sin (x+y)=sin x+sin y


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Question Number 161698 by cortano last updated on 21/Dec/21

$\sin (x + y) = \sin x + \sin y$
Commented by Rasheed.Sindhi last updated on 21/Dec/21

$(x, 0), (0, y), (0, 0) a r e o b v i o u s s o l u t i o n s$
Commented by cortano last updated on 21/Dec/21

$a n d {\begin{cases} x = 2 n π \\ y = 2 n π \end{cases}$
Commented by Rasheed.Sindhi last updated on 21/Dec/21

$(x, y) = (0, 0) = (0 + 2 n π, 0 + 2 n π)$ $= (2 n π, 2 n π)$ $A c t u a l l y i n g e n e r a l$ $(x, y) = (2 m π, 2 n π) w h e r e m, n \in Z$
Commented by mr W last updated on 21/Dec/21

$f o l l o w i n g s o l u t i o n s i t h i n k$ ${\begin{cases} x = 2 k π \\ y = a n y v a l u e \end{cases}$ ${\begin{cases} x = a n y v a l u e \\ y = 2 k π \end{cases}$ ${x + y = 2 k π$
Commented by mr W last updated on 21/Dec/21

$y e s, t h a n k s!$ $i h a d a m i s t a k e i n m y w o r k i n g . n o w$ ${i t}^{'} s f i x e d .$ $y = - x i s i n c l u d e d i n c a s e 3 .$
Commented by Rasheed.Sindhi last updated on 21/Dec/21

$A l s o$ $y = - x$
	Answered by mr W last updated on 21/Dec/21
	$sin x cos y+cos x sin y=sin x+sin y sin x(1−cos y)=sin y (cos x−1) 2sin (x/2) cos (x/2)(2 sin^2 (y/2))=2 sin (y/2) cos (y/2) (−2sin^2 (x/2)) (cos (x/2) sin (y/2)+cos (y/2) sin (x/2))sin (x/2)sin (y/2)=0 ⇒sin ((x+y)/2) sin (x/2) sin (y/2)=0 case 1: sin (x/2)=0 { (((x/2)=kπ ⇒x=2kπ)),((y=any value)) :} case 2: sin (y/2)=0 { (((y/2)=kπ ⇒y=2kπ)),((x=any value)) :} case 3: sin ((x+y)/2)=0 ((x+y)/2)=kπ x+y=2kπ$
	$\sin x \cos y + \cos x \sin y = \sin x + \sin y$ $\sin x (1 - \cos y) = \sin y (\cos x - 1)$ $2 \sin \frac{x}{2} \cos \frac{x}{2} (2 \sin^{2} \frac{y}{2}) = 2 \sin \frac{y}{2} \cos \frac{y}{2} (- {2 \sin}^{2} \frac{x}{2})$ $(\cos \frac{x}{2} \sin \frac{y}{2} + \cos \frac{y}{2} \sin \frac{x}{2}) \sin \frac{x}{2} \sin \frac{y}{2} = 0$ $\Rightarrow \sin \frac{x + y}{2} \sin \frac{x}{2} \sin \frac{y}{2} = 0$ $c a s e 1 : \sin \frac{x}{2} = 0$ ${\begin{cases} \frac{x}{2} = k π \Rightarrow x = 2 k π \\ y = a n y v a l u e \end{cases}$ $c a s e 2 : \sin \frac{y}{2} = 0$ ${\begin{cases} \frac{y}{2} = k π \Rightarrow y = 2 k π \\ x = a n y v a l u e \end{cases}$ $c a s e 3 : \sin \frac{x + y}{2} = 0$ $\frac{x + y}{2} = k π$ $x + y = 2 k π$
	Answered by Rasheed.Sindhi last updated on 22/Dec/21

	$\sin (x + y) = \sin x + \sin y$ $\begin{matrix} \sin a + \sin b = 2 \sin \frac{a + b}{2} \cos \frac{a - b}{2} \end{matrix}$ $\sin 2 (\frac{x + y}{2}) = 2 \sin \frac{x + y}{2} \cos \frac{x - y}{2}$ $2 \sin \frac{x + y}{2} \cos \frac{x + y}{2} - 2 \sin \frac{x + y}{2} \cos \frac{x - y}{2} = 0$ $2 \sin \frac{x + y}{2} (\cos \frac{x + y}{2} - \cos \frac{x - y}{2}) = 0$ $\sin \frac{x + y}{2} = 0 ∣ \cos \frac{x + y}{2} - \cos \frac{x - y}{2} = 0$ $\frac{x + y}{2} = n π ∣ \cos \frac{x + y}{2} = \cos \frac{x - y}{2}$ $x + y = 2 n π ∣ \frac{x + y}{2} = \frac{x - y}{2}, \frac{y - x}{2}$ $\frac{x + y}{2} = \frac{x - y}{2} ∣ \frac{x + y}{2} = \frac{y - x}{2}$ $y = 0 for any x ∣ x = 0 for any y$ $y = 2 n π for any x ∣ x = 2 m π for any y$
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