Inverse trig functions with tan-1

[mickellowery]

Homework Statement

sin(tan^-1(x))

Homework Equations

The Attempt at a Solution

y=tan^-1(x)
tan(y)=x
sec²(y)= 1+tan²(y)
sec(y)=\sqrt{1+x^2}
This is where I'm getting stuck. I know that I have to say that the sin(y)= whatever, but I'm not sure how to tie the sin sec and tan together. Is there a trig identity or should I make tan= \frac{sin}{cos}?

 

[Char. Limit]

What exactly are you trying to do here? Are you trying to differentiate the first expression? It's not clear.

 

[Mark44]

Homework Statement

sin(tan^-1(x))

Homework Equations

The Attempt at a Solution

y=tan^-1(x)
tan(y)=x
sec²(y)= 1+tan²(y)
sec(y)=\sqrt{1+x^2}

Draw a right triangle, labelled according to y = tan^-1(x) or equivalently, tan(y) = x/1. One acute angle should be labelled y. The side opposite should be labelled x and the side adjacent should be labelled 1. From this you can figure out the hypotenuse.

What then is sin(y)?

This is where I'm getting stuck. I know that I have to say that the sin(y)= whatever, but I'm not sure how to tie the sin sec and tan together. Is there a trig identity or should I make tan= \frac{sin}{cos}?

 

[mickellowery]

Sorry, the problem says to simplify the expression. The final answer is supposed to be \frac{x}{\sqrt{1+x^2}}

 

[Dick]

Homework Statement

sin(tan^-1(x))

Homework Equations

The Attempt at a Solution

y=tan^-1(x)
tan(y)=x
sec²(y)= 1+tan²(y)
sec(y)=\sqrt{1+x^2}
This is where I'm getting stuck. I know that I have to say that the sin(y)= whatever, but I'm not sure how to tie the sin sec and tan together. Is there a trig identity or should I make tan= \frac{sin}{cos}?

sec(y)=1/cos(y). So you've got cos(y)=1/sqrt(1+x^2). To get sin(y) use sin^2(y)=1-cos^2(y).