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)?

2 Answers
Jul 20, 2018

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

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(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} .