$
Commented by Rasheed.Sindhi last updated on 21/Dec/21
(x,0),(0,y),(0,0) are obvious solutions
$$\left({x},\mathrm{0}\right),\left(\mathrm{0},{y}\right),\left(\mathrm{0},\mathrm{0}\right)\:{are}\:{obvious}\:{solutions} \\ $$
Commented by cortano last updated on 21/Dec/21
and { ((x=2nπ)),((y=2nπ)) :}
$${and}\:\begin{cases}{{x}=\mathrm{2}{n}\pi}\\{{y}=\mathrm{2}{n}\pi}\end{cases} \\ $$
Commented by Rasheed.Sindhi last updated on 21/Dec/21
(x,y)=(0,0)=(0+2nπ,0+2nπ) =(2nπ,2nπ) Actually in general (x,y)=(2mπ,2nπ) where m,n∈Z
$$\left({x},{y}\right)=\left(\mathrm{0},\mathrm{0}\right)=\left(\mathrm{0}+\mathrm{2}{n}\pi,\mathrm{0}+\mathrm{2}{n}\pi\right) \\ $$$$=\left(\mathrm{2}{n}\pi,\mathrm{2}{n}\pi\right) \\ $$$${Actually}\:{in}\:{general} \\ $$$$\left({x},{y}\right)=\left(\mathrm{2}{m}\pi,\mathrm{2}{n}\pi\right)\:{where}\:{m},{n}\in\mathbb{Z} \\ $$
Commented by mr W last updated on 21/Dec/21
following solutions i think { ((x=2kπ)),((y=any value)) :} { ((x=any value)),((y=2kπ)) :} {x+y=2kπ
$${following}\:{solutions}\:{i}\:{think} \\ $$$$\begin{cases}{{x}=\mathrm{2}{k}\pi}\\{{y}={any}\:{value}}\end{cases} \\ $$$$\begin{cases}{{x}={any}\:{value}}\\{{y}=\mathrm{2}{k}\pi}\end{cases} \\ $$$$\left\{{x}+{y}=\mathrm{2}{k}\pi\right. \\ $$
Commented by mr W last updated on 21/Dec/21
yes, thanks! i had a mistake in my working.now it′s fixed. y=−x is included in case 3.
$${yes},\:{thanks}! \\ $$$${i}\:{had}\:{a}\:{mistake}\:{in}\:{my}\:{working}.{now} \\ $$$${it}'{s}\:{fixed}. \\ $$$${y}=−{x}\:{is}\:{included}\:{in}\:{case}\:\mathrm{3}. \\ $$
Commented by Rasheed.Sindhi last updated on 21/Dec/21
Also y=−x
$${Also} \\ $$$${y}=−{x} \\ $$
Answered by mr W last updated on 21/Dec/21
sin x cos y+cos x sin y=sin x+sin y sin x(1−cos y)=sin y (cos x−1) 2sin (x/2) cos (x/2)(2 sin^2 (y/2))=2 sin (y/2) cos (y/2) (−2sin^2 (x/2)) (cos (x/2) sin (y/2)+cos (y/2) sin (x/2))sin (x/2)sin (y/2)=0 ⇒sin ((x+y)/2) sin (x/2) sin (y/2)=0 case 1: sin (x/2)=0 { (((x/2)=kπ ⇒x=2kπ)),((y=any value)) :} case 2: sin (y/2)=0 { (((y/2)=kπ ⇒y=2kπ)),((x=any value)) :} case 3: sin ((x+y)/2)=0 ((x+y)/2)=kπ x+y=2kπ
$$\mathrm{sin}\:{x}\:\mathrm{cos}\:{y}+\mathrm{cos}\:{x}\:\mathrm{sin}\:{y}=\mathrm{sin}\:{x}+\mathrm{sin}\:{y} \\ $$$$\mathrm{sin}\:{x}\left(\mathrm{1}−\mathrm{cos}\:{y}\right)=\mathrm{sin}\:{y}\:\left(\mathrm{cos}\:{x}−\mathrm{1}\right) \\ $$$$\mathrm{2sin}\:\frac{{x}}{\mathrm{2}}\:\mathrm{cos}\:\frac{{x}}{\mathrm{2}}\left(\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\frac{{y}}{\mathrm{2}}\right)=\mathrm{2}\:\mathrm{sin}\:\frac{{y}}{\mathrm{2}}\:\mathrm{cos}\:\frac{{y}}{\mathrm{2}}\:\left(−\mathrm{2sin}^{\mathrm{2}} \:\frac{{x}}{\mathrm{2}}\right) \\ $$$$\left(\mathrm{cos}\:\frac{{x}}{\mathrm{2}}\:\mathrm{sin}\:\frac{{y}}{\mathrm{2}}+\mathrm{cos}\:\frac{{y}}{\mathrm{2}}\:\mathrm{sin}\:\frac{{x}}{\mathrm{2}}\right)\mathrm{sin}\:\frac{{x}}{\mathrm{2}}\mathrm{sin}\:\frac{{y}}{\mathrm{2}}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{sin}\:\frac{{x}+{y}}{\mathrm{2}}\:\mathrm{sin}\:\frac{{x}}{\mathrm{2}}\:\mathrm{sin}\:\frac{{y}}{\mathrm{2}}=\mathrm{0} \\ $$$${case}\:\mathrm{1}:\:\mathrm{sin}\:\frac{{x}}{\mathrm{2}}=\mathrm{0} \\ $$$$\begin{cases}{\frac{{x}}{\mathrm{2}}={k}\pi\:\Rightarrow{x}=\mathrm{2}{k}\pi}\\{{y}={any}\:{value}}\end{cases} \\ $$$${case}\:\mathrm{2}:\:\mathrm{sin}\:\frac{{y}}{\mathrm{2}}=\mathrm{0} \\ $$$$\begin{cases}{\frac{{y}}{\mathrm{2}}={k}\pi\:\Rightarrow{y}=\mathrm{2}{k}\pi}\\{{x}={any}\:{value}}\end{cases} \\ $$$${case}\:\mathrm{3}:\:\mathrm{sin}\:\frac{{x}+{y}}{\mathrm{2}}=\mathrm{0} \\ $$$$\frac{{x}+{y}}{\mathrm{2}}={k}\pi \\ $$$${x}+{y}=\mathrm{2}{k}\pi \\ $$
Answered by Rasheed.Sindhi last updated on 22/Dec/21
sin (x+y)=sin x+sin y determinant (((sin a+sin b=2sin((a+b)/2) cos((a−b)/2)))) sin2(((x+y)/2))=2sin((x+y)/2) cos((x−y)/2) 2sin((x+y)/2)cos((x+y)/2)−2sin((x+y)/2) cos((x−y)/2)=0 2sin((x+y)/2)(cos((x+y)/2)−cos((x−y)/2))=0 sin((x+y)/2)=0∣cos((x+y)/2)−cos((x−y)/2)=0 ((x+y)/2)=nπ ∣ cos((x+y)/2)=cos((x−y)/2) x+y=2nπ ∣((x+y)/2)=((x−y)/2) ,((y−x)/2) ((x+y)/2)=((x−y)/2) ∣((x+y)/2)= ((y−x)/2) y=0 for any x ∣ x=0 for any y y=2nπ for any x ∣ x=2mπ for any y
$$\:\:\mathrm{sin}\:\left({x}+{y}\right)=\mathrm{sin}\:{x}+\mathrm{sin}\:{y} \\ $$$$\begin{array}{|c|}{\mathrm{sin}\:{a}+\mathrm{sin}\:{b}=\mathrm{2sin}\frac{{a}+{b}}{\mathrm{2}}\:\mathrm{cos}\frac{{a}−{b}}{\mathrm{2}}}\\\hline\end{array} \\ $$$$\:\:\mathrm{sin2}\left(\frac{{x}+{y}}{\mathrm{2}}\right)=\mathrm{2sin}\frac{{x}+{y}}{\mathrm{2}}\:\mathrm{cos}\frac{{x}−{y}}{\mathrm{2}}\:\: \\ $$$$\:\mathrm{2sin}\frac{{x}+{y}}{\mathrm{2}}\mathrm{cos}\frac{{x}+{y}}{\mathrm{2}}−\mathrm{2sin}\frac{{x}+{y}}{\mathrm{2}}\:\mathrm{cos}\frac{{x}−{y}}{\mathrm{2}}=\mathrm{0} \\ $$$$\mathrm{2sin}\frac{{x}+{y}}{\mathrm{2}}\left(\mathrm{cos}\frac{{x}+{y}}{\mathrm{2}}−\mathrm{cos}\frac{{x}−{y}}{\mathrm{2}}\right)=\mathrm{0} \\ $$$$\mathrm{sin}\frac{{x}+{y}}{\mathrm{2}}=\mathrm{0}\mid\mathrm{cos}\frac{{x}+{y}}{\mathrm{2}}−\mathrm{cos}\frac{{x}−{y}}{\mathrm{2}}=\mathrm{0} \\ $$$$\frac{{x}+{y}}{\mathrm{2}}={n}\pi\:\:\mid\:\mathrm{cos}\frac{{x}+{y}}{\mathrm{2}}=\mathrm{cos}\frac{{x}−{y}}{\mathrm{2}} \\ $$$${x}+{y}=\mathrm{2}{n}\pi\:\mid\frac{{x}+{y}}{\mathrm{2}}=\frac{{x}−{y}}{\mathrm{2}}\:,\frac{{y}−{x}}{\mathrm{2}} \\ $$$$\frac{{x}+{y}}{\mathrm{2}}=\frac{{x}−{y}}{\mathrm{2}}\:\mid\frac{{x}+{y}}{\mathrm{2}}=\:\frac{{y}−{x}}{\mathrm{2}} \\ $$$${y}=\mathrm{0}\:\mathrm{for}\:\mathrm{any}\:\mathrm{x}\:\:\mid\:\:{x}=\mathrm{0}\:\mathrm{for}\:\mathrm{any}\:\mathrm{y} \\ $$$${y}=\mathrm{2}{n}\pi\:\mathrm{for}\:\mathrm{any}\:\mathrm{x}\:\:\mid\:\:{x}=\mathrm{2}{m}\pi\:\mathrm{for}\:\mathrm{any}\:\mathrm{y} \\ $$$$ \\ $$

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