How do i find cos[arctan(3/5)] ?

I know I need to set up some kind of triangle diagram, but I don't know how or what to do next. I am not allowed to use a calculator.

Thanks to anyone that tries to help me.

--Cori

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- August 26th 2007, 10:43 PMliz155How do I find cos[arctan(3/5)] ?
How do i find cos[arctan(3/5)] ?

I know I need to set up some kind of triangle diagram, but I don't know how or what to do next. I am not allowed to use a calculator.

Thanks to anyone that tries to help me.

--Cori - August 26th 2007, 10:52 PMJhevon
Let

Thus we can set up a right-triangle and label one of the acute angles "theta" and fill in the sides accordingly. (tangent = opposite/adjacent). we can use Pythagoras' theorem to find the length of the hypotenuse, which here is

Thus we can say, ?

It shouldn't be too hard for you to finish, right? - August 26th 2007, 11:03 PMliz155
Since cos= adjacent over hypotenuse...Would the answer be 5/sqrt(34)?

Thanks so much for your help and the great picture!!!:) - August 26th 2007, 11:04 PMJhevon
- August 26th 2007, 11:13 PMliz155
Thanks again! I really understand it now.:)

- August 26th 2007, 11:14 PMJhevon
- August 27th 2007, 07:00 AMKrizalid