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

1. ## 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

2. 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?

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

4. Originally Posted by liz155

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

5. Thanks again! I really understand it now.

6. 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

7. 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.

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### cos(arcctg (-4/3))

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