Find $\tan(x+y)$ given $\sin x=3\sin(x+2y)$

Question

If $\sin x=3\sin(x+2y)$, then find $\tan(x+y)$

My reference gives the solution $-2\tan y$, but how do I prove it ?

My Attempt $$ 3=\frac{\sin(x+2y)}{\sin x}\implies3\sin x=\sin x.\cos2y+\cos x.\sin2y\\ 3\sin x=2\sin x.\cos^2y-\sin x+2\cos x.\sin y.\cos y=2\cos y.\sin(x+y)-\sin x\\ \implies 4\sin x=2\cos y.\sin(x+y)\implies\sin(x+y)=2.\frac{\sin x}{\cos y}\\ \cos(x+y)=\frac{\sin x}{\sin y}\implies\boxed{\tan(x+y)=2\tan y} $$

It seems like the solution is +ve, how can it be a negative solution as given in my reference ?

Answer 1

$$ \frac{\sin(x+2y)}{\sin x}=\frac{1}{3}\\ \frac{\sin(x+2y)+\sin x}{\sin(x+2y)-\sin x}=\frac{1+3}{1-3}=-2\\ \frac{2\sin(x+y)\cos y}{2\cos(x+y)\sin y}=-2\implies\frac{\tan(x+y)}{\tan y}=-2\\ \implies\tan(x+y)=-2\tan y $$

Answer 2

Hint:

As we need to eliminate $x$

let $x+y=k,x=k-y$

$$\sin(y-k)=3\sin(k-y+2y)$$

Method$\#1:$

Expand $\sin(y-k),\sin(y+k)$ and group $\sin k,\cos k$

Method$\#2:$

$$\dfrac{\sin(y-k)}{\sin(y+k)}=\dfrac31$$

Apply Componendo dividendo

Answer 3

Rewrite the given equality as $$\sin\left(\left(x+y\right)-y\right) = 3\sin\left(\left(x+y\right)+y\right).$$ Now use the formula for $\sin\left(A+B\right)$ on both sides and simplify.

Answer 4

Let indicate $x+y=z$, then

$$\sin x =3\sin(x+2y) \iff \sin (z-y)=3\sin (z+y)$$

$$\sin z\cos y-\cos z\sin y=3\sin z \cos y + 3 \cos z \sin y$$

$$-2\sin z\cos y= 4 \cos z \sin y \implies \tan z=\tan (x+y)=-2\tan y$$