# If theta is eliminated from the equation x=acos(theta-alpha) and y=bcos(theta-beta) then prove that (x/a)^2+(y/b)^2-(2xy)/(ab)cos(alpha-beta)=sin^2(alpha-beta)?

Jul 20, 2018

Given $x = a \cos \left(\theta - \alpha\right)$ and $y = b \cos \left(\theta - \beta\right)$

then

$\frac{x}{a} = \cos \left(\theta - \alpha\right)$ and $\frac{y}{b} = \cos \left(\theta - \beta\right)$

Inserting in LHS we get

$L H S = {\left(\frac{x}{a}\right)}^{2} + {\left(\frac{y}{b}\right)}^{2} - \frac{2 x y}{a b} \cos \left(\alpha - \beta\right)$

$= {\cos}^{2} \left(\theta - \alpha\right) + {\cos}^{2} \left(\theta - \beta\right) - 2 \cos \left(\theta - \alpha\right) \cos \left(\theta - \beta\right) \cos \left(\alpha - \beta\right)$

$= \frac{1}{2} \left(1 + \cos 2 \left(\theta - \alpha\right)\right) + \frac{1}{2} \left(1 + \cos 2 \left(\theta - \beta\right)\right) - \left(\cos \left(2 \theta - \alpha - \beta\right) + \cos \left(\alpha - \beta\right)\right) \cos \left(\alpha - \beta\right)$

$= 1 + \frac{1}{2} \left(\cos 2 \left(\theta - \alpha\right) + \cos 2 \left(\theta - \beta\right)\right) - \left(\cos \left(2 \theta - \alpha - \beta\right) + \cos \left(\alpha - \beta\right)\right) \cos \left(\alpha - \beta\right)$

$= 1 + \frac{1}{2} \left(2 \cos \left(2 \theta - \alpha - \beta\right) \cos \left(\alpha - \beta\right)\right) - \left(\cos \left(2 \theta - \alpha - \beta\right) + \cos \left(\alpha - \beta\right)\right) \cos \left(\alpha - \beta\right)$

$= 1 + \cancel{\cos \left(2 \theta - \alpha - \beta\right) \cos \left(\alpha - \beta\right)} - \cancel{\cos \left(2 \theta - \alpha - \beta\right) \cos \left(\alpha - \beta\right)} - {\cos}^{2} \left(\alpha - \beta\right)$

$= 1 - {\cos}^{2} \left(\alpha - \beta\right)$

$= {\sin}^{2} \left(\alpha - \beta\right) = R H S$

Jul 20, 2018

I add here what I see in this family of ellipses given by the .second degree equation f(x, y; a, b, alpha, beta)
$= {x}^{2} / {a}^{2} + {y}^{2} / {b}^{2} - 2 \frac{x y}{a b} \cos \left(\alpha - \beta\right) - {\sin}^{2} \left(\alpha - \beta\right) = 0$

#### Explanation:

I add here what I see in this 4-parameter family of ellipses.

The elimination of $\theta$ is already well done. Now, it is for the

$\left\mid \frac{x}{a} \right\mid = \left\mid \cos \right\mid \left(\theta - \alpha\right) \le 1.$

$\left\mid \frac{y}{a} \right\mid = \left\mid \cos \right\mid \left(\theta - \beta\right) \le 1.$

The second degree equation equation

f(x, y; a, b, alpha, beta)

$= {x}^{2} / {a}^{2} + {y}^{2} / {b}^{2} - 2 \frac{x y}{a b} \cos \left(\alpha - \beta\right) - {\sin}^{2} \left(\alpha - \beta\right) = 0$

represents a family of ellipses, bracing the rectangle

$x = \pm a \mathmr{and} y = \pm b$.

The Choice $a = 4 , b = 3 , \alpha = \frac{\pi}{2} \mathmr{and} \beta = \frac{\pi}{4}$ gives

${x}^{2} / 16 + {y}^{2} / 9 - \frac{x y}{6 \sqrt{2}} = \frac{1}{2}$ that represents an ellipse., graph{(x^2/16 +y^2/9 -(xy)/(6sqrt2)- 1/2)(x^2-16)(y^2-9)=0}

If $\alpha - \beta$ = an odd multiple of $\frac{\pi}{2}$, the

equation

becomes ${x}^{2} / {a}^{2} + {y}^{2} / {b}^{2} = 1$

Degenerate cases:

It represents the diagonals of the enveloping rectangle, with ends

as vertices of the rectangle.a straight line

$\frac{x}{a} \pm \frac{y}{b} = 0$

for $\alpha - \beta = k \pi$

Another form of the equation, without $\theta$, is :

$\arccos \left(\frac{x}{a}\right) \pm \arcsin \left(\frac{y}{b}\right) = k \pi \pm \left(\beta - \alpha\right\}$.
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