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

• Aug 26th 2007, 09:43 PM
liz155
How 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
• Aug 26th 2007, 09:52 PM
Jhevon
Quote:

Originally Posted by liz155
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

Let $\displaystyle \arctan \left( \frac {3}{5} \right) = \theta \implies \tan \theta = \frac {3}{5}$

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 $\displaystyle \sqrt {34}$

Thus we can say, $\displaystyle \cos \left[ \arctan \left( \frac {3}{5} \right) \right] = \cos \theta =$ ?

It shouldn't be too hard for you to finish, right?
• Aug 26th 2007, 10:03 PM
liz155

Thanks so much for your help and the great picture!!!:)
• Aug 26th 2007, 10:04 PM
Jhevon
Quote:

Originally Posted by liz155

Thanks so much for your help and the great picture!!!:)

correct :)
• Aug 26th 2007, 10:13 PM
liz155
Thanks again! I really understand it now.:)
• Aug 26th 2007, 10:14 PM
Jhevon
Quote:

Originally Posted by liz155
Thanks again! I really understand it now.:)

you're welcome.

note the technique, it works on a lot of problems of this nature. with some their are alternatives though, you can use trig identities to work them out, but let's save those for another time
• Aug 27th 2007, 06:00 AM
Krizalid
Quote:

Originally Posted by liz155
How do i find cos[arctan(3/5)] ?

Also, you can try to prove that $\displaystyle \cos x=\frac1{\sqrt{1+\tan^2x}}\implies\cos(\arctan x)=\frac1{\sqrt{1+x^2}}$

Plugging $\displaystyle x=\frac35$ you have the asked.