# Arc values not on the unit circle

• Oct 13th 2011, 09:28 AM
Nervous
Arc values not on the unit circle
How do you find cos (arctan4/3 - arcsin3/5)?
• Oct 13th 2011, 10:36 AM
skeeter
Re: Arc values not on the unit circle
Quote:

Originally Posted by Nervous
How do you find acrcos (arctan4/3 - arcsin3/5)?

I think you really mean the value of cos[arctan(4/3)-arcsin(3/5)] ... correct?
• Oct 13th 2011, 10:40 AM
Nervous
Re: Arc values not on the unit circle
Yes, I'll fix that.
• Oct 13th 2011, 12:23 PM
HallsofIvy
Re: Arc values not on the unit circle
I'm a bit confused by the title: "Arcvalues not on the unit circle". arctan(4/3) and arcsin(3/5) definitely are on the unit circle. I think you mean simply that those are not any of the "easy" values you learn- arcsin(\sqrt{2}/2}, cos(1/2), etc.

However, it should be very easy to see that if we have a right triangle with one leg of length 3 and and another of length 4, then the hypotenuse has length 5. So, since sine is "opposite side over hypotenuse" while tangent is "opposite side over near side", it should be just as easy to see that "arctan(4/3)" and "arctan(3/5)" are opposite angles in such a triangle. Cosine of their sum would be very easy to find! For their difference, us the fact that cos(a- b)= cos(a)cos(b)+ sin(a)sin(b) and its easy to find sine and cosine of those two angles.
• Oct 13th 2011, 12:59 PM
Nervous
Re: Arc values not on the unit circle
Thanks.