# Trigonometric Relation

I was wanting a simpler solution to the problem below (I'll leave my solution below). Who has a simpler solution, could you please introduce it here?

Let \begin{equation*} \sin(u)+\sin(v)+\sin(w)=-\sin(u+v+w) \end{equation*} Then the equality below is true: \begin{equation*} \sin(v+w)+\sin(u+v)+\sin(u+w)=-\sin(v+w)\sin(u+v)\sin(u+w) \end{equation*}

$$Proof$$ Observe that if $\displaystyle x+y+z=-xyz$, so we have:

$$\frac{(1+y^{2})(1+z^{2})}{4yz}+\frac{(1+x^{2})(1+z^{2})}{4xz}+\frac{(1+x^{2})(1+y^{2})}{4xy}=-1$$ Let's prove the above equality: $$\frac{1+y^{2}+z^{2}+y^{2}z^{2}}{4yz}+\frac{1+x^{2}+z^{2}+x^{2}z^{2}}{4xz}+\frac{1+x^{2}+y^{2}+x^{2}y^{2}}{4xy}$$ Observe that $\ \displaystyle x+y+z=-xyz \Rightarrow \frac{1}{xy}=-\frac{z}{x+y+z},\frac{1}{yz}=-\frac{x}{x+y+z},\frac{1}{xz}=-\frac{y}{x+y+z} \\ \\$

We get: $$-\frac{(1+y^{2}+z^{2}+y^{2}z^{2})x}{4(x+y+z)}-\frac{(1+x^{2}+z^{2}+x^{2}z^{2})y}{4(x+y+z)}-\frac{(1+x^{2}+y^{2}+x^{2}y^{2})z}{4(x+y+z)}\Rightarrow$$

$$\frac{-x-xy^{2}-xz^{2}-xy^{2}z^{2}}{4(x+y+z)}+\frac{-y-yx^{2}-yz^{2}-yx^{2}z^{2}}{4(x+y+z)}+\frac{-z-zx^{2}-zy^{2}-zx^{2}y^{2}}{4(x+y+z)}\Rightarrow$$

$$\frac{-x-xy^{2}-xz^{2}-xyz(yz)}{4(x+y+z)}+\frac{-y-yx^{2}-yz^{2}-xyz(xz)}{4(x+y+z)}+\frac{-z-zx^{2}-zy^{2}-xyz(xy)}{4(x+y+z)}\Rightarrow$$

$$\frac{-x-xy^{2}-xz^{2}+(x+y+z)yz}{4(x+y+z)}+\frac{-y-yx^{2}-yz^{2}+(x+y+z)xz}{4(x+y+z)}+\frac{-z-zx^{2}-zy^{2}+(x+y+z)xy}{4(x+y+z)}\Rightarrow$$

$$\frac{-(x+y+z)+3xyz}{4(x+y+z)}=\frac{-(x+y+z)-3(x+y+z)}{4(x+y+z)}=-1$$

So, if $\displaystyle x+y+z=-xyz$, is true:

$\\ \\ \displaystyle \frac{x}{1+x^{2}}+\frac{y}{1+y^{2}}+\frac{z}{1+z^{2}}=-\frac{4xyz}{(1+x^{2})(1+y^{2})(1+z^{2})}\\$

Replace $\displaystyle x=\tan(\alpha),y=\tan(\beta), z=\tan(\gamma)$, our relationship gets rewritten as:

$\displaystyle \tan(\alpha)+\tan(\beta)+\tan(\gamma)=-\tan(\alpha)\tan(\beta)\tan(\gamma)$,

on the other hand we have:

$\\ \\ \displaystyle \sin(\alpha)\cos(\alpha)+\sin(\beta)\cos(\beta)+\sin(\gamma)\cos(\gamma)=-4\sin(\alpha)\cos(\alpha)\sin(\beta)\cos(\beta)\sin(\gamma)\cos(\gamma)\\$ Multiplying the two sides by 2 and using the sine of the double arch finally, we will have:

$\\ \\ \displaystyle \sin(2\alpha)+\sin(2\beta)+\sin(2\gamma)=-\sin(2\alpha)\sin(2\beta)\sin(2\gamma)\\ \\$

Substituting for tangents:

