# Tan(t) = x/a => sin(t) = x/sqrt( a^2 + x^2) ?

1. Jul 13, 2008

### TristanH

tan(t) = x/a => sin(t) = x/sqrt( a^2 + x^2) ???

Folks,

In the solution guide to Kline's calculus book, he gives the following in working a problem:

tan(t) = x/a => sin(t) = x/sqrt( a^2 + x^2)

Can someone explain why this is the case?

Thanks,

Tristan

2. Jul 13, 2008

### rock.freak667

Re: tan(t) = x/a => sin(t) = x/sqrt( a^2 + x^2) ???

tan(t)=Opposite/Adjacent. Draw a right-angled triangle with the side opposite to the angle ,t, as x and the adjacent side to it as a. Then use pythagoras' theorem.

3. Jul 13, 2008

### TristanH

Re: tan(t) = x/a => sin(t) = x/sqrt( a^2 + x^2) ???

I just figured this out on my own. For anyone who is curious:

tan(t) = x/a means the opposite and adjacent legs of a right triangle with angle t, are x and a respectively. By the Pythagorean Therom, this implies that the hypotenuse is sqrt(a^2 + x^2). Since sin t = opp/hyp, sin t = x/sqrt( a^2 + x^2). QED

4. Jul 16, 2008

### Gib Z

Re: tan(t) = x/a => sin(t) = x/sqrt( a^2 + x^2) ???

lol ?

$\int \frac{1}{\sqrt{1-x^2}} dx$ Dec 27, 2016