Answer :
To prove that [tex]\(\tan \frac{\alpha - \beta}{2} = \pm \sqrt{\frac{4 - x^2}{x^2 + y^2}}\)[/tex], given [tex]\(\sin \alpha + \sin \beta = x\)[/tex] and [tex]\(\cos \alpha + \cos \beta = y\)[/tex], follow these steps:
1. Use Trigonometric Identities:
We start with the given trigonometric sum identities:
[tex]\[ \sin \alpha + \sin \beta = 2 \sin \left( \frac{\alpha + \beta}{2} \right) \cos \left( \frac{\alpha - \beta}{2} \right) \][/tex]
[tex]\[ \cos \alpha + \cos \beta = 2 \cos \left( \frac{\alpha + \beta}{2} \right) \cos \left( \frac{\alpha - \beta}{2} \right) \][/tex]
2. Assign These to the Given Values:
According to the problem:
[tex]\[ \sin \alpha + \sin \beta = x \][/tex]
[tex]\[ \cos \alpha + \cos \beta = y \][/tex]
3. Relate to Sum and Product Formulas:
Substitute the identities into the given equations:
[tex]\[ 2 \sin \left( \frac{\alpha + \beta}{2} \right) \cos \left( \frac{\alpha - \beta}{2} \right) = x \][/tex]
[tex]\[ 2 \cos \left( \frac{\alpha + \beta}{2} \right) \cos \left( \frac{\alpha - \beta}{2} \right) = y \][/tex]
4. Simplify the Equations:
Divide both sides of each equation by 2:
[tex]\[ \sin \left( \frac{\alpha + \beta}{2} \right) \cos \left( \frac{\alpha - \beta}{2} \right) = \frac{x}{2} \][/tex]
[tex]\[ \cos \left( \frac{\alpha + \beta}{2} \right) \cos \left( \frac{\alpha - \beta}{2} \right) = \frac{y}{2} \][/tex]
5. Introduce New Terms for Simplicity:
Let:
[tex]\[ A = \cos \left( \frac{\alpha - \beta}{2} \right) \][/tex]
[tex]\[ B = \sin \left( \frac{\alpha + \beta}{2} \right) \][/tex]
[tex]\[ C = \cos \left( \frac{\alpha + \beta}{2} \right) \][/tex]
Therefore, the equations become:
[tex]\[ B A = \frac{x}{2} \][/tex]
[tex]\[ C A = \frac{y}{2} \][/tex]
6. Divide the Equations:
Divide the first equation by the second:
[tex]\[ \frac{B A}{C A} = \frac{\frac{x}{2}}{\frac{y}{2}} \][/tex]
This simplifies to:
[tex]\[ \frac{B}{C} = \frac{x}{y} \][/tex]
Thus:
[tex]\[ \tan \left( \frac{\alpha + \beta}{2} \right) = \frac{x}{y} \][/tex]
7. Use the Pythagorean Identity:
We know:
[tex]\[ B^2 + C^2 = \left( \sin \left( \frac{\alpha + \beta}{2} \right) \right)^2 + \left( \cos \left( \frac{\alpha + \beta}{2} \right) \right)^2 = 1 \][/tex]
Therefore:
[tex]\[ \left( \frac{x}{2A} \right)^2 + \left( \frac{y}{2A} \right)^2 = 1 \][/tex]
8. Simplify This Expression:
[tex]\[ \frac{x^2}{4A^2} + \frac{y^2}{4A^2} = 1 \][/tex]
[tex]\[ \frac{x^2 + y^2}{4A^2} = 1 \][/tex]
Solving for [tex]\(A\)[/tex]:
[tex]\[ 4A^2 = x^2 + y^2 \][/tex]
[tex]\[ A^2 = \frac{x^2 + y^2}{4} \][/tex]
Therefore:
[tex]\[ A = \frac{\sqrt{x^2 + y^2}}{2} \][/tex]
9. Use the [tex]\(\tan \frac{\alpha - \beta}{2}\)[/tex] Identity:
We know:
[tex]\[ \sin^2 \left( \frac{\alpha - \beta}{2} \right) + \cos^2 \left( \frac{\alpha - \beta}{2} \right) = 1 \][/tex]
So:
[tex]\[ \sin^2 \left( \frac{\alpha - \beta}{2} \right) = 1 - \cos^2 \left( \frac{\alpha - \beta}{2} \right) \][/tex]
[tex]\[ \sin^2 \left( \frac{\alpha - \beta}{2} \right) = 1 - \left( \frac{\sqrt{x^2 + y^2}}{2} \right)^2 \][/tex]
[tex]\[ \sin^2 \left( \frac{\alpha - \beta}{2} \right) = 1 - \frac{x^2 + y^2}{4} \][/tex]
[tex]\[ \sin^2 \left( \frac{\alpha - \beta}{2} \right) = \frac{4 - (x^2 + y^2)}{4} \][/tex]
[tex]\[ \sin \left( \frac{\alpha - \beta}{2} \right) = \frac{\sqrt{4 - x^2 - y^2}}{2} \][/tex]
10. Find [tex]\(\tan \left( \frac{\alpha - \beta}{2} \right)\)[/tex]:
Thus:
[tex]\[ \tan \left( \frac{\alpha - \beta}{2} \right) = \frac{\sin \left( \frac{\alpha - \beta}{2} \right)}{\cos \left( \frac{\alpha - \beta}{2} \right)} \][/tex]
[tex]\[ = \frac{\frac{\sqrt{4 - x^2 - y^2}}{2}}{\frac{\sqrt{x^2 + y^2}}{2}} \][/tex]
[tex]\[ = \pm \sqrt{\frac{4 - x^2 - y^2}{x^2 + y^2}} \][/tex]
