Answer: sin(2tan^-1(u/6)) = (2u) / [(u² + 36) * √(u² + 36)]
Step-by-step explanation: We can use the trigonometric identity:
tan(2θ) = (2 tan θ) / (1 - tan² θ)
to rewrite sin(2tan^-1(u/6)) as an algebraic expression.
Step 1: Let θ = tan^-1(u/6). Then we have:
tan θ = u/6
Step 2: Substitute θ into the formula for tan(2θ):
tan(2θ) = (2 tan θ) / (1 - tan² θ)
tan(2 tan^-1(u/6)) = (2 tan(tan^-1(u/6))) / [1 - tan²(tan^-1(u/6))]
tan(2 tan^-1(u/6)) = (2u/6) / [1 - (u/6)²]
tan(2 tan^-1(u/6)) = (u/3) / [(36 - u²) / 36]
Step 3: Simplify the expression by using the Pythagorean identity:
1 + tan² θ = sec² θ
tan² θ = sec² θ - 1
1 - tan² θ = 1 / sec² θ
tan(2 tan^-1(u/6)) = (u/3) / [(36 - u²) / 36]
tan(2 tan^-1(u/6)) = (u/3) * (6 / √(36 - u²))²
tan(2 tan^-1(u/6)) = (u/3) * (36 / (36 - u²))
Step 4: Rewrite the expression in terms of sine.
Recall that:
tan θ = sin θ / cos θ
sin θ = tan θ * cos θ
cos θ = 1 / √(1 + tan² θ)
Using this identity, we can rewrite the expression for tan(2tan^-1(u/6)) as:
sin(2tan^-1(u/6)) = tan(2tan^-1(u/6)) * cos(2tan^-1(u/6))
sin(2tan^-1(u/6)) = [(u/3) * (36 / (36 - u²))] * [1 / √(1 + [(u/6)²])]
simplify to get:
sin(2tan^-1(u/6)) = (2u) / [(u² + 36) * √(u² + 36)]