Answer :
Sure, let's prove that if [tex]\(\tan \theta = \frac{2xy}{x^2 - y^2}\)[/tex], then [tex]\(\sin \theta = \frac{2xy}{x^2 + y^2}\)[/tex].
Step-by-Step Proof:
We start with the given identity for [tex]\(\tan \theta\)[/tex]:
[tex]\[ \tan \theta = \frac{2xy}{x^2 - y^2} \][/tex]
We know from trigonometric identities that:
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \][/tex]
Let [tex]\(\sin \theta = \sin \theta\)[/tex] and [tex]\(\cos \theta = \cos \theta\)[/tex]. Then:
[tex]\[ \frac{\sin \theta}{\cos \theta} = \frac{2xy}{x^2 - y^2} \][/tex]
Cross-multiplying gives us:
[tex]\[ \sin \theta (x^2 - y^2) = 2xy \cos \theta \][/tex]
To express [tex]\(\sin \theta\)[/tex], we need another trigonometric identity. We use the Pythagorean identity:
[tex]\[ \sin^2 \theta + \cos^2 \theta = 1 \][/tex]
Our goal is to find [tex]\(\sin \theta\)[/tex]. Let's isolate [tex]\(\sin \theta\)[/tex] in terms of [tex]\(\tan \theta\)[/tex]:
First, squaring both sides of [tex]\(\tan \theta\)[/tex]:
[tex]\[ \tan^2 \theta = \left(\frac{2xy}{x^2 - y^2}\right)^2 \][/tex]
We know that [tex]\(\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta}\)[/tex]:
[tex]\[ \left(\frac{\sin \theta}{\cos \theta}\right)^2 = \left(\frac{2xy}{x^2 - y^2}\right)^2 \][/tex]
This simplifies to:
[tex]\[ \frac{\sin^2 \theta}{\cos^2 \theta} = \frac{4x^2 y^2}{(x^2 - y^2)^2} \][/tex]
Let [tex]\(A = \frac{4x^2 y^2}{(x^2 - y^2)^2}\)[/tex]. Thus,
[tex]\[ \frac{\sin^2 \theta}{\cos^2 \theta} = A \][/tex]
And using the Pythagorean identity:
[tex]\[ \sin^2 \theta = A \cos^2 \theta \][/tex]
We substitute back for [tex]\(\cos^2 \theta\)[/tex]:
[tex]\[ \cos^2 \theta = 1 - \sin^2 \theta \][/tex]
So:
[tex]\[ \sin^2 \theta = A (1 - \sin^2 \theta) \][/tex]
Therefore:
[tex]\[ \sin^2 \theta = A - A \sin^2 \theta \][/tex]
Solving for [tex]\(\sin^2 \theta\)[/tex]:
[tex]\[ \sin^2 \theta(1 + A) = A \][/tex]
And:
[tex]\[ \sin^2 \theta = \frac{A}{1 + A} \][/tex]
Recalling the value of [tex]\(A\)[/tex]:
[tex]\[ A = \frac{4 x^2 y^2}{(x^2 - y^2)^2} \][/tex]
Substitute [tex]\(A\)[/tex]:
[tex]\[ \sin^2 \theta = \frac{\frac{4 x^2 y^2}{(x^2 - y^2)^2}}{1 + \frac{4 x^2 y^2}{(x^2 - y^2)^2}} \][/tex]
Now simplify the denominator:
[tex]\[ \sin^2 \theta = \frac{4 x^2 y^2}{(x^2 - y^2)^2 + 4 x^2 y^2} \][/tex]
Simplify the terms in the denominator:
[tex]\[ \sin^2 \theta = \frac{4 x^2 y^2}{x^4 - 2 x^2 y^2 + y^4 + 4 x^2 y^2} \][/tex]
This results in:
[tex]\[ \sin^2 \theta = \frac{4 x^2 y^2}{x^4 + 2 x^2 y^2 + y^4} \][/tex]
Notice that [tex]\(x^4 + 2 x^2 y^2 + y^4\)[/tex] can be factored as:
[tex]\[ \sin^2 \theta = \frac{4 x^2 y^2}{(x^2 + y^2)^2} \][/tex]
So:
[tex]\[ \sin^2 \theta = \left(\frac{2 x y}{x^2 + y^2}\right)^2 \][/tex]
Taking the square root of both sides:
[tex]\[ \sin \theta = \frac{2 x y}{x^2 + y^2} \][/tex]