Finally, we have proven:
[tex]\[ \tan \left( \frac{\alpha - \beta}{2} \right) = \pm \sqrt{\frac{4 - x^2}{x^2 + y^2}} \][/tex]
1. Use Trigonometric Identities:
We start with the given trigonometric sum identities:
[tex]\[ \sin \alpha + \sin \beta = 2 \sin \left( \frac{\alpha + \beta}{2} \right) \cos \left( \frac{\alpha - \beta}{2} \right) \][/tex]
[tex]\[ \cos \alpha + \cos \beta = 2 \cos \left( \frac{\alpha + \beta}{2} \right) \cos \left( \frac{\alpha - \beta}{2} \right) \][/tex]
2. Assign These to the Given Values:
According to the problem:
[tex]\[ \sin \alpha + \sin \beta = x \][/tex]
[tex]\[ \cos \alpha + \cos \beta = y \][/tex]
3. Relate to Sum and Product Formulas:
Substitute the identities into the given equations:
[tex]\[ 2 \sin \left( \frac{\alpha + \beta}{2} \right) \cos \left( \frac{\alpha - \beta}{2} \right) = x \][/tex]
[tex]\[ 2 \cos \left( \frac{\alpha + \beta}{2} \right) \cos \left( \frac{\alpha - \beta}{2} \right) = y \][/tex]
4. Simplify the Equations:
Divide both sides of each equation by 2:
[tex]\[ \sin \left( \frac{\alpha + \beta}{2} \right) \cos \left( \frac{\alpha - \beta}{2} \right) = \frac{x}{2} \][/tex]
[tex]\[ \cos \left( \frac{\alpha + \beta}{2} \right) \cos \left( \frac{\alpha - \beta}{2} \right) = \frac{y}{2} \][/tex]
5. Introduce New Terms for Simplicity:
Let:
[tex]\[ A = \cos \left( \frac{\alpha - \beta}{2} \right) \][/tex]
[tex]\[ B = \sin \left( \frac{\alpha + \beta}{2} \right) \][/tex]
[tex]\[ C = \cos \left( \frac{\alpha + \beta}{2} \right) \][/tex]
Therefore, the equations become:
[tex]\[ B A = \frac{x}{2} \][/tex]
[tex]\[ C A = \frac{y}{2} \][/tex]
6. Divide the Equations:
Divide the first equation by the second:
[tex]\[ \frac{B A}{C A} = \frac{\frac{x}{2}}{\frac{y}{2}} \][/tex]
This simplifies to:
[tex]\[ \frac{B}{C} = \frac{x}{y} \][/tex]
Thus:
[tex]\[ \tan \left( \frac{\alpha + \beta}{2} \right) = \frac{x}{y} \][/tex]
7. Use the Pythagorean Identity:
We know:
[tex]\[ B^2 + C^2 = \left( \sin \left( \frac{\alpha + \beta}{2} \right) \right)^2 + \left( \cos \left( \frac{\alpha + \beta}{2} \right) \right)^2 = 1 \][/tex]
Therefore:
[tex]\[ \left( \frac{x}{2A} \right)^2 + \left( \frac{y}{2A} \right)^2 = 1 \][/tex]
8. Simplify This Expression:
[tex]\[ \frac{x^2}{4A^2} + \frac{y^2}{4A^2} = 1 \][/tex]
[tex]\[ \frac{x^2 + y^2}{4A^2} = 1 \][/tex]
Solving for [tex]\(A\)[/tex]:
[tex]\[ 4A^2 = x^2 + y^2 \][/tex]
[tex]\[ A^2 = \frac{x^2 + y^2}{4} \][/tex]
Therefore:
[tex]\[ A = \frac{\sqrt{x^2 + y^2}}{2} \][/tex]
9. Use the [tex]\(\tan \frac{\alpha - \beta}{2}\)[/tex] Identity:
We know:
[tex]\[ \sin^2 \left( \frac{\alpha - \beta}{2} \right) + \cos^2 \left( \frac{\alpha - \beta}{2} \right) = 1 \][/tex]
So:
[tex]\[ \sin^2 \left( \frac{\alpha - \beta}{2} \right) = 1 - \cos^2 \left( \frac{\alpha - \beta}{2} \right) \][/tex]
[tex]\[ \sin^2 \left( \frac{\alpha - \beta}{2} \right) = 1 - \left( \frac{\sqrt{x^2 + y^2}}{2} \right)^2 \][/tex]
[tex]\[ \sin^2 \left( \frac{\alpha - \beta}{2} \right) = 1 - \frac{x^2 + y^2}{4} \][/tex]
[tex]\[ \sin^2 \left( \frac{\alpha - \beta}{2} \right) = \frac{4 - (x^2 + y^2)}{4} \][/tex]
[tex]\[ \sin \left( \frac{\alpha - \beta}{2} \right) = \frac{\sqrt{4 - x^2 - y^2}}{2} \][/tex]
10. Find [tex]\(\tan \left( \frac{\alpha - \beta}{2} \right)\)[/tex]:
Thus:
[tex]\[ \tan \left( \frac{\alpha - \beta}{2} \right) = \frac{\sin \left( \frac{\alpha - \beta}{2} \right)}{\cos \left( \frac{\alpha - \beta}{2} \right)} \][/tex]
[tex]\[ = \frac{\frac{\sqrt{4 - x^2 - y^2}}{2}}{\frac{\sqrt{x^2 + y^2}}{2}} \][/tex]
[tex]\[ = \pm \sqrt{\frac{4 - x^2 - y^2}{x^2 + y^2}} \][/tex]
Finally, we have proven:
[tex]\[ \tan \left( \frac{\alpha - \beta}{2} \right) = \pm \sqrt{\frac{4 - x^2}{x^2 + y^2}} \][/tex]
