# Trig relation

1. Nov 15, 2007

### miccol999

Given tanA=y/x..................(1)

Can anyone tell me how you get the following relations:

=>sinA=ay/sqrt(x^2+y^2)..................(2)
=>cosA=ax/sqrt(x^2+y^2).................(3)

where a=(+/-)1

I know tanA=sinA/cosA and sin^2(A)+cos^2(A)=1...and I can see by substituting (2) and (3) into (1) it works, but I really can't work out how to come up with them!! I know I'm probably overlooking something quite obvious but its late+I'm not trusting my own judgement atm!!!

Last edited: Nov 15, 2007
2. Nov 15, 2007

### hage567

You've got the right approach. Can you show all of your steps so we can pick out where you're going wrong?

3. Nov 15, 2007

### Dick

tan(A)=sin(A)/cos(A)=y/x. So y*cos(A)=x*sin(A). Now put in sin(A)=+/-sqrt(1-cos(A)^2) and solve for cos(A) by squaring both sides. Ditto for sin(A).

Last edited: Nov 16, 2007
4. Nov 16, 2007

### HallsofIvy

Staff Emeritus
Given Tan(A)= x/y, the first thing I would do is draw a right triangle having angle A, "opposite side" of length x, "near side" of length y, and then calculate the length of the hypotenuse. Once you have done that, the other trig functions fall into place.