Calculus help I have to rewrite sin(tan-1 u) without using trig or inverse trig fucncitons.?

Question

My solution is √(u^2 + 1) / (u^2 + 1)

or is it u√(u2 + 1) / (u2 + 1)? or was I correct.

Thanks

Answer 1

Consider a right-angled triangle with base length t, upright length u and hypotenuse length h. Scale the triangle so that t=1 (this doesn't change the angles).

The angle between the base and the hypotenuse is tan-1 u.

The sine of this angle is u/h. But h is √(1+u^2).

So the answer is u/√(1+u^2).

Answer 2

Here's another way -

Let tan^-1(u) = x. Therefore, u = tan(x).

Knowing that sin^2(x) + cos^2(x) = 1,
divide through by sin^2(x) to give :

1 + 1 / tan^2(x) = 1 / sin^2(x)

So, 1 + 1 / u^2 = 1 / sin^2(x)

or (1 + u^2) / u^2 = 1 / sin^2(x)

Take the reciprocal of both sides :
sin^2(x) = u^2 / (1 + u^2)

Take the square root of both sides (positive value only) :

sin(x) = u / sqrt(1 + u^2) {= sin(tan^-1(u)}

Edit : except I think I used a trig function!! - Oh! well.

Answer 3

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Answer 4

Who's correct?
Notice that Puggy (usually spot on!!) has slipped up in using
opp/hyp for tan,
when it should be opp/adj

Try x = 1. Your answer and sangmo's both give 1/√2, which is correct because tan^(-1)[1] = π/4,
and sin(π/4) = 1/√2

Try x = √3.
tan^(-1)[√3] = π/3
and
sin(π/3) = √3/2, which verifies sangmo's answer,
u/√(u^2 + 1)
I'd give him the points.

Answer 5

sin(tanˉ¹(u))

To solve this, let t = tanˉ¹(u). Then, taking the tan of both sides,

tan(t) = u

By SOHCAHTOA,

tan(t) = u/1 = opposite/hypotenuse. Therefore,

opposite = u
hypotenuse = 1, and by Pythagoras,
adjacent = √(hypotenuse² - opposite²)
adjacent = √(1 - u²)

Therefore,

sin(t) = sin(tanˉ¹(u)) = opposite/hypotenuse
= u / √(1 - u²)