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

**liz155** · August 26th 2007, 09:43 PM

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

**Jhevon** · August 26th 2007, 09:52 PM

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

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 $\sqrt {34}$

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

It shouldn't be too hard for you to finish, right?

**liz155** · August 26th 2007, 10:03 PM

Since cos= adjacent over hypotenuse...Would the answer be 5/sqrt(34)?

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

**Jhevon** · August 26th 2007, 10:04 PM

Originally Posted by liz155

Since cos= adjacent over hypotenuse...Would the answer be 5/sqrt(34)?

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

correct

**liz155** · August 26th 2007, 10:13 PM

Thanks again! I really understand it now.

**Jhevon** · August 26th 2007, 10:14 PM

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

**Krizalid** · August 27th 2007, 06:00 AM

Originally Posted by liz155

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

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

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

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