Thus, we have proven:
[tex]\[ \sin \theta = \frac{2xy}{x^2 + y^2} \][/tex]
Step-by-Step Proof:
We start with the given identity for [tex]\(\tan \theta\)[/tex]:
[tex]\[ \tan \theta = \frac{2xy}{x^2 - y^2} \][/tex]
We know from trigonometric identities that:
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \][/tex]
Let [tex]\(\sin \theta = \sin \theta\)[/tex] and [tex]\(\cos \theta = \cos \theta\)[/tex]. Then:
[tex]\[ \frac{\sin \theta}{\cos \theta} = \frac{2xy}{x^2 - y^2} \][/tex]
Cross-multiplying gives us:
[tex]\[ \sin \theta (x^2 - y^2) = 2xy \cos \theta \][/tex]
To express [tex]\(\sin \theta\)[/tex], we need another trigonometric identity. We use the Pythagorean identity:
[tex]\[ \sin^2 \theta + \cos^2 \theta = 1 \][/tex]
Our goal is to find [tex]\(\sin \theta\)[/tex]. Let's isolate [tex]\(\sin \theta\)[/tex] in terms of [tex]\(\tan \theta\)[/tex]:
First, squaring both sides of [tex]\(\tan \theta\)[/tex]:
[tex]\[ \tan^2 \theta = \left(\frac{2xy}{x^2 - y^2}\right)^2 \][/tex]
We know that [tex]\(\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta}\)[/tex]:
[tex]\[ \left(\frac{\sin \theta}{\cos \theta}\right)^2 = \left(\frac{2xy}{x^2 - y^2}\right)^2 \][/tex]
This simplifies to:
[tex]\[ \frac{\sin^2 \theta}{\cos^2 \theta} = \frac{4x^2 y^2}{(x^2 - y^2)^2} \][/tex]
Let [tex]\(A = \frac{4x^2 y^2}{(x^2 - y^2)^2}\)[/tex]. Thus,
[tex]\[ \frac{\sin^2 \theta}{\cos^2 \theta} = A \][/tex]
And using the Pythagorean identity:
[tex]\[ \sin^2 \theta = A \cos^2 \theta \][/tex]
We substitute back for [tex]\(\cos^2 \theta\)[/tex]:
[tex]\[ \cos^2 \theta = 1 - \sin^2 \theta \][/tex]
So:
[tex]\[ \sin^2 \theta = A (1 - \sin^2 \theta) \][/tex]
Therefore:
[tex]\[ \sin^2 \theta = A - A \sin^2 \theta \][/tex]
Solving for [tex]\(\sin^2 \theta\)[/tex]:
[tex]\[ \sin^2 \theta(1 + A) = A \][/tex]
And:
[tex]\[ \sin^2 \theta = \frac{A}{1 + A} \][/tex]
Recalling the value of [tex]\(A\)[/tex]:
[tex]\[ A = \frac{4 x^2 y^2}{(x^2 - y^2)^2} \][/tex]
Substitute [tex]\(A\)[/tex]:
[tex]\[ \sin^2 \theta = \frac{\frac{4 x^2 y^2}{(x^2 - y^2)^2}}{1 + \frac{4 x^2 y^2}{(x^2 - y^2)^2}} \][/tex]
Now simplify the denominator:
[tex]\[ \sin^2 \theta = \frac{4 x^2 y^2}{(x^2 - y^2)^2 + 4 x^2 y^2} \][/tex]
Simplify the terms in the denominator:
[tex]\[ \sin^2 \theta = \frac{4 x^2 y^2}{x^4 - 2 x^2 y^2 + y^4 + 4 x^2 y^2} \][/tex]
This results in:
[tex]\[ \sin^2 \theta = \frac{4 x^2 y^2}{x^4 + 2 x^2 y^2 + y^4} \][/tex]
Notice that [tex]\(x^4 + 2 x^2 y^2 + y^4\)[/tex] can be factored as:
[tex]\[ \sin^2 \theta = \frac{4 x^2 y^2}{(x^2 + y^2)^2} \][/tex]
So:
[tex]\[ \sin^2 \theta = \left(\frac{2 x y}{x^2 + y^2}\right)^2 \][/tex]
Taking the square root of both sides:
[tex]\[ \sin \theta = \frac{2 x y}{x^2 + y^2} \][/tex]
Thus, we have proven:
[tex]\[ \sin \theta = \frac{2xy}{x^2 + y^2} \][/tex]
