1. ## trigonometric equality

$\displaystyle \begin{array}{l} \frac{{\sin x + \sin y + \sin z}}{{\sin (x + y + z)}} = \frac{{\cos x + \cos y + \cos z}}{{\cos (x + y + z)}} = 2 \\ \\ find \\ \sin x\sin y + \sin y\sin z + \sin x\sin z \\ \end{array}$

2. This is a very interesting problem. It has either too much symmetry, or not enough, depending on how you're looking at it.

One thing I would highly recommend: abbreviate! I'm going to abbreviate as follows:

$\displaystyle \sin(x)=s(x)$
$\displaystyle \cos(x)=c(x).$

If you square both equations and add together in the appropriate way, you can get it down to

$\displaystyle \displaystyle{s(x)s(y)+s(x)s(z)+s(y)s(z)+c(x)c(y)+ c(x)c(z)+c(y)c(z)=\frac{1}{2}}.$

Unfortunately, if you plug in a test case, you find out that the sin terms do not necessarily make up half of the sum. Here's one test case:

$\displaystyle x=0.1$

$\displaystyle y=1.011189$

$\displaystyle z=-1.07805,$

approximately. All of these are in radians. The sin terms add up to $\displaystyle -0.75$, and the cosine terms add up to $\displaystyle 1.25.$

So there's a specific case, plus some general information about the problem. Nothing more is coming to mind at the moment. Your result should be -3/4. But now you have to prove it!

3. Hello, fxs12!

Askbeet has an excellent game plan,
. . but I can't finish the problem.

$\displaystyle \dfrac{{\sin x + \sin y + \sin z}}{{\sin (x + y + z)}} \;=\; \dfrac{{\cos x + \cos y + \cos z}}{{\cos (x + y + z)}} \;=\; 2$

$\displaystyle \text{Find: }\,\sin x\sin y + \sin y\sin z + \sin x\sin z$

We have: .$\displaystyle \begin{array}{cccc} \sin x + \sin y + \sin z &=& 2\sin(x+y+z) & [1] \\ \cos x + \cos y + \cos z &=& 2\cos(x+y+z) & [2] \end{array}$

$\displaystyle \begin{array}{ccccc}\text{Square [1]:} & (\sin x + \sin y + \sin z)^2 &=& 4\sin^2(x+y+z) & [3] \\ \text{Square [2]:} & (\cos x + \cos y + \cos z)^2 &=& 4\cos^2(x+y+z) & [4]\end{array}$

$\displaystyle \underbrace{\begin{Bmatrix}\sin^2x \\ \cos^2x\end{Bmatrix}}_{\text{This is 1}} + \underbrace{\begin{Bmatrix}\sin^2y \\ \cos^2y\end{Bmatrix}}_{\text{This is 1}} + \underbrace{\begin{Bmatrix}\sin^2z \\ \cos^2z\end{Bmatrix}}_{\text{This is 1}} + 2\begin{Bmatrix}\sin x\sin y + \sin y\sin z + \sin x\sin z \\ \cos x\cos y + \cos y\cos z + \cos x\cos z \end{Bmatrix}$

. . . . . . . $\displaystyle =\; 4\underbrace{\bigg[\sin^2(x+y+z) + \cos^2(x+y+z)\bigg]}_{\text{This is 1}}$

We have:

. . $\displaystyle 1 + 1 + 1 + 2\begin{Bmatrix}\sin x\sin y +\sin y\sin z + \sin x\sin z \\ \cos x\cos y + \cos y\cos z + \cos x\cos z\end{Bmatrix} \;=\;4$

. . $\displaystyle \begin{Bmatrix}\sin x\sin y + \sin y\sin z + \sin x\sin z \\ \cos x\cos y + \cos y\cos z + \cos x\cos z\end{Bmatrix} \;=\;\frac{1}{2}$

And I can't go any further . . .

4. ...

cos(x-y)+cos(y-z)+cos(x-z)=1/2

5. I also have proved:

sin(x-y)/sinx siny + sin(y-z)/siny sinz + sin(x-z)/sinx sinz = 0

$\displaystyle sinx=a$
$\displaystyle siny=b$
$\displaystyle sinz=c$

$\displaystyle cosx=\sqrt{1-a^2}$
$\displaystyle cosy=\sqrt{1-b^2}$
$\displaystyle cosz=\sqrt{1-c^2}$

$\displaystyle tan(x+y+z)=\frac{tanx+tany+tanz-tanxtanytanz}{1-tanxtany-tanytanz-tanztanx}$

$\displaystyle tanx=\frac{a}{\sqrt{1-a^2}}$

$\displaystyle tany=\frac{b}{\sqrt{1-b^2}}$

$\displaystyle tanx=\frac{c}{\sqrt{1-c^2}}$

Hence:

$\displaystyle \dfrac{{\sin x + \sin y + \sin z}}{{\sin (x + y + z)}} \;=\; \dfrac{{\cos x + \cos y + \cos z}}{{\cos (x + y + z)}} \;=\; 2$

becomes:

$\displaystyle \frac{a+b+c}{\sqrt{1-a^2}+\sqrt{1-b^2}+\sqrt{1-c^2}}=tan(x+y+z)=2$

when we need to find:$\displaystyle ab+bc+ac$