Rewrite sin(tan^-1(u/sqrt(25-u^2))) as an algebraic expression in u.

Answer 1

u/5=sine of the angle whose tangent is u/sqr(25-u^2)

construct a right triangle, base = sqr(25-u^2), height=u, hypotenuse= sqr(u^2+25-u^2)=5

sine of the angle opposite u = u/5 = height/hypotenuse

Answer 2

Because tangent is the trig ratio opp/adj then "tan^-1(u/sqrt(25-u^2))" is talking about the angle for which the opposite is u and the adjacent is √(25 - u²) so like this:

To take the sine, you must know the hypotenuse (because sin(θ) = opp/hyp)

To find the hypotenuse, use the Pythagorean Theorem:

hyp² = u² + (√(25 - u²))²

hyp² = u² + 25 - u²

hyp² = 25

hyp = 5

So the sine of that angle is u/5