# Homework Help: Inverse trig functions with tan-1

1. Jul 1, 2010

### mickellowery

1. The problem statement, all variables and given/known data
sin(tan-1(x))

2. Relevant equations

3. 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= $$\frac{sin}{cos}$$?

2. Jul 1, 2010

### Char. Limit

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

3. Jul 1, 2010

### Staff: Mentor

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)?

4. Jul 1, 2010

### mickellowery

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

5. Jul 1, 2010

### Dick

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).