# Dependent equations, with Trigonometry.

algebra-precalculustrigonometry

My questions is: do the first two equations imply the third ?

$$\left(x+\frac{1}{x}\right)\sin\alpha=\frac{y}{z}+\frac{z}{y}+\cos^2\alpha$$
$$\left(y+\frac{1}{y}\right)\sin\alpha=\frac{z}{x}+\frac{x}{z}+\cos^2\alpha$$
$$\left(z+\frac{1}{z}\right)\sin\alpha= \frac{x}{y}+\frac{y}{x}+\cos^2\alpha$$

This is an old trigonometry problem stated as follows:

Prove that the equations are not independent, and that
they are equivalent to

$$x+y+z=\frac{1}{x}+ \frac{1}{y}+\frac{1}{z}=-\sin\alpha$$

The problem can be solved by subtracing the first two to get

$$\sin\alpha= \frac{z^2-xy}{z(xy-1)}$$

And by subtracting the last two,

$$\sin\alpha= \frac{x^2-zy}{x(zy-1)}$$
Note that we are making use of all three equations.

Setting the expressions equal we derive

$$x+y+z=\frac{1}{x}+ \frac{1}{y}+\frac{1}{z}$$

To get the last part

Let

$$S=x+y+z$$ and $$R=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$$
our condition is $$R=S$$
then if we add all three equations,

$$(R+S)\sin\alpha=RS-3+3\cos^2\alpha$$

and setting $$R=S$$, this factors as

$$(\sin\alpha+R)(3\sin\alpha-R)$$
For some reason we reject one solution.

Substitution will verify the equations.

Note that this solution makes use of all three equations.
They are dependent in the sense that they imply a relation on the coefficients.

Can the problem be solved using only two of the equations ?

Theoretically one can take the original expression above for $$\sin\alpha$$ and substitute in the first, this should give a relation. But I cannot make the algebra work, it is horrible.

Also how does some one come up with such a question ? Do they reverse engineer the answer ? I dont have any insight into the process of making such a problem.

do the first two equations imply the third ?

No, take $$x=\frac 12,y=4,z=2$$ and $$\alpha=\frac{\pi}{2}$$ for which the first two equations hold and the third doesn't.

Subtracting $$\left(y+\frac{1}{y}\right)\sin\alpha=\frac{z}{x}+\frac{x}{z}+1-\sin^2\alpha\tag1$$ from $$\left(x+\frac{1}{x}\right)\sin\alpha=\frac{y}{z}+\frac{z}{y}+1-\sin^2\alpha\tag2$$ gives $$\frac{(x-y)(xy-1)}{xy}\sin\alpha=\frac{(x-y)(z^2-xy)}{xyz}$$

Assuming that $$x\not=y$$, one gets $$\sin\alpha=\frac{z^2-xy}{z(xy-1)}\tag3$$ under the condition that $$xy\not=1\tag4$$

Plugging $$(3)$$ into $$(1)$$ gives $$\frac{(x y - z) (x y + z) (x^2 y z + x y^2 z + x y z^2 - x y-yz - zx)}{x y z^2 (x y - 1)^2}=0\tag5$$

Also, $$\left(z+\frac{1}{z}\right)\sin\alpha-\bigg(\frac{x}{y}+\frac{y}{x}+\cos^2\alpha\bigg)$$ $$=\frac{(x^2yz +xy^2z -x y z^2 +xy -yz- zx ) (x^2 y z + x y^2 z + x y z^2 - x y-yz - zx)}{x y z^2 (x y - 1)^2}\tag6$$

So, if one can get $$(x,y,z,\alpha)$$ such that $$\begin{cases}x\not=y \\\\xy\not=1 \\\\xy=\pm z \\\\x^2 y z + x y^2 z + x y z^2 - x y-yz - zx\not=0 \\\\x^2yz +xy^2z -x y z^2 +xy -yz- zx\not=0 \\\\\sin\alpha=\dfrac{z^2-xy}{z(xy-1)} \end{cases}$$ then one can say that the answer is no to your question.

Here, I take $$\alpha=\dfrac{\pi}{2}$$ to have $$z^2-xy=zxy-z\iff (z-xy)(z+1)=0$$ Also, taking $$z+1\not=0$$, it is enough to find $$(x,y,z)$$ such that $$\begin{cases}xy=z \\\\x\not=y \\\\xy\not=1 \\\\z(xz + zy + z^2 - 1-y -x)\not=0 \\\\z(xz +zy -z^2 +1 -y- x)\not=0 \\\\z\not=-1 \end{cases}$$

Taking $$z=2$$, one gets $$\begin{cases}xy=2 \\\\x\not=y \\\\x + y + 3\not=0 \\\\x +y -3\not=0 \end{cases}$$

So, $$(x,y)=(\frac 12,4)$$ works.

Added : I realized that one can write $$(5)$$ as

$$\frac{(x y - z) (x y + z) }{z(x y - 1)^2}\bigg(x+y+z-\frac 1x-\frac 1y-\frac 1z\bigg)=0$$

and $$(6)$$ as $$\left(z+\frac{1}{z}\right)\sin\alpha-\bigg(\frac{x}{y}+\frac{y}{x}+\cos^2\alpha\bigg)$$ $$=\frac{xy}{(x y - 1)^2}\bigg(x +y -z -\frac 1x- \frac 1y+\frac 1z\bigg)\bigg(x+y+z-\frac 1x-\frac 1y-\frac 1z\bigg)$$