Deriving Relations from tanA=y/x

[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!

 

[hage567]

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

 

[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).

 

[HallsofIvy]

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.