To determine the correct relation(s) between α \alpha α β \beta β
2 ( cos β − cos α ) + cos α cos β = 1 2(\cos \beta - \cos \alpha) + \cos \alpha \cos \beta = 1 2 ( cos β − cos α ) + cos α cos β = 1
Let x = tan ( α 2 ) x = \tan\left(\frac{\alpha}{2}\right) x = tan ( 2 α ) y = tan ( β 2 ) y = \tan\left(\frac{\beta}{2}\right) y = tan ( 2 β )
cos α = 1 − x 2 1 + x 2 and cos β = 1 − y 2 1 + y 2 \cos \alpha = \frac{1-x^2}{1+x^2} \quad \text{and} \quad \cos \beta = \frac{1-y^2}{1+y^2} cos α = 1 + x 2 1 − x 2 and cos β = 1 + y 2 1 − y 2
Substitute these expressions into the original equation:
2 ( 1 − y 2 1 + y 2 − 1 − x 2 1 + x 2 ) + ( 1 − x 2 1 + x 2 ) ( 1 − y 2 1 + y 2 ) = 1 2\left(\frac{1-y^2}{1+y^2} - \frac{1-x^2}{1+x^2}\right) + \left(\frac{1-x^2}{1+x^2}\right)\left(\frac{1-y^2}{1+y^2}\right) = 1 2 ( 1 + y 2 1 − y 2 − 1 + x 2 1 − x 2 ) + ( 1 + x 2 1 − x 2 ) ( 1 + y 2 1 − y 2 ) = 1
Multiplying the entire equation by ( 1 + x 2 ) ( 1 + y 2 ) (1+x^2)(1+y^2) ( 1 + x 2 ) ( 1 + y 2 )
2 [ ( 1 − y 2 ) ( 1 + x 2 ) − ( 1 − x 2 ) ( 1 + y 2 ) ] + ( 1 − x 2 ) ( 1 − y 2 ) = ( 1 + x 2 ) ( 1 + y 2 ) 2\left[(1-y^2)(1+x^2) - (1-x^2)(1+y^2)\right] + (1-x^2)(1-y^2) = (1+x^2)(1+y^2) 2 [ ( 1 − y 2 ) ( 1 + x 2 ) − ( 1 − x 2 ) ( 1 + y 2 ) ] + ( 1 − x 2 ) ( 1 − y 2 ) = ( 1 + x 2 ) ( 1 + y 2 )
Expanding each term individually:
For the first term in brackets:
( 1 − y 2 + x 2 − x 2 y 2 ) − ( 1 + y 2 − x 2 − x 2 y 2 ) = 2 x 2 − 2 y 2 (1 - y^2 + x^2 - x^2y^2) - (1 + y^2 - x^2 - x^2y^2) = 2x^2 - 2y^2 ( 1 − y 2 + x 2 − x 2 y 2 ) − ( 1 + y 2 − x 2 − x 2 y 2 ) = 2 x 2 − 2 y 2 4 x 2 − 4 y 2 4x^2 - 4y^2 4 x 2 − 4 y 2
Expanding the second term on the left-hand side:
( 1 − x 2 ) ( 1 − y 2 ) = 1 − x 2 − y 2 + x 2 y 2 (1-x^2)(1-y^2) = 1 - x^2 - y^2 + x^2y^2 ( 1 − x 2 ) ( 1 − y 2 ) = 1 − x 2 − y 2 + x 2 y 2
Expanding the right-hand side:
( 1 + x 2 ) ( 1 + y 2 ) = 1 + x 2 + y 2 + x 2 y 2 (1+x^2)(1+y^2) = 1 + x^2 + y^2 + x^2y^2 ( 1 + x 2 ) ( 1 + y 2 ) = 1 + x 2 + y 2 + x 2 y 2
Substituting these back into the simplified equation gives:
( 4 x 2 − 4 y 2 ) + ( 1 − x 2 − y 2 + x 2 y 2 ) = 1 + x 2 + y 2 + x 2 y 2 (4x^2 - 4y^2) + (1 - x^2 - y^2 + x^2y^2) = 1 + x^2 + y^2 + x^2y^2 ( 4 x 2 − 4 y 2 ) + ( 1 − x 2 − y 2 + x 2 y 2 ) = 1 + x 2 + y 2 + x 2 y 2
Subtracting 1 + x 2 y 2 1 + x^2y^2 1 + x 2 y 2
3 x 2 − 5 y 2 = x 2 + y 2 3x^2 - 5y^2 = x^2 + y^2 3 x 2 − 5 y 2 = x 2 + y 2
Rearranging the terms:
2 x 2 − 6 y 2 = 0 ⟹ x 2 − 3 y 2 = 0 2x^2 - 6y^2 = 0 \implies x^2 - 3y^2 = 0 2 x 2 − 6 y 2 = 0 ⟹ x 2 − 3 y 2 = 0
Factoring the quadratic relation:
( x + 3 y ) ( x − 3 y ) = 0 (x + \sqrt{3}y)(x - \sqrt{3}y) = 0 ( x + 3 y ) ( x − 3 y ) = 0
Substituting back x = tan ( α 2 ) x = \tan\left(\frac{\alpha}{2}\right) x = tan ( 2 α ) y = tan ( β 2 ) y = \tan\left(\frac{\beta}{2}\right) y = tan ( 2 β )
( tan ( α 2 ) + 3 tan ( β 2 ) ) ( tan ( α 2 ) − 3 tan ( β 2 ) ) = 0 \left(\tan\left(\frac{\alpha}{2}\right) + \sqrt{3}\tan\left(\frac{\beta}{2}\right)\right)\left(\tan\left(\frac{\alpha}{2}\right) - \sqrt{3}\tan\left(\frac{\beta}{2}\right)\right) = 0 ( tan ( 2 α ) + 3 tan ( 2 β ) ) ( tan ( 2 α ) − 3 tan ( 2 β ) ) = 0
This yields two valid relations:
tan ( α 2 ) + 3 tan ( β 2 ) = 0 \tan\left(\frac{\alpha}{2}\right) + \sqrt{3}\tan\left(\frac{\beta}{2}\right) = 0 tan ( 2 α ) + 3 tan ( 2 β ) = 0 tan ( α 2 ) − 3 tan ( β 2 ) = 0 \tan\left(\frac{\alpha}{2}\right) - \sqrt{3}\tan\left(\frac{\beta}{2}\right) = 0 tan ( 2 α ) − 3 tan ( 2 β ) = 0
(Note: As indicated in the official exam evaluation, full marks were awarded to all candidates for this question).