Question

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`)?

Answer 1

Given `x=acos(theta-alpha)` and `y=bcos(theta-beta)`

then

`x/a=cos(theta-alpha)` and `y/b=cos(theta-beta)`

Inserting in LHS we get

`LHS=(x/a)^2+(y/b)^2-(2xy)/(ab)cos(alpha-beta)`

`=cos^2(theta-alpha)+cos^2(theta-beta)-2cos(theta-alpha)cos(theta-beta)cos(alpha-beta)`

`=1/2(1+cos2(theta-alpha))+1/2(1+cos2(theta-beta))-(cos(2theta-alpha-beta)+cos(alpha-beta))cos(alpha-beta)`

`=1+1/2(cos2(theta-alpha)+cos2(theta-beta))-(cos(2theta-alpha-beta)+cos(alpha-beta))cos(alpha-beta)`

`=1+1/2(2cos(2theta-alpha-beta)cos(alpha-beta))-(cos(2theta-alpha-beta)+cos(alpha-beta))cos(alpha-beta)`

`=1+cancel(cos(2theta-alpha-beta)cos(alpha-beta))-cancel(cos(2theta-alpha-beta)cos(alpha-beta))-cos^2(alpha-beta)`

`=1-cos^2(alpha-beta)`

`=sin^2(alpha-beta) =RHS`

Answer 2

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(xy)/(ab)cos ( alpha - beta ) - sin^2 ( alpha - beta ) = 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

interested readers to know more about these equations.

`abs(x /a) = abs cos( theta - alpha ) <= 1.`

`abs (y / a) = abs cos( theta - beta ) <= 1.`

The second degree equation equation

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

`= x^2/a^2 + y^2/b^2 -2(xy)/(ab)cos ( alpha - beta ) - sin^2 ( alpha - beta ) = 0`

represents a family of ellipses, bracing the rectangle

`x = +- a and y = +- b`.

The Choice `a = 4 , b = 3, alpha = pi/2 and beta = pi/4` gives

`x^2/16 +y^2/9 -(xy)/(6sqrt2)= 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 `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

`x/a +-y/b = 0`

for `alpha - beta = kpi`

Another form of the equation, without `theta`, is :

`arccos (x/a) +- arcsin(y/b) = kpi +-(beta - alpha}`.