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#)?

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#

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} #.