$\\ \\ \displaystyle \tan(\alpha)+\tan(\beta)+\tan(\gamma)=-\tan(\alpha)\tan(\beta)\tan(\gamma) \\ \Rightarrow \tan(\alpha)+\tan(\beta)+\tan(\gamma)+\tan(\alpha)\tan(\beta)\tan(\gamma)=0\Rightarrow \\ \sec(\alpha)\sec(\beta)\sec(\gamma)(\sin(\alpha)\cos(\beta)\cos(\gamma)+\sin(\beta)\cos(\alpha)\cos(\gamma)+\sin(\gamma)\cos(\alpha)\cos(\beta)+\sin(\alpha)\sin(\beta)\sin(\gamma))=0 \\ \Rightarrow \sec(\alpha)\sec(\beta)\sec(\gamma)(2\sin(\alpha)\cos(\beta)\cos(\gamma)+2\sin(\beta)\cos(\alpha)\cos(\gamma)+2\sin(\gamma)\cos(\alpha)\cos(\beta)+2\sin(\alpha)\sin(\beta)\sin(\gamma))=0 \\ \Rightarrow \sec(\alpha)\sec(\beta)\sec(\gamma) \times \\ \{ \{[\cos(\alpha) \sin(\beta) - \sin(\alpha) \cos(\beta)]\cos(\gamma)+[\sin(\alpha) \sin(\beta) + \cos(\alpha) \cos(\beta) ]\sin(\gamma)\}+\{[\sin(\alpha) \cos(\beta) - \cos(\alpha) \sin(\beta) ]\cos(\gamma)+[\sin(\alpha) \sin(\beta) + \cos(\alpha) \cos(\beta) ]\sin(\gamma)\}+\{[\sin(\alpha) \cos(\beta) + \cos(\alpha) \sin(\beta)]\cos(\gamma)-[\cos(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta) ]\sin(\gamma)\}+\{[\sin(\alpha) \cos(\beta) + \cos(\alpha) \sin(\beta)]\cos(\gamma)+[\cos(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta) ]\sin(\gamma)\}\}=0 \Rightarrow \sec(\alpha)\sec(\beta)\sec(\gamma) \{[\sin(-\alpha+\beta)\cos(\gamma)+\cos(-\alpha+\beta)\sin(\gamma)]+[\sin(\alpha-\beta)\cos(\gamma)+\cos(\alpha-\beta)\sin(\gamma)]+[\sin(\alpha+\beta)\cos(\gamma)-\cos(\alpha+\beta)\sin(\gamma)]+[\sin(\alpha+\beta)\cos(\gamma)+\cos(\alpha+\beta)\sin(\gamma)]\}= 0 \\ \Rightarrow \sec(\alpha)\sec(\beta)\sec(\gamma)\{\sin(-\alpha+\beta+\gamma)+\sin(\alpha-\beta+\gamma)+\sin(\alpha+\beta-\gamma)+\sin(\alpha+\beta+\gamma)\}=0\\ \\$

The only form of this equality equal to zero is if the equality below occurs: $$\sin(\alpha+\beta-\gamma)+\sin(\alpha-\beta+\gamma)+\sin(-\alpha+\beta+\gamma)+\sin(\alpha+\beta+\gamma)=0$$

And this implies that

$$\sin(\alpha+\beta-\gamma)+\sin(\alpha-\beta+\gamma)+\sin(-\alpha+\beta+\gamma)=-\sin(\alpha+\beta+\gamma)$$ In other words what we have proved is that if the equality below is true:

$$\sin(\alpha+\beta-\gamma)+\sin(\alpha-\beta+\gamma)+\sin(-\alpha+\beta+\gamma)=-\sin(\alpha+\beta+\gamma)$$ So is true: $$\sin(2\alpha)+\sin(2\beta)+\sin(2\gamma)=-\sin(2\alpha)\sin(2\beta)\sin(2\gamma)$$ Replacing $\displaystyle \alpha=\frac{v+w}{2},\beta=\frac{u+w}{2},\gamma=\frac{u+v}{2}$, therefore:

$$\sin(u)+\sin(v)+\sin(w)=-\sin(u+v+w)$$

And: $$\sin(v+w)+\sin(u+v)+\sin(u+w)=-\sin(v+w)\sin(u+v)\sin(u+w)$$

As desired.

Edit:

Replacing $\displaystyle u=\frac{\pi}{2}-x,v=\frac{\pi}{2}-y,w=\frac{\pi}{2}-z$, we get:

Let $$\cos(x)+\cos(y)+\cos(z)=\cos(x+y+z)$$ Then the equality below is true: $$\sin(x+y)+\sin(x+z)+\sin(y+z)=-\sin(x+y)\sin(x+z)\sin(y+z)$$

• It wasn't immediately clear what the question was with your presentation, so I modified it slightly to make this more obvious. If it wasn't helpful, let me know. – Semiclassical Dec 14 '16 at 19:13
• @Semiclassical thanks! – Simply Beautiful Art Dec 14 '16 at 19:15
• @Semiclassical thanks! – Israel Meireles Chrisostomo Dec 14 '16 at 23